SCHEME OF WORK
WEEKS TOPICS
1. Position, Distance and Displacement.
2. Derivation of equation of Linear motion, Motion under Gravity, calculation using these Equations.
3. Projectiles and falling bodies.
4. Scalar and vector Quantities¬¬-Concept of scalar and vector quantities, vector representation etc.
5. Pressure-Definition, Application to hydraulic press, Car Brakes. pressure and dept h etc.
6. Equilibrium if Forces-principle of moment, conditions for equilibrium of a Rigid Bodies etc.
7. Newton Laws of Motion-Conservation of Linear momentum and energy.
8. Simple Harmonic Motion-definition, speed, amplitude, displacement, acceleration,etc.
9. Energy of simple harmonic motion and forced vibration, Resonance.
10. Machines-Types, Examples and Calculation.
11. Laboratory exercises
12. Revision.
1ST TERM
WEEK 1
POSITION, DISTANCE AND DISPLACEMENT.
CONTENTS.
POSITION
COORDINATES SYSTEM
DISTANCE
DISPLACEMENT
POSITION
The position of an object in space or on a plane is the point at which the object can be located with reference to a given point (the origin).
COORDINATES SYSTEM
1. RECTANGULAR CARTESIAN COORDINATE SYSTEM
This is a system that consists of two or three intersecting lines mutually perpendicular and which serves as a reference frame that guides one in locating the position of a point in a plane or in space.
This system also assigns direction(with arrow head) to these reference lines(called the coordinate axes) and make the distances measured from the point of intersection(known as the origin)positive along OX, OY and OZ and that measured on the opposite direction, negative.
3:Dimensional coordinate system diagram.
https://youtu.be/mgx0kT5UbKk
PLANE.
A plane is a geometric figure defined by two reference frame or 2-dimensional coordinate system.
SPACE.
A space is defined by three reference frames or 3-dimensional coordinate system.
DISTANCE
This is a measure of the separation between two points. It has magnitude but no direction, hence. it is a scalar quantity
DETERMINATION OF DISTANCE BETWEEN TWO POINTS
This is the gap between any two positions in space. It denoted by S and measured in metre (m). It is a scalar quantity and is calculated as the product of average speed and time.
Thus, distance = average speed x time. The distance between points A and B in shown in the Cartesian(x – y) plane below.
If two points A and B located in a plane are defined by two ordered pair of values(x[sub]1[/sub] y[sub]1[/sub]) and (x[sub]2[/sub] y[sub]2[/sub]) or assumed to be in space
where they are defined by (x[sub]1[/sub] y[sub]1[/sub] z[sub]1[/sub]) and (X[sub]2[/sub], Y[sub]2[/sub] Z[sub]2[/sub])
the distance between them can be determined by applying the relation.
https://youtu.be/0IOEPcAHgi4
AB = [ ( x2 – x1)[sup]2[/sup] + ( y2 - y1)[sup]2[/sup]][sup]½[/sup]
Example 1 : What is the distance between point A(9,6) and point B(5,3)
SOLUTION
Y2 ...........................B(x2,y2).
Y1 ............................A(x1,y1)
X1 ------------------------- x2
AB[sup]2[/sup] = (9 – 5)[sup]2[/sup] + (6 – 3)[sup]2[/sup]
AB[sup]2[/sup] = (4)[sup]2[/sup] + (3)[sup]2[/sup]
= 16 + 9
AB[sup]2[/sup] = 25
AB = 25
AB = 5
EXAMPLES OF DISTANCE COVERED ON THE FIELD IN A SPECIFIC TIME.
https://youtu.be/yXcN8fqTyf4
DISPLACEMENT.
Displacement is the distance covered in a specified direction. It is a vector quantity, that has the same unit as distance.
Displacement is the distance covered in a specific direction. It is a vector quantity measured in metre (m).
Displacement = Velocity x time taken
Velocity is defined as the rate of change of distance moved with time in a specific direction
Velocity = Displacement = change in displacement
time change in time
However speed is used in place of velocity and vice versa. The S.I unit for velocity is m/s.
DIFFERENCES BETWEEN DISTANCE AND DISPLACEMENT
DISTANCE ----------------------------------------------------- DISPLACEMENT
Distance is the gap between any two position in space---------------------- Displacement is the distance moved in a specific direction
Distance is a scalar quantity ------------------------------Displacement is a vector quantity.
Distance = Average speed x time------------------------------ Displacement = Average velocity x time
It is denoted by S ------------------------------It is denoted by x
https://youtu.be/edm8uy7O9NY
SIMILARITIES BETWEEN DISTANCE AND DISPLACEMENT
(i) They are both measured in metre (m).
(ii) They both have magnitude
THE USE OF BEARING TO INDICATE DIRECTION AND DISPLACEMENT
The bearing of an object from the origin is the angle which it makes with the North Pole in the clockwise sense. It is specified in two ways: (a) the use of cardinal points: N – North, S – South, W – West and E – East. (b) The use of three- digit notation.
Note : bearings which are located by cardinal points are with respect or reference to the North and South.
https://youtu.be/gBw2g60MY-8
https://youtu.be/21BwUNDOQno
EVALUATION.
1. Define the word displacement.
2. In tabular form, differentiate between distance and displacement.
3. With the aid of a diagram, explain the term plane and space.
4. Briefly describe how the position of a point can be located in space using rectangular Cartesian coordinate system?.
5. Calculate the distance between points A(2,3) and B(-5,1).
6. Calculate the distance between points J(-2,-4) and K(-5,-10) in space.
7. Why is displacement regarded as vector quantity?.
8. Differentiate between distance and displacement.
READING ASSIGNMENT.
New school physics for s s s-M W ANYAKOHA. Pages 121-126.
GENERAL EVALUATION
1. What are the distances of the points below?
(i) D(3,8) and V(0,5)
(ii) R(-6,4) and T(4,8)
(iii) F(1,8) and U(9,3) and W(4,0)
(iv) X(6,1) and S(2,5)
2. State 7 fundamental quantities.
3. for the fundamental quantities stated above give their respective units.
ASSIGNMENT.
1. In the diagram below, the position of A is
(A)2,3 (B)3,3 (c)3,4 (D)4,3.
2, To locate a point in a plane or space, we can use.
(A)Bearing system. (B) Centrifugal (C) Centripetal. (D) None of the above
3.Displacement can be classified as
(A) Scalar quantity (B)Vector quantity (C)Both scalar and vector quantities. (D)All of the above.
4. Determine the distance between S(3,4,-5)and T(2,1,0). (A)5.8 (B) 5.9 (C) 6.0 (D) 6.2.
5. Distance can be measured by.
(A) Tape rule (B) Eureka can (C) Lever balance. (D) Stop watch.
THEORY.
1. Sketch clearly using scale indicators, the position of a point P (4,-5) and Q( -4, 10) with reference to a point Q(O,0). Determine the distance between P and Q.
2. Distinguish between distance and displacement. Which of the terms is a vector and why?.
3. A car moves from a point P to a point Q, a distance of 17 metres in the direction, 0400. It then changes direction and moves to a point R, in the direction 1600. If the point R is due west of P, find the distance of point R from Q.
4. A butterfly moves from a point X to a point Y, a distance of 15 metres in the direction, 0450. It then changes direction and moves to a point Z, in the direction 1550. If the point Z is due east of X, find the distance of the point Z from the point Y.
CONTENTS.
POSITION
COORDINATES SYSTEM
DISTANCE
DISPLACEMENT
POSITION
The position of an object in space or on a plane is the point at which the object can be located with reference to a given point (the origin).
COORDINATES SYSTEM
1. RECTANGULAR CARTESIAN COORDINATE SYSTEM
This is a system that consists of two or three intersecting lines mutually perpendicular and which serves as a reference frame that guides one in locating the position of a point in a plane or in space.
This system also assigns direction(with arrow head) to these reference lines(called the coordinate axes) and make the distances measured from the point of intersection(known as the origin)positive along OX, OY and OZ and that measured on the opposite direction, negative.
3:Dimensional coordinate system diagram.
https://youtu.be/mgx0kT5UbKk
PLANE.
A plane is a geometric figure defined by two reference frame or 2-dimensional coordinate system.
SPACE.
A space is defined by three reference frames or 3-dimensional coordinate system.
DISTANCE
This is a measure of the separation between two points. It has magnitude but no direction, hence. it is a scalar quantity
DETERMINATION OF DISTANCE BETWEEN TWO POINTS
This is the gap between any two positions in space. It denoted by S and measured in metre (m). It is a scalar quantity and is calculated as the product of average speed and time.
Thus, distance = average speed x time. The distance between points A and B in shown in the Cartesian(x – y) plane below.
If two points A and B located in a plane are defined by two ordered pair of values(x[sub]1[/sub] y[sub]1[/sub]) and (x[sub]2[/sub] y[sub]2[/sub]) or assumed to be in space
where they are defined by (x[sub]1[/sub] y[sub]1[/sub] z[sub]1[/sub]) and (X[sub]2[/sub], Y[sub]2[/sub] Z[sub]2[/sub])
the distance between them can be determined by applying the relation.
https://youtu.be/0IOEPcAHgi4
AB = [ ( x2 – x1)[sup]2[/sup] + ( y2 - y1)[sup]2[/sup]][sup]½[/sup]
Example 1 : What is the distance between point A(9,6) and point B(5,3)
SOLUTION
Y2 ...........................B(x2,y2).
Y1 ............................A(x1,y1)
X1 ------------------------- x2
AB[sup]2[/sup] = (9 – 5)[sup]2[/sup] + (6 – 3)[sup]2[/sup]
AB[sup]2[/sup] = (4)[sup]2[/sup] + (3)[sup]2[/sup]
= 16 + 9
AB[sup]2[/sup] = 25
AB = 25
AB = 5
EXAMPLES OF DISTANCE COVERED ON THE FIELD IN A SPECIFIC TIME.
https://youtu.be/yXcN8fqTyf4
DISPLACEMENT.
Displacement is the distance covered in a specified direction. It is a vector quantity, that has the same unit as distance.
Displacement is the distance covered in a specific direction. It is a vector quantity measured in metre (m).
Displacement = Velocity x time taken
Velocity is defined as the rate of change of distance moved with time in a specific direction
Velocity = Displacement = change in displacement
time change in time
However speed is used in place of velocity and vice versa. The S.I unit for velocity is m/s.
DIFFERENCES BETWEEN DISTANCE AND DISPLACEMENT
DISTANCE ----------------------------------------------------- DISPLACEMENT
Distance is the gap between any two position in space---------------------- Displacement is the distance moved in a specific direction
Distance is a scalar quantity ------------------------------Displacement is a vector quantity.
Distance = Average speed x time------------------------------ Displacement = Average velocity x time
It is denoted by S ------------------------------It is denoted by x
https://youtu.be/edm8uy7O9NY
SIMILARITIES BETWEEN DISTANCE AND DISPLACEMENT
(i) They are both measured in metre (m).
(ii) They both have magnitude
THE USE OF BEARING TO INDICATE DIRECTION AND DISPLACEMENT
The bearing of an object from the origin is the angle which it makes with the North Pole in the clockwise sense. It is specified in two ways: (a) the use of cardinal points: N – North, S – South, W – West and E – East. (b) The use of three- digit notation.
Note : bearings which are located by cardinal points are with respect or reference to the North and South.
https://youtu.be/gBw2g60MY-8
https://youtu.be/21BwUNDOQno
EVALUATION.
1. Define the word displacement.
2. In tabular form, differentiate between distance and displacement.
3. With the aid of a diagram, explain the term plane and space.
4. Briefly describe how the position of a point can be located in space using rectangular Cartesian coordinate system?.
5. Calculate the distance between points A(2,3) and B(-5,1).
6. Calculate the distance between points J(-2,-4) and K(-5,-10) in space.
7. Why is displacement regarded as vector quantity?.
8. Differentiate between distance and displacement.
READING ASSIGNMENT.
New school physics for s s s-M W ANYAKOHA. Pages 121-126.
GENERAL EVALUATION
1. What are the distances of the points below?
(i) D(3,8) and V(0,5)
(ii) R(-6,4) and T(4,8)
(iii) F(1,8) and U(9,3) and W(4,0)
(iv) X(6,1) and S(2,5)
2. State 7 fundamental quantities.
3. for the fundamental quantities stated above give their respective units.
ASSIGNMENT.
1. In the diagram below, the position of A is
(A)2,3 (B)3,3 (c)3,4 (D)4,3.
2, To locate a point in a plane or space, we can use.
(A)Bearing system. (B) Centrifugal (C) Centripetal. (D) None of the above
3.Displacement can be classified as
(A) Scalar quantity (B)Vector quantity (C)Both scalar and vector quantities. (D)All of the above.
4. Determine the distance between S(3,4,-5)and T(2,1,0). (A)5.8 (B) 5.9 (C) 6.0 (D) 6.2.
5. Distance can be measured by.
(A) Tape rule (B) Eureka can (C) Lever balance. (D) Stop watch.
THEORY.
1. Sketch clearly using scale indicators, the position of a point P (4,-5) and Q( -4, 10) with reference to a point Q(O,0). Determine the distance between P and Q.
2. Distinguish between distance and displacement. Which of the terms is a vector and why?.
3. A car moves from a point P to a point Q, a distance of 17 metres in the direction, 0400. It then changes direction and moves to a point R, in the direction 1600. If the point R is due west of P, find the distance of point R from Q.
4. A butterfly moves from a point X to a point Y, a distance of 15 metres in the direction, 0450. It then changes direction and moves to a point Z, in the direction 1550. If the point Z is due east of X, find the distance of the point Z from the point Y.
WEEK 2
DERIVATION OF EQUATONS OF LINEAR MOTION.
CONTENTS.
Basic definition
Derivation of equations of linear motion
Motion under gravity
Basic definition.
There are four major terms associated with motion in a straight line . These are speed (v) or velocity (v), distance (s) or displacement (s), acceleration (a) and time (t).
DISPLACEMENT: This is the distance traveled in a specified direction. For example, if a body moves a distance of 50m northwards, it is a vector quantity while distance is a scalar quantity.
Distance indicates how far an object has moved. It’s a scalar quantity.
The rate at which a body covers a distance is called the SPEED of the body.
Thus, speed = distance
Time ( m/s , km/hr)
VELOCITY is the rate of change of displacement with time. When a body moves with equal displacement in equal interval of time, no matter how small the time intervals may be, the velocity is said to be uniform or constant.
ACCELERATION: is the rate of change of velocity with time . When the velocity increases in time , the rate of change of velocity is termed acceleration but when the velocity decreases with time ,it is called RETARDATION .Retardation is a negative acceleration.
When the rate of change of velocity with time is constant, we have uniform
acceleration
https://youtu.be/ZM8ECpBuQYE
EVALUATION
Sketch the velocity—time graph for a body that starts from rest and accelerates uniformly to a certain velocity If it maintains this for a given period before its eventual deceleration. Indicate the following:
1 Uniform acceleration, retardation
2 Total distance travelled
DERIVATION OF EQUATIONS OF LINEAR MOTION.
v= Final velocity
u = Initial velocity
a = Acceleration
t = Time
s = Distance
v= u + at --------------------- (i)
https://youtu.be/xViRvJxTu6k
FIRST EQUATION
If a body at rest or already moving with an initial velocity, u m/s, begins to accelerate at a m/s2, after a particular time, t sec, it will obtain a final velocity v m/s.
The acceleration ,a , will be defined by
a= change in velocity/time interval
a= final velocity - initial velocity/time interval
Substituting letters for word we have,
a=v-u/t
cross-multiplying we have,
v - u = at
Rearranging
V = u + at................................................(i)
This is the first equation of motion
SECOND EQUATION OF MOTION
The second equation of motion can be derived if we consider that a body moving with uniform or constant acceleration must have had an initial velocity u, before attaining a final velocity, v.
Therefore the average velocity is equal to the sum of the initial velocity, u and the final velocity , divided by two.
Thus, Average velocity u + v/2
Substituting the first equation of motion, v = u + at into the above we obtain
Average velocity = u + u + at/2
= 2u + at/2
= 2u/2 + at/2
= u + 1/2at
Average velocity = displacement (distance)/2 = s/t
= s/t = u + 1/2at2
Cross multiplying,
S= t(u + 1/2at)
S = ut + 1/2at2 ..............................................................(ii)
This is second equation of motion.
THIRD EQUATION OF MOTION
A combination of first and second equation of motion yields the third equation of motion as follows:
Square both sides of first equation of motion, v = u + at
(v)2 = (u + at )2
V2 = (u + at) (u + at)
V2 = u2 + uat + uat + (at)2
V2 = u2 + 2uat + a2t2
Factor out 2a from the last two terms of the right hand side of the above equation.
V2 = u2 + 2a/2a(2uat + a2t2)
V2 = u2 + 2a(2uat/2a + a2t2/2a)
V2 = u2 + 2a(ut + 1/2at2)
Substituting the second equation of motion s = ut + 1/2at2 into the above, we obtain
V2 = u2 + 2as .............................................................(3)
This is the third equation of motion.
The three equations of motion with uniform acceleration are:
(1) V = u + at
(2) S = ut + 1/2at2
(3) V2 = u2 + 2as
Where u = initial velocity in m/s or ms-1
V = final velocity in m or ms-1
t = time in sec
s= distance in m
a= uniform acceleration /deceleration in m/s2 or ms-2
https://youtu.be/viicuX8bZDA
https://youtu.be/GX5zToM_Vvg
CALCULATIONS USING THE EQUATION OF MOTION.
A car moves from rest with an acceleration of 0.2 m/s2. Find its velocity when it has covered distance of 50m
u= 0m/s
a= 0.2m/s2
s= 50m
v = ?
v2 =u2 + 2as = (0)2 + 2 ( 0.2 x 50) = 20
v= √20 = 2√5m/s
A car travels with a uniform velocity of 108km/hr .How far does it travels in ½ minutes?
Solution
https://youtu.be/h7aeEiBCxrk
PROBLEM SOLVING USING EQUATION OF MOTION
Example 1: A particle accelerates uniformly from rest at 6.0m/s2 for 8s and then decelerates uniformly to rest in the next 5s. Determine the magnitude of the deceleration.
Solution
You are to calculate the particle's deceleration. It decelerated to rest from a particular velocity which was attained during acceleration. So, you are to calculate the velocity attained as follows:
Initial velocity, u = 0, acceleration, a=6m/s2, t= 8s, velocity attained, v=?
Substituting into the most appropriate equation, v =u + at
V = 0 + 6 x 8
V = 48m/s
Next the particle decelerated to rest from a velocity of 48m/s, therefore, the deceleration is calculated thus;
Initial velocity, u = 48m/s; final velocity, v = 0, time, t = 5s, deceleration, a = ?
Substituting into the most appropriate equation,
V = u + at
0 = 48 + a x 5
0 = 48 + 5a
-5a = 48
A = - 9.6ms-2
Example 2: A particle starts from rest and moves with a uniform acceleration of 4m/s2. What is its velocity after covering a distance of 8m.
SOLUTION
Initial velocity, u = 0, acceleration, a = 4m/s2, distance, s = 8m, final velocity, v = ?
The equation containing u, a, s, and v is the 3rd equation of motion, v2 = u2 + 2as.
Substituting in to most appropriate equation,
V2 = 0 + 2 x 4 x 8
V2 = 64
V = 64
https://youtu.be/6KbS05evvDc
CLASS ACTIVITIES.
(1) A train slopes from 108 km/hr with a uniform retardation of 5 m/s[sup]2[/sup] . How long will it take to reach 18 km/hr and what is the distance covered ?
(2) An orange fruit drops to the ground from the top of a tree 45m tall .How long does it take to reach the ground (g= 10m/s2)?
(3) A car moving with a speed of 90 km/h was brought uniformly to rest by the application of brake in 10s . How far did the car travel after the far did the car travel after the brakes were applied .calculate the distance it covers in the last one second its motion.
(4) A body which is uniformly retarded comes to rest in 5s after travelling a distance of 10m. What is its initial velocity?
FURTHER ACTIVITIES
A car starts from rest and accelerates uniformly until it reaches a velocity of 30m/s after 5secs . It travels with this uniform velocity for 15secs and it is then brought to rest in 10 secs with a uniform acceleration. Determine
(a) the acceleration of the car
(b) the retardation
(c) the distance covered after 5 secs
(d) the total distance covered.
SOLUTION
CLASS WORK
A lorry starts from rest and accelerates uniformly until it reaches a velocity of 50 m/s after 10 secs . It travels with uniform velocity for 15 secs and is brought to rest I 5secs with a uniform retardation .
Calculate :
a) The acceleration of the lorry
b) The retardation
c) The total distance covered
d) The average speed of the lorry
MOTION UNDER GRAVITY
A body moving with a uniform acceleration in space does so under the influence of gravity with a constant acceleration . (g = 10 m/s2 ). In dealing with vertical motion under gravity , the following points must be noted
All objects dropped near the surface of the Earth in the absence of air resistance fall toward the Earth with the same nearly constant acceleration We denote the magnitude of free-fall acceleration as g.
The magnitude of free-fall acceleration decreases with increasing altitude. Furthermore, slight variations occur with latitude. At the surface of the Earth the magnitude is approximately 9.8 m/s². The vector is directed downward toward the centre of the Earth.
Free-fall acceleration is an important example of straight-line motion with constant acceleration.
When air resistance is negligible, even a feather and an apple fall with the same acceleration, regardless of their masses.
a= +g is positive for a downward motion but (-g) negative for an upward motion
The velocity v = 0 at maximum height for a vertically projected object
The initial velocity u = 0 for a body dropped from rest above the ground .
For a rebounding body the height h above the ground is zero
The time of fall of two objects of different masses has nothing to do with their masses but is dependent on the distance and acceleration due to gravity as shown below
S= ut + ½ gt2
S = ½ gt2 (u=0)
t= √2s/g
The above relationship can also be used to determine the value of acceleration due to gravity. If we plot s against t, it will give us a parabolic curve.
But the graph of s against t2 will give us a straight line through the origin with slope ½ g from which g can be computed
Case One. For a body projected from a tower or plane of height h.
1. The body covers both horizontal Sx ( also known as the range , R) and vertical Sy ( height) distance.
Gravity has no effect on the horizontal distance covered but on the vertical distance, hence
Sx = R= ut…………………… #
Sy = ut + 1/2gt2 but u = 0
Sy = ½ gt2 ……………………. #
Case two: for a body thrown vertically upward from the ground to a maximum height h and back to the ground.
At maximum height v = 0
Time taken to maximum height is same time taken from maximum height to the ground.
Time to maximum height (t)
V = u – gt
0 = u – gt
u = gt…………….. #
maximum height attained
V2 = u2 - 2ghmax
0 = u2 – 2ghmax
u2 = 2ghmax ………………….. #
Case three for a body projected from the top of a tower to a maximum height h
At maximum height h, v=0
Time (t) to maximum height
V = u - gt
0 = u –gt
U = gt
Time from maximum height h to the ground
Total distance travelled = h + h1
h + h1 = ut + 1/2gt2
https://youtu.be/Sw2P1hpwpqU
CALCULATIONS
1. A ball is thrown vertically into the air with an initial velocity , u . What is the greatest height reached
Solution
V2 = u2 + 2as
U= u , a = -g , v = 0
02 = u2 + 2 (-g) s
2gs = u2
s = u2 /2g
2. A ball is released from a height of 20m .Calculate
(i) the time it takes to fall
(ii) the velocity with which it hits the ground
a= +g u=0 s =20m t = ?
t = √2s/g
t = √ 2 x20 /10
t = 2 secs
v = u + gt
v= gt
v = 10 x2
v = 20 m/s
ASSIGNMENT
1. A body starts from rest and accelerate uniformly at 5m/s2 until it attain a velocity of 25m/s .Calculate the time taken to attain this velocity (A) 2S (b) 3s (c) 5s (d) 6s.
2. A particle accelerates uniformly from rest at 6m/s2 for 8secs and then decelerates uniformly to rest in the next 5 secs. Determine magnitude of the deceleration (a) 9.6 m/s2 (b) -9.6 m/s2 (c) 6.9 m/s2 (d) – 6.9 m/s2
3. A car takes off from rest and covers a distance of 80m on a straight road in10secs Calculate its acceleration (a) 160 m/s2 (b) 16 m/s2 (c) 1.6 m/s2 (d) 0.16 m/s2
4. An object is released from rest at a height of 20m. Calculate the time it takes to fall to the ground ( g= 10m/s2) (a) 1s (b) 2s (c) 3s (d) 4s.
5. A body accelerates uniformly rest at the rate of 3m/s2 for 8 secs. Calculate the distance it covers. (a) 24m (b) 48m (c) 72m (d) 96m.
THEORY
1. A particle start from rest and moves with constant acceleration of 0.5m/s2 . Calculate the time taken by the particle to cover a distance of 25m .
A particle accelerate uniformly from rest at 6m/s2 for 8secs and then decelerates uniformly to rest in the next 7secs .Determine the magnitude of the deceleration.
2. A motorist, travelling at 120km/hr sees a broken down truck at the middle of the road and immediately applies his brakes and comes to a stop with uniform retardation in 20s. What distance does the car travel after the brakes were applied?
READING ASSIGNMENT
New Sch. Physics for Senior Sec. Schls. Pages 130-134.
GENERAL EVALUATION
1.List 5 apparatus for measuring the mass of a body.
2. List 5 apparatus for measuring the length of a body.
3. Generate the two equation of motion from the definition of law
CONTENTS.
Basic definition
Derivation of equations of linear motion
Motion under gravity
Basic definition.
There are four major terms associated with motion in a straight line . These are speed (v) or velocity (v), distance (s) or displacement (s), acceleration (a) and time (t).
DISPLACEMENT: This is the distance traveled in a specified direction. For example, if a body moves a distance of 50m northwards, it is a vector quantity while distance is a scalar quantity.
Distance indicates how far an object has moved. It’s a scalar quantity.
The rate at which a body covers a distance is called the SPEED of the body.
Thus, speed = distance
Time ( m/s , km/hr)
VELOCITY is the rate of change of displacement with time. When a body moves with equal displacement in equal interval of time, no matter how small the time intervals may be, the velocity is said to be uniform or constant.
ACCELERATION: is the rate of change of velocity with time . When the velocity increases in time , the rate of change of velocity is termed acceleration but when the velocity decreases with time ,it is called RETARDATION .Retardation is a negative acceleration.
When the rate of change of velocity with time is constant, we have uniform
acceleration
https://youtu.be/ZM8ECpBuQYE
EVALUATION
Sketch the velocity—time graph for a body that starts from rest and accelerates uniformly to a certain velocity If it maintains this for a given period before its eventual deceleration. Indicate the following:
1 Uniform acceleration, retardation
2 Total distance travelled
DERIVATION OF EQUATIONS OF LINEAR MOTION.
v= Final velocity
u = Initial velocity
a = Acceleration
t = Time
s = Distance
v= u + at --------------------- (i)
https://youtu.be/xViRvJxTu6k
FIRST EQUATION
If a body at rest or already moving with an initial velocity, u m/s, begins to accelerate at a m/s2, after a particular time, t sec, it will obtain a final velocity v m/s.
The acceleration ,a , will be defined by
a= change in velocity/time interval
a= final velocity - initial velocity/time interval
Substituting letters for word we have,
a=v-u/t
cross-multiplying we have,
v - u = at
Rearranging
V = u + at................................................(i)
This is the first equation of motion
SECOND EQUATION OF MOTION
The second equation of motion can be derived if we consider that a body moving with uniform or constant acceleration must have had an initial velocity u, before attaining a final velocity, v.
Therefore the average velocity is equal to the sum of the initial velocity, u and the final velocity , divided by two.
Thus, Average velocity u + v/2
Substituting the first equation of motion, v = u + at into the above we obtain
Average velocity = u + u + at/2
= 2u + at/2
= 2u/2 + at/2
= u + 1/2at
Average velocity = displacement (distance)/2 = s/t
= s/t = u + 1/2at2
Cross multiplying,
S= t(u + 1/2at)
S = ut + 1/2at2 ..............................................................(ii)
This is second equation of motion.
THIRD EQUATION OF MOTION
A combination of first and second equation of motion yields the third equation of motion as follows:
Square both sides of first equation of motion, v = u + at
(v)2 = (u + at )2
V2 = (u + at) (u + at)
V2 = u2 + uat + uat + (at)2
V2 = u2 + 2uat + a2t2
Factor out 2a from the last two terms of the right hand side of the above equation.
V2 = u2 + 2a/2a(2uat + a2t2)
V2 = u2 + 2a(2uat/2a + a2t2/2a)
V2 = u2 + 2a(ut + 1/2at2)
Substituting the second equation of motion s = ut + 1/2at2 into the above, we obtain
V2 = u2 + 2as .............................................................(3)
This is the third equation of motion.
The three equations of motion with uniform acceleration are:
(1) V = u + at
(2) S = ut + 1/2at2
(3) V2 = u2 + 2as
Where u = initial velocity in m/s or ms-1
V = final velocity in m or ms-1
t = time in sec
s= distance in m
a= uniform acceleration /deceleration in m/s2 or ms-2
https://youtu.be/viicuX8bZDA
https://youtu.be/GX5zToM_Vvg
CALCULATIONS USING THE EQUATION OF MOTION.
A car moves from rest with an acceleration of 0.2 m/s2. Find its velocity when it has covered distance of 50m
u= 0m/s
a= 0.2m/s2
s= 50m
v = ?
v2 =u2 + 2as = (0)2 + 2 ( 0.2 x 50) = 20
v= √20 = 2√5m/s
A car travels with a uniform velocity of 108km/hr .How far does it travels in ½ minutes?
Solution
https://youtu.be/h7aeEiBCxrk
PROBLEM SOLVING USING EQUATION OF MOTION
Example 1: A particle accelerates uniformly from rest at 6.0m/s2 for 8s and then decelerates uniformly to rest in the next 5s. Determine the magnitude of the deceleration.
Solution
You are to calculate the particle's deceleration. It decelerated to rest from a particular velocity which was attained during acceleration. So, you are to calculate the velocity attained as follows:
Initial velocity, u = 0, acceleration, a=6m/s2, t= 8s, velocity attained, v=?
Substituting into the most appropriate equation, v =u + at
V = 0 + 6 x 8
V = 48m/s
Next the particle decelerated to rest from a velocity of 48m/s, therefore, the deceleration is calculated thus;
Initial velocity, u = 48m/s; final velocity, v = 0, time, t = 5s, deceleration, a = ?
Substituting into the most appropriate equation,
V = u + at
0 = 48 + a x 5
0 = 48 + 5a
-5a = 48
A = - 9.6ms-2
Example 2: A particle starts from rest and moves with a uniform acceleration of 4m/s2. What is its velocity after covering a distance of 8m.
SOLUTION
Initial velocity, u = 0, acceleration, a = 4m/s2, distance, s = 8m, final velocity, v = ?
The equation containing u, a, s, and v is the 3rd equation of motion, v2 = u2 + 2as.
Substituting in to most appropriate equation,
V2 = 0 + 2 x 4 x 8
V2 = 64
V = 64
https://youtu.be/6KbS05evvDc
CLASS ACTIVITIES.
(1) A train slopes from 108 km/hr with a uniform retardation of 5 m/s[sup]2[/sup] . How long will it take to reach 18 km/hr and what is the distance covered ?
(2) An orange fruit drops to the ground from the top of a tree 45m tall .How long does it take to reach the ground (g= 10m/s2)?
(3) A car moving with a speed of 90 km/h was brought uniformly to rest by the application of brake in 10s . How far did the car travel after the far did the car travel after the brakes were applied .calculate the distance it covers in the last one second its motion.
(4) A body which is uniformly retarded comes to rest in 5s after travelling a distance of 10m. What is its initial velocity?
FURTHER ACTIVITIES
A car starts from rest and accelerates uniformly until it reaches a velocity of 30m/s after 5secs . It travels with this uniform velocity for 15secs and it is then brought to rest in 10 secs with a uniform acceleration. Determine
(a) the acceleration of the car
(b) the retardation
(c) the distance covered after 5 secs
(d) the total distance covered.
SOLUTION
CLASS WORK
A lorry starts from rest and accelerates uniformly until it reaches a velocity of 50 m/s after 10 secs . It travels with uniform velocity for 15 secs and is brought to rest I 5secs with a uniform retardation .
Calculate :
a) The acceleration of the lorry
b) The retardation
c) The total distance covered
d) The average speed of the lorry
MOTION UNDER GRAVITY
A body moving with a uniform acceleration in space does so under the influence of gravity with a constant acceleration . (g = 10 m/s2 ). In dealing with vertical motion under gravity , the following points must be noted
All objects dropped near the surface of the Earth in the absence of air resistance fall toward the Earth with the same nearly constant acceleration We denote the magnitude of free-fall acceleration as g.
The magnitude of free-fall acceleration decreases with increasing altitude. Furthermore, slight variations occur with latitude. At the surface of the Earth the magnitude is approximately 9.8 m/s². The vector is directed downward toward the centre of the Earth.
Free-fall acceleration is an important example of straight-line motion with constant acceleration.
When air resistance is negligible, even a feather and an apple fall with the same acceleration, regardless of their masses.
a= +g is positive for a downward motion but (-g) negative for an upward motion
The velocity v = 0 at maximum height for a vertically projected object
The initial velocity u = 0 for a body dropped from rest above the ground .
For a rebounding body the height h above the ground is zero
The time of fall of two objects of different masses has nothing to do with their masses but is dependent on the distance and acceleration due to gravity as shown below
S= ut + ½ gt2
S = ½ gt2 (u=0)
t= √2s/g
The above relationship can also be used to determine the value of acceleration due to gravity. If we plot s against t, it will give us a parabolic curve.
But the graph of s against t2 will give us a straight line through the origin with slope ½ g from which g can be computed
Case One. For a body projected from a tower or plane of height h.
1. The body covers both horizontal Sx ( also known as the range , R) and vertical Sy ( height) distance.
Gravity has no effect on the horizontal distance covered but on the vertical distance, hence
Sx = R= ut…………………… #
Sy = ut + 1/2gt2 but u = 0
Sy = ½ gt2 ……………………. #
Case two: for a body thrown vertically upward from the ground to a maximum height h and back to the ground.
At maximum height v = 0
Time taken to maximum height is same time taken from maximum height to the ground.
Time to maximum height (t)
V = u – gt
0 = u – gt
u = gt…………….. #
maximum height attained
V2 = u2 - 2ghmax
0 = u2 – 2ghmax
u2 = 2ghmax ………………….. #
Case three for a body projected from the top of a tower to a maximum height h
At maximum height h, v=0
Time (t) to maximum height
V = u - gt
0 = u –gt
U = gt
Time from maximum height h to the ground
Total distance travelled = h + h1
h + h1 = ut + 1/2gt2
https://youtu.be/Sw2P1hpwpqU
CALCULATIONS
1. A ball is thrown vertically into the air with an initial velocity , u . What is the greatest height reached
Solution
V2 = u2 + 2as
U= u , a = -g , v = 0
02 = u2 + 2 (-g) s
2gs = u2
s = u2 /2g
2. A ball is released from a height of 20m .Calculate
(i) the time it takes to fall
(ii) the velocity with which it hits the ground
a= +g u=0 s =20m t = ?
t = √2s/g
t = √ 2 x20 /10
t = 2 secs
v = u + gt
v= gt
v = 10 x2
v = 20 m/s
ASSIGNMENT
1. A body starts from rest and accelerate uniformly at 5m/s2 until it attain a velocity of 25m/s .Calculate the time taken to attain this velocity (A) 2S (b) 3s (c) 5s (d) 6s.
2. A particle accelerates uniformly from rest at 6m/s2 for 8secs and then decelerates uniformly to rest in the next 5 secs. Determine magnitude of the deceleration (a) 9.6 m/s2 (b) -9.6 m/s2 (c) 6.9 m/s2 (d) – 6.9 m/s2
3. A car takes off from rest and covers a distance of 80m on a straight road in10secs Calculate its acceleration (a) 160 m/s2 (b) 16 m/s2 (c) 1.6 m/s2 (d) 0.16 m/s2
4. An object is released from rest at a height of 20m. Calculate the time it takes to fall to the ground ( g= 10m/s2) (a) 1s (b) 2s (c) 3s (d) 4s.
5. A body accelerates uniformly rest at the rate of 3m/s2 for 8 secs. Calculate the distance it covers. (a) 24m (b) 48m (c) 72m (d) 96m.
THEORY
1. A particle start from rest and moves with constant acceleration of 0.5m/s2 . Calculate the time taken by the particle to cover a distance of 25m .
A particle accelerate uniformly from rest at 6m/s2 for 8secs and then decelerates uniformly to rest in the next 7secs .Determine the magnitude of the deceleration.
2. A motorist, travelling at 120km/hr sees a broken down truck at the middle of the road and immediately applies his brakes and comes to a stop with uniform retardation in 20s. What distance does the car travel after the brakes were applied?
READING ASSIGNMENT
New Sch. Physics for Senior Sec. Schls. Pages 130-134.
GENERAL EVALUATION
1.List 5 apparatus for measuring the mass of a body.
2. List 5 apparatus for measuring the length of a body.
3. Generate the two equation of motion from the definition of law
WEEK 3
PROJECTILES AND FALLING BODIES
CONTENTS
Terms associated with projectiles
Equation of projectile motion
Uses of projectile
A PROJECTILE is an object or body launched into the air and allowed to move on its own or move freely under gravity.
A projectile motion is one that follows a curved or parabolic path .It is due to two independent motions at right angle to each other These motions are
i. a horizontal constant velocity
ii. a vertical free fall due to gravity
Projectile is a two dimensional motion of an object thrown obliquely into the air, the part followed by a projectile is called a trajectory
Examples of projectile motion are the motion of;
i A throw rubber ball rebouncing from a wall
ii. An athlete doing the high jump
ii A stone released from a catapult
iv A bullet fired from a gum
v. A cricket ball thrown against a vertical wall.
TERMS ASSOCIATED WITH PROJECTILE
Time of flight ( T ):The time of flight of a projectile is the time required for the projectile to get to maximum height and return to the same level from which it projected.
2.The maximum height (H):is defined as the highest vertical distance reached and is measured from the horizontal projection plane.
3.The range ( R): is the horizontal distance from the point of projection of a particle to the point where the particle hit the projection plane again.
U[sub]y[/sub] = U sin θ (vertical component)
U[sub]x[/sub] = U cos θ (Horizontal component)
t = time to reach the greatest height (s)
V = u + at v =o, a = -g
θ= u sin – gt
t = [sup]U sin θ[/sup]/[sub]g[/sub] ………………………………………….. 1
T = Time of flight (s)
https://youtu.be/8NLzuURxFwY
DETERMINATION OF TIME OF FLIGHTCT, RANGE,(R) AND MAXTMUM HEIGHT.
Assuming that Q is the point where the particles meet the target. Let T be the time of flight at Q, the vertical displacement is zero
Vertically S=U sin θ t - ½gt[sup]2[/sup]
0 = U sin θ – ½ gt[sup]2[/sup]
½ gt[sup]2[/sup] = U sin θ t
T = [sup]2 U sin θ[/sup]/[sub]g[/sub] ....................... 2
Horizontally, considering the range covered
R= [sup]U2 sin 2θ[/sup]/[sub]g[/sub] ………………………………. 3
For max . range θ = 45º
Sin2θ = sin 2 (45) = sin 90º = 1
R= [sup]U2[/sup]/[sub]g [/sub]
R max = [sup]U2[/sup]/[sub]g[/sub] ……………………………………………. 4
For maximum height H ,
V2 = U2 sin2θ - 2g H
At max height H, V=0
H = [sup]U2 sin2θ[/sup]/[sub]2g[/sub] …………………………………………… 5
USE OF PROJECTILES
1. To launch missiles in modern warfare
2. To give athletes maximum take off speed at meets
In artillery warfare, in order to strike a specified target, the bomb must be released when the target appears at the angle of depression p given by :
Tan φ =[sup]1[/sup]/[sub]u[/sub] [sup]√gh[/sup]/[sub]2 [/sub]
https://youtu.be/pZZt357pk-I
EXAMPLES
1. A bomber on a military mission is flying horizontally at a height of zoom above the ground at 60kmmin-1
2. lt drops a bomb on a target on the ground. determine the acute angle between the vertical and the line joining the bomber and the tangent at the instant. the bomb is released
Horizontal velocity of bomber = 60km/min= 10[sup]3[/sup] ms[sup]-1[/sup]
Bomb falls with a vertical acceleration of g = 10m/s
At the release of the bomb, it moves with a horizontal velocity equals that of the aircraft i.e. 1000m/s
Considering the vertical motion of the bomb we have
h =ut+½ gt[sup]2[/sup](u=o)
h =½gt[sup]2[/sup]
where t is the time the bomb takes to reach the ground :. 300=½gt[sup]2[/sup]
t2= 600
t=10 √6 sec, considering the horizontal motion we have that horizontal distance moved by the bomb in time t is given by
s =horizontal velocity x time
= 1000 x10√6 = 2.449x10[sup]4[/sup] m
but tanθ = s = [sup]2.449 x 10[sup]4[/sup][/sup] /[sub]3,000 x 3,000[/sub]
θ =83.02º
2.A stone is shot out from a catapult with an initial velocity of 30m at an elevation of 60,find
a. the time of flight
b. the maximum height attained
c the range
T = [sup]2U sin θ[/sup]/[sub]g[/sub]
T= [sup]2 x 30 sin 60º[/sup]/[sub]10[/sub]
T= 5.2s
The maximum height,
3. A body is projected horizontally with a velocity of 60m/s from the top of a mast 120m above the grand, calculate
(i) Time of flight, and
(ii) Range
a. s =ut+½gt[sup]2[/sup]
a=g, u=o
120= ½ (10)t[sup]2[/sup]
t2 = 24
t = 24
t =4.9s
(b) Range =u cosθ x T.
but in this case θ =o
cos o =1
R =ut
= 60x 4.9
=294m
s =ut+½gt[sup]2[/sup]
a=g, u=o
120= ½ (10)t[sup]2[/sup]
t2 = 24
t = 24
t =4.9s
4 A stone is projected horizontally with a speed of 10m/s from the top of a tower and with what speed does the stone strike the ground?
T = [sup]√ 2H[/sup]/[sub]g [/sub] =[sup]√2x50[/sup]/[sub]10[/sub] =10
R = ut = 10√10 m
V[sup]2[/sup]=u[sup]2[/sup] + 2gh
=10[sup]2[/sup]+2x10x50
=100+100
=200
=10m/s
5. A projectile is fired at an angle of 60 with the horizontal with an initial velocity of 80m/s. Calculate:
i the time of flight
ii. the maximum height attained and the time taken to reach the height
iii. the velocity of projection 2 seconds after being fired (g = 10m/s)
(iii) Vy = U sin θ – gt
Vy = 80 sin 60 – 20 = 49.28m/s
Ux = U cos θ
Ux = 80 cos 60 = 40 m/s
U² = U²y + U²x
= 49.28² + 40² = √1600+ 2420 = 63.41 m/s
https://youtu.be/zuFs17Nf41c
ASSIGNMENT
1 A ball is projected horizontally from the top of a hill with a velocity of 30m/s. if it reaches the ground 5 seconds later, the height of the hill is:
(a).200m (b). 65m (c ).250m (d).100m.
2.The maximum height of a projectile projected with an angle of to the horizontal and an initial velocity of U is given by
(a) U sin2 θ/g (b) U2 sin θ/2g (c) U2 sin θ/g (d) 2U2sin2θ/g (e) 2U sin2 θ /g
3. A stone is projected at an angle 60 and an initial velocity of 20m/s determine the time of flight (a) 34.6s (b) 3.46s (c) 1.73s (d) 17.3s (e) 6.92s
4.The range of a projectile projected at θ to the horizontal with a velocity U is given by
(a) U sin2 θ/g (b) U2 sin θ/2g (c) U2 sin θ/g (d) 2U2sin2θ/g (e) 2U sin2 θ /g
5.For a projectile the maximum range is obtained when the angle of projection is;
a). 60 b) .30 c).45 d).75 e).90
THEORY
1. A gun fires a shell at an angle of elevation of 30 with a velocity of 2x10m what are the horizontal and vertical components of the velocity? What is the range of the shell? How high will it rise?
2. A stone propelled from a catapult with a speed of 50m/s attains a height of 100m. Calculate. (a)the time of flight (b). the angle of projection (c). the range attained.
READING ASSIGNMENT
New Sch. Physics for Senior Sec. Schls. Pages 137-144.
GENERAL EVALUATION
A stone of mass 0.4Kg is attached to a string of length 2.5m and its is spin around by a boy at 5rad/s. calculate
1. The force necessary for this motion.
2. The linear velocity with the stone moves.
MAIN TOPIC: APPROXIMATION AND PERCENTAGE ERROR.
REFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to: Solve problems on approximation and percentage error
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Hardly do we take any measurement and get exact results. The results often got are inaccurate. The importance being stressed here is not the slight error made in the measurement but to decide the degree of such error. This degree or level of error always measured in percentage will then guide any decision maker on the acceptable level of error.
The error that may be made in any measurement could either be positive or negative, e.g. the actual length of a rope is 15.6m. If the rope is measured by two boys as 13.4m and 16.9m respectively, the errors are as follows:
First boy's error = 15.6 - 13.4 = 2.2m
Second boy's error = 16.9 ' 15.6 = 1.3m
Whether the error is against (as in first boy) or in favour (as in second boy), error is error, therefore, the following formula is used:
Percentage error = [sup]Absolute error[/sup]/[sub]Actual value[/sub] x 100
https://youtu.be/hR3hOyk_Qjw
Example 1: A rope of length 15cm was measured by a girl to be 14.4cm. Find the percentage error.
Solution
The actual error is 15 - 14.4 = 0.6cm
Percentage error = 0.6/15 x 100%
= 4%
Example 2: The length of a wire is 6.35, a student measured it as 6.65. What is the percentage error to 1decimal place?
Solution
The actual error is 6.65 - 6.35 = 2.5
Percentage error = 2.5/6.35 x 100%
= 4.7%
Example 3: A man underestimated his expenses by 6.5%,but actually spent #400.00. What was his estimate?
Solution
6.5% of #400.00 = #26.00
Therefore, the actual estimate is 400.00 - 26.00
#374.00
Example 4: A man underestimated his expenses by 1 2/3%, but actually spent #6500.00. What is his estimate?
Solution
1 2/3% = 5/3 = 1.66%
1.66 x 6500
100 1
1.66 x 65
#107.9
Therefore the actual estimate = #6500.00 - #107.9
= #6392.1
Example 5: An error of 4% was made in finding the length of a rope that was actually 25m. By how many metres was the measurement wrong?
Solution
4% of 25 = 1
The actual length of the rope is 25m
The error is 1
Therefore, the measurement was wrong with 1m
https://youtu.be/hR3hOyk_Qjw
EVALUATION:
1. A candidate was to subtract 15 from a certain number, but mistakenly added 25 and his answer was 145. Find the percentage error.
If the age of a man 64 years is written as 71 years, calculate error to 3 significant figures.
2. A man underestimated his expenses by 4.8%but actually spent #2355.00. What was his estimate?
3.(i) What is the percentage error in an area of a lawn that actually measures 750m2 but found to be 690m2
(ii) The percentage error in the measurement of the length of a rope was 6%. If the measurement was 35m, find the actual length of the rope to 1 decimal place.
ASSIGNMENT:
Find the percentage error in a piece of wood that was measured to be 1.26m whose actual length was 1.24m
CONTENTS
Terms associated with projectiles
Equation of projectile motion
Uses of projectile
A PROJECTILE is an object or body launched into the air and allowed to move on its own or move freely under gravity.
A projectile motion is one that follows a curved or parabolic path .It is due to two independent motions at right angle to each other These motions are
i. a horizontal constant velocity
ii. a vertical free fall due to gravity
Projectile is a two dimensional motion of an object thrown obliquely into the air, the part followed by a projectile is called a trajectory
Examples of projectile motion are the motion of;
i A throw rubber ball rebouncing from a wall
ii. An athlete doing the high jump
ii A stone released from a catapult
iv A bullet fired from a gum
v. A cricket ball thrown against a vertical wall.
TERMS ASSOCIATED WITH PROJECTILE
Time of flight ( T ):The time of flight of a projectile is the time required for the projectile to get to maximum height and return to the same level from which it projected.
2.The maximum height (H):is defined as the highest vertical distance reached and is measured from the horizontal projection plane.
3.The range ( R): is the horizontal distance from the point of projection of a particle to the point where the particle hit the projection plane again.
U[sub]y[/sub] = U sin θ (vertical component)
U[sub]x[/sub] = U cos θ (Horizontal component)
t = time to reach the greatest height (s)
V = u + at v =o, a = -g
θ= u sin – gt
t = [sup]U sin θ[/sup]/[sub]g[/sub] ………………………………………….. 1
T = Time of flight (s)
https://youtu.be/8NLzuURxFwY
DETERMINATION OF TIME OF FLIGHTCT, RANGE,(R) AND MAXTMUM HEIGHT.
Assuming that Q is the point where the particles meet the target. Let T be the time of flight at Q, the vertical displacement is zero
Vertically S=U sin θ t - ½gt[sup]2[/sup]
0 = U sin θ – ½ gt[sup]2[/sup]
½ gt[sup]2[/sup] = U sin θ t
T = [sup]2 U sin θ[/sup]/[sub]g[/sub] ....................... 2
Horizontally, considering the range covered
R= [sup]U2 sin 2θ[/sup]/[sub]g[/sub] ………………………………. 3
For max . range θ = 45º
Sin2θ = sin 2 (45) = sin 90º = 1
R= [sup]U2[/sup]/[sub]g [/sub]
R max = [sup]U2[/sup]/[sub]g[/sub] ……………………………………………. 4
For maximum height H ,
V2 = U2 sin2θ - 2g H
At max height H, V=0
H = [sup]U2 sin2θ[/sup]/[sub]2g[/sub] …………………………………………… 5
USE OF PROJECTILES
1. To launch missiles in modern warfare
2. To give athletes maximum take off speed at meets
In artillery warfare, in order to strike a specified target, the bomb must be released when the target appears at the angle of depression p given by :
Tan φ =[sup]1[/sup]/[sub]u[/sub] [sup]√gh[/sup]/[sub]2 [/sub]
https://youtu.be/pZZt357pk-I
EXAMPLES
1. A bomber on a military mission is flying horizontally at a height of zoom above the ground at 60kmmin-1
2. lt drops a bomb on a target on the ground. determine the acute angle between the vertical and the line joining the bomber and the tangent at the instant. the bomb is released
Horizontal velocity of bomber = 60km/min= 10[sup]3[/sup] ms[sup]-1[/sup]
Bomb falls with a vertical acceleration of g = 10m/s
At the release of the bomb, it moves with a horizontal velocity equals that of the aircraft i.e. 1000m/s
Considering the vertical motion of the bomb we have
h =ut+½ gt[sup]2[/sup](u=o)
h =½gt[sup]2[/sup]
where t is the time the bomb takes to reach the ground :. 300=½gt[sup]2[/sup]
t2= 600
t=10 √6 sec, considering the horizontal motion we have that horizontal distance moved by the bomb in time t is given by
s =horizontal velocity x time
= 1000 x10√6 = 2.449x10[sup]4[/sup] m
but tanθ = s = [sup]2.449 x 10[sup]4[/sup][/sup] /[sub]3,000 x 3,000[/sub]
θ =83.02º
2.A stone is shot out from a catapult with an initial velocity of 30m at an elevation of 60,find
a. the time of flight
b. the maximum height attained
c the range
T = [sup]2U sin θ[/sup]/[sub]g[/sub]
T= [sup]2 x 30 sin 60º[/sup]/[sub]10[/sub]
T= 5.2s
The maximum height,
3. A body is projected horizontally with a velocity of 60m/s from the top of a mast 120m above the grand, calculate
(i) Time of flight, and
(ii) Range
a. s =ut+½gt[sup]2[/sup]
a=g, u=o
120= ½ (10)t[sup]2[/sup]
t2 = 24
t = 24
t =4.9s
(b) Range =u cosθ x T.
but in this case θ =o
cos o =1
R =ut
= 60x 4.9
=294m
s =ut+½gt[sup]2[/sup]
a=g, u=o
120= ½ (10)t[sup]2[/sup]
t2 = 24
t = 24
t =4.9s
4 A stone is projected horizontally with a speed of 10m/s from the top of a tower and with what speed does the stone strike the ground?
T = [sup]√ 2H[/sup]/[sub]g [/sub] =[sup]√2x50[/sup]/[sub]10[/sub] =10
R = ut = 10√10 m
V[sup]2[/sup]=u[sup]2[/sup] + 2gh
=10[sup]2[/sup]+2x10x50
=100+100
=200
=10m/s
5. A projectile is fired at an angle of 60 with the horizontal with an initial velocity of 80m/s. Calculate:
i the time of flight
ii. the maximum height attained and the time taken to reach the height
iii. the velocity of projection 2 seconds after being fired (g = 10m/s)
(iii) Vy = U sin θ – gt
Vy = 80 sin 60 – 20 = 49.28m/s
Ux = U cos θ
Ux = 80 cos 60 = 40 m/s
U² = U²y + U²x
= 49.28² + 40² = √1600+ 2420 = 63.41 m/s
https://youtu.be/zuFs17Nf41c
ASSIGNMENT
1 A ball is projected horizontally from the top of a hill with a velocity of 30m/s. if it reaches the ground 5 seconds later, the height of the hill is:
(a).200m (b). 65m (c ).250m (d).100m.
2.The maximum height of a projectile projected with an angle of to the horizontal and an initial velocity of U is given by
(a) U sin2 θ/g (b) U2 sin θ/2g (c) U2 sin θ/g (d) 2U2sin2θ/g (e) 2U sin2 θ /g
3. A stone is projected at an angle 60 and an initial velocity of 20m/s determine the time of flight (a) 34.6s (b) 3.46s (c) 1.73s (d) 17.3s (e) 6.92s
4.The range of a projectile projected at θ to the horizontal with a velocity U is given by
(a) U sin2 θ/g (b) U2 sin θ/2g (c) U2 sin θ/g (d) 2U2sin2θ/g (e) 2U sin2 θ /g
5.For a projectile the maximum range is obtained when the angle of projection is;
a). 60 b) .30 c).45 d).75 e).90
THEORY
1. A gun fires a shell at an angle of elevation of 30 with a velocity of 2x10m what are the horizontal and vertical components of the velocity? What is the range of the shell? How high will it rise?
2. A stone propelled from a catapult with a speed of 50m/s attains a height of 100m. Calculate. (a)the time of flight (b). the angle of projection (c). the range attained.
READING ASSIGNMENT
New Sch. Physics for Senior Sec. Schls. Pages 137-144.
GENERAL EVALUATION
A stone of mass 0.4Kg is attached to a string of length 2.5m and its is spin around by a boy at 5rad/s. calculate
1. The force necessary for this motion.
2. The linear velocity with the stone moves.
MAIN TOPIC: APPROXIMATION AND PERCENTAGE ERROR.
REFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to: Solve problems on approximation and percentage error
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Hardly do we take any measurement and get exact results. The results often got are inaccurate. The importance being stressed here is not the slight error made in the measurement but to decide the degree of such error. This degree or level of error always measured in percentage will then guide any decision maker on the acceptable level of error.
The error that may be made in any measurement could either be positive or negative, e.g. the actual length of a rope is 15.6m. If the rope is measured by two boys as 13.4m and 16.9m respectively, the errors are as follows:
First boy's error = 15.6 - 13.4 = 2.2m
Second boy's error = 16.9 ' 15.6 = 1.3m
Whether the error is against (as in first boy) or in favour (as in second boy), error is error, therefore, the following formula is used:
Percentage error = [sup]Absolute error[/sup]/[sub]Actual value[/sub] x 100
https://youtu.be/hR3hOyk_Qjw
Example 1: A rope of length 15cm was measured by a girl to be 14.4cm. Find the percentage error.
Solution
The actual error is 15 - 14.4 = 0.6cm
Percentage error = 0.6/15 x 100%
= 4%
Example 2: The length of a wire is 6.35, a student measured it as 6.65. What is the percentage error to 1decimal place?
Solution
The actual error is 6.65 - 6.35 = 2.5
Percentage error = 2.5/6.35 x 100%
= 4.7%
Example 3: A man underestimated his expenses by 6.5%,but actually spent #400.00. What was his estimate?
Solution
6.5% of #400.00 = #26.00
Therefore, the actual estimate is 400.00 - 26.00
#374.00
Example 4: A man underestimated his expenses by 1 2/3%, but actually spent #6500.00. What is his estimate?
Solution
1 2/3% = 5/3 = 1.66%
1.66 x 6500
100 1
1.66 x 65
#107.9
Therefore the actual estimate = #6500.00 - #107.9
= #6392.1
Example 5: An error of 4% was made in finding the length of a rope that was actually 25m. By how many metres was the measurement wrong?
Solution
4% of 25 = 1
The actual length of the rope is 25m
The error is 1
Therefore, the measurement was wrong with 1m
https://youtu.be/hR3hOyk_Qjw
EVALUATION:
1. A candidate was to subtract 15 from a certain number, but mistakenly added 25 and his answer was 145. Find the percentage error.
If the age of a man 64 years is written as 71 years, calculate error to 3 significant figures.
2. A man underestimated his expenses by 4.8%but actually spent #2355.00. What was his estimate?
3.(i) What is the percentage error in an area of a lawn that actually measures 750m2 but found to be 690m2
(ii) The percentage error in the measurement of the length of a rope was 6%. If the measurement was 35m, find the actual length of the rope to 1 decimal place.
ASSIGNMENT:
Find the percentage error in a piece of wood that was measured to be 1.26m whose actual length was 1.24m
WEEK 4
SCALAR AND VECTOR QUANTITIES.
CONTENTS
Concept of scalar and vector quantities.
Vector representation, addition of vectors.
Resolution of vectors and resultant.
CONCEPT OF SCALAR AND VECTOR QUANTITIES.
Physical quantities are divided into two types
i Scalar quantity
ii Vector quantity
CONCEPT OF SCALAR: A quantity that has only a magnitude but no direction is called a scalar quantity. Examples: length, area, volume, mass of a body, energy, speed, etc.
A scalar is one which has only magnitude (size ) but no direction e.g. distance, speed , temperature , volume , work , energy, power , mass ,electric potential , gravitational potential electric charge .
Scalar quantities are added according to the ordinary rules of arithmetic for example , a mark of 50 added to a mark of 40 produces a mark of 90 –no directional property .But a force of 50N combined with a force of 40N may produce 90N if they are acting in same direction. But if in opposite direction would produce a different result. This vectors are combined or added by a special law the parallelogram law of addition of vectors.
https://youtu.be/Pj8Zh0A-uLU
CONCEPT OF VECTOR: Vector quantity is that quantity which has both a magnitude and a direction. Examples are; Displacement, force, momentum, acceleration, velocity, weight, magnetic flux, electric fields and gravitational fields etc.
The magnitude is shown by the length and the direction by the arrow representing the position or direction. The resultant of a scalar quantity is simply added algebraically. To find the resultant of two or more vector quantities, simple algebraic addition is not applicable. The addition is done by applying the parallelogram's law or the polygon's law of vectors.
https://drive.google.com/file/d/1q0vlmT ... sp=sharing
https://youtu.be/V_CJXlgar8Y
VECTOR REPRESENTATION.
Vector representation may be referred to as a way of representing vector either graphically or analytically.
A vector quantity can be graphically represented by a line drawn so that the length of the line denotes the magnitude of the quantity . The direction of the line indicates the direction in which the vector quantity act and it is shown by an arrow head . E.g a distance of 5km west represented by 5cm length of line where 1km = 1cm
https://drive.google.com/file/d/1JX85P- ... sp=sharing
ADDITION AND SUBTRACTION OF VECTORS
Two or more vectors acting on a body in a specified direction can be combined to produce a single vector having the same effect .The single vector is called the resultant.
For example:
(a) Two forces Y and X with magnitude of 3N and 4N respectively acting along the same direction will produce a resultant of 7N (algebraic sum of the two vectors).
(c) If the two vectors are inclined at an angle less than 900 or more than 900 , the resultant cannot be obtained by Pythagoras theorem but by vector addition,. Parallelogram law of vector, trigonometric or scale drawings can be used to calculate the magnitude and direction of the resultant
https://youtu.be/VFRW0f0XUU8
VECTORS AT RIGHT ANGLES
i. Parallelogram law of vectors states that if two vectors are represented in magnitude and direction by adjacent sides of a parallelogram , the resultant is represented in magnitude and direction by the diagonal of the parallelogram drawn from the common point
ii. If the two vectors are inclined at an angle less than 900 , the scale drawing or trigonometric method can used . In using scale drawing (graphical ) methods, a convenient scale is chosen ( if the magnitude of the forces given is large ) and then draw the lengths corresponding to the magnitude of the forces. A Protractor is used to draw the angle in between the forces. The parallelogram is completed and the resultant and its fraction obtained
https://youtu.be/iOipHqLV3OU
RESOLUTION OF VECTORS
A single vector can be resolved into two vectors called components. A vector F represented as the diagonal of the parallelogram can be resolved into its component later taken as the adjacent sides of the parallelogram.
THE RESULATNT OF MORE THAN TWO VECTORS
To find the resultant of more than two vectors, we resolve each vector in two perpendicular direction s add all the horizontal components X, and all the vertical components, Y.
For example, consider four forces acting on a body as shown below
https://youtu.be/RxV06Oj5sMc
EVALUATION
1 Calculate the resultant of five coplanar forces of values10N, 12N , 16N , 20N , 15N on an object as shown below
2. (i) Find by means of vector diagram, the resultant of two forces of 7N and 3N acting perpendicular to each other.
(ii) Two forces, 3N and 4N act on a body in the direction due North and East respectively. Calculate their equilibrium.
3. Two forces whose resultant is 10n, are perpendicular to each other. If one of them makes an angle of 60 with the resultant, calculate its magnitude.
ASSIGNMENT:
1. Which of the following is not a vector quantity (a) speed (b) velocity (c) force (d) acceleration (e) Electric field
2. Which of the following is not a scalar quantity ( a) density (b) weight (c) speed (d) mass (e) temperature
3. Two forces , whose resultant is 100N are perpendicular to each other .If one of the makes an angle of 60 with the resultant , calculate its magnitude ( sin60 = 0.8660 , cos 60 = 0.500) (a) 200N (b) 173.2N (c) 115 .5N (d) 86.6 N
4. A boy pulls his toy on a smooth horizontal surface with a rope inclined at 60 to the horizontal .If the effective force pulling the toy along the tension in rope (a) 2.5 N (b) 4.33N (c) 5.0 N (d) 8.66N (e) 10.0N
5. A boy is pulling a load of 150N with a string inclined at an angle of 30 to the horizontal . If the tension in the lift the load off the ground is ( sin 30 = ½ , cos 30 = √3/2 and tan30 = 1/√3 ) (a) 255N (b) 202.5N (c) 105 √3/2 N (d) 75N (e) 52.5N
THEORY
1. Two forces of magnitude 12N and 9N act at right angle to each other f ind the resulrant?
2. Four forces act as shown below.
Calculate their resultant
READING ASSIGNMENT
New Sch. Physics for Senior Sec. Schls. Pages 346---356
GENERAL EVALUATION:
1. A body of mass 3.0Kg is acted upon by a force of 24N, if the frictional force on the body is 13N.Calculate the acceleration of the body.
2. For the body in question 1 above, what distance would it move if the force was applied for a period of 7s?
3. A ship moves due north at 100kh-1 on a river flowing due east at 25km-1. Calculate the magnitude of the resultant velocity of the ship.
CONTENTS
Concept of scalar and vector quantities.
Vector representation, addition of vectors.
Resolution of vectors and resultant.
CONCEPT OF SCALAR AND VECTOR QUANTITIES.
Physical quantities are divided into two types
i Scalar quantity
ii Vector quantity
CONCEPT OF SCALAR: A quantity that has only a magnitude but no direction is called a scalar quantity. Examples: length, area, volume, mass of a body, energy, speed, etc.
A scalar is one which has only magnitude (size ) but no direction e.g. distance, speed , temperature , volume , work , energy, power , mass ,electric potential , gravitational potential electric charge .
Scalar quantities are added according to the ordinary rules of arithmetic for example , a mark of 50 added to a mark of 40 produces a mark of 90 –no directional property .But a force of 50N combined with a force of 40N may produce 90N if they are acting in same direction. But if in opposite direction would produce a different result. This vectors are combined or added by a special law the parallelogram law of addition of vectors.
https://youtu.be/Pj8Zh0A-uLU
CONCEPT OF VECTOR: Vector quantity is that quantity which has both a magnitude and a direction. Examples are; Displacement, force, momentum, acceleration, velocity, weight, magnetic flux, electric fields and gravitational fields etc.
The magnitude is shown by the length and the direction by the arrow representing the position or direction. The resultant of a scalar quantity is simply added algebraically. To find the resultant of two or more vector quantities, simple algebraic addition is not applicable. The addition is done by applying the parallelogram's law or the polygon's law of vectors.
https://drive.google.com/file/d/1q0vlmT ... sp=sharing
https://youtu.be/V_CJXlgar8Y
VECTOR REPRESENTATION.
Vector representation may be referred to as a way of representing vector either graphically or analytically.
A vector quantity can be graphically represented by a line drawn so that the length of the line denotes the magnitude of the quantity . The direction of the line indicates the direction in which the vector quantity act and it is shown by an arrow head . E.g a distance of 5km west represented by 5cm length of line where 1km = 1cm
https://drive.google.com/file/d/1JX85P- ... sp=sharing
ADDITION AND SUBTRACTION OF VECTORS
Two or more vectors acting on a body in a specified direction can be combined to produce a single vector having the same effect .The single vector is called the resultant.
For example:
(a) Two forces Y and X with magnitude of 3N and 4N respectively acting along the same direction will produce a resultant of 7N (algebraic sum of the two vectors).
(c) If the two vectors are inclined at an angle less than 900 or more than 900 , the resultant cannot be obtained by Pythagoras theorem but by vector addition,. Parallelogram law of vector, trigonometric or scale drawings can be used to calculate the magnitude and direction of the resultant
https://youtu.be/VFRW0f0XUU8
VECTORS AT RIGHT ANGLES
i. Parallelogram law of vectors states that if two vectors are represented in magnitude and direction by adjacent sides of a parallelogram , the resultant is represented in magnitude and direction by the diagonal of the parallelogram drawn from the common point
ii. If the two vectors are inclined at an angle less than 900 , the scale drawing or trigonometric method can used . In using scale drawing (graphical ) methods, a convenient scale is chosen ( if the magnitude of the forces given is large ) and then draw the lengths corresponding to the magnitude of the forces. A Protractor is used to draw the angle in between the forces. The parallelogram is completed and the resultant and its fraction obtained
https://youtu.be/iOipHqLV3OU
RESOLUTION OF VECTORS
A single vector can be resolved into two vectors called components. A vector F represented as the diagonal of the parallelogram can be resolved into its component later taken as the adjacent sides of the parallelogram.
THE RESULATNT OF MORE THAN TWO VECTORS
To find the resultant of more than two vectors, we resolve each vector in two perpendicular direction s add all the horizontal components X, and all the vertical components, Y.
For example, consider four forces acting on a body as shown below
https://youtu.be/RxV06Oj5sMc
EVALUATION
1 Calculate the resultant of five coplanar forces of values10N, 12N , 16N , 20N , 15N on an object as shown below
2. (i) Find by means of vector diagram, the resultant of two forces of 7N and 3N acting perpendicular to each other.
(ii) Two forces, 3N and 4N act on a body in the direction due North and East respectively. Calculate their equilibrium.
3. Two forces whose resultant is 10n, are perpendicular to each other. If one of them makes an angle of 60 with the resultant, calculate its magnitude.
ASSIGNMENT:
1. Which of the following is not a vector quantity (a) speed (b) velocity (c) force (d) acceleration (e) Electric field
2. Which of the following is not a scalar quantity ( a) density (b) weight (c) speed (d) mass (e) temperature
3. Two forces , whose resultant is 100N are perpendicular to each other .If one of the makes an angle of 60 with the resultant , calculate its magnitude ( sin60 = 0.8660 , cos 60 = 0.500) (a) 200N (b) 173.2N (c) 115 .5N (d) 86.6 N
4. A boy pulls his toy on a smooth horizontal surface with a rope inclined at 60 to the horizontal .If the effective force pulling the toy along the tension in rope (a) 2.5 N (b) 4.33N (c) 5.0 N (d) 8.66N (e) 10.0N
5. A boy is pulling a load of 150N with a string inclined at an angle of 30 to the horizontal . If the tension in the lift the load off the ground is ( sin 30 = ½ , cos 30 = √3/2 and tan30 = 1/√3 ) (a) 255N (b) 202.5N (c) 105 √3/2 N (d) 75N (e) 52.5N
THEORY
1. Two forces of magnitude 12N and 9N act at right angle to each other f ind the resulrant?
2. Four forces act as shown below.
Calculate their resultant
READING ASSIGNMENT
New Sch. Physics for Senior Sec. Schls. Pages 346---356
GENERAL EVALUATION:
1. A body of mass 3.0Kg is acted upon by a force of 24N, if the frictional force on the body is 13N.Calculate the acceleration of the body.
2. For the body in question 1 above, what distance would it move if the force was applied for a period of 7s?
3. A ship moves due north at 100kh-1 on a river flowing due east at 25km-1. Calculate the magnitude of the resultant velocity of the ship.
WEEK 5
PRESSURE
CONTENTS
Definition
Applications
Pressure and depth
Atmospheric pressure
DEFINTION:
Pressure is defined as the force acting normally or perpendicularly per unit area. The s.i. unit of pressure is N/m[sup]2[/sup] or Pascal (Pa).
P=F/A.
From the relation above, we can see that the smaller the area the larger is the pressure. For this reason, little or apparently no force is needed while using a sharp knife as against blunt one and also a pointed heel will exert a greater force than a flat heel due to the small area involved.
APPLICATIONS.
According to Pascal principle, pressure is transmitted equally throughout the length of a liquid in an enclosure. This principle of transmission of pressure in fluids has many important applications, e.g. hydraulic press, car brakes, pumps, syringe, siphon, etc.
The principle of hydraulic press is to transmit equal pressure to all parts of the liquid at the same level. Two cylinders are joined by a connecting tube. The base of one cylinder is smaller than the other. Tight piston is fitted into each cylinder with liquid between them. When a little force is applied on the small piston of a cross-sectional area; pressure is transmitted through the liquid to the second piston in a large cylinder.
Rubber Sucker: used for attaching notices to shop windows.
Siphon: The siphon is a bent tube used in transferring liquid from one vessel to another. It is fixed in a way to make the transferred liquid flow.
Pressure is also one of the factors that enables this process to work .Example of where siphon used is in a petrol or diesel tank. When petrol is to be taken out from the petrol tank to the outside keg, a siphon is used. The pressure at the end of the siphon inside the petrol tank must be greater than that outside , i.e the one inside the keg so that the liquid petrol will flow from the petrol tank to the available keg
It is a bent tube used in transferring liquid from a higher level to a lower level by taking advantage of the difference in pressure height.
Hydraulic Press: this operates by transmitting an equal pressure of a liquid equally in all directions when a small force is applied.
If cross sectional area of small piston=a.
Cross sectional area of large piston=A
and force applied at small area=f
Then, pressure transmitted P=F/A.
Force impacted on large piston due to pressure P=PA.
=FA/a.
Car Brake: Car break is an hydraulic device. The principle involved is the use of little force on a pedal to the breaking system. While the pedal is still pressed, a piston in a master cylinder is opened to perform the function. The fluid transmits pressure equally to the four wheels, making the breaking system balance.
It operates by transmitting equally in all directions same pressure when a force is applied at the brake pedal. Larger force on brake pads due to the pressure makes it possible for stoppage to occur at the same time.
Syringe: The syringe consists of a barrel and a tight fitted piston. When the nozzle is dipped into water, or if it is fitted with a needle and the needle is passed into a liquid (as used in hospitals), on drawing the piston upwards. The air pressure above liquid surface forces the liquid into the lower barrel, and when the piston is pushed down, the piston press tightly to the wall of the barrel and pushes the liquid out.
The syringe is commonly used in the hospital. It used the principle of bicycle pump where a piston is moved forward and backward in a very tight tube. When the piston at the holding point is drawn back, pressure pushes liquid into the tube and it is placed inside a bowl of wall before pulling. After pulling backward, liquid is seen inside the tube because, the atmospheric pressure has pushed the liquid inside the tube. If the piston is pushed forward, the pressure is moved out of the tube which makes the liquid inside the tube to move out through the applied pressure.
Common pump(or lift pump): The common pump or lift pump used for lifting water operates using the air pressure, it consists of a barrel with a valve that opens upward and a tight fitted piston with a valve that also opens upwards.
Pumps are devices generally used to compress or force fluids (liquid or gas) into a given space or to evacuate from it.
Examples of pumps are: bicycle pumps, water pumps, lift pumps, force pumps etc
EVALUATION.
1. Explain the term pressure. No force is required when using sharp knife as against blunt knife, explain.
2. State three applications of the principle of transmission of pressure in liquids.
3. Explain the action of simple mercury barometer and Fortin barometer.
ATMOSPHERIC PRESSURE.
Atmospheric pressure is the pressure exerted by the surrounding air. The atmospheric pressure at sea level is about 1.0x10[sup]5[/sup]
N/M[sup]2[/sup] (1 bar) or 760mmHg. It decreases in magnitude as the altitude increases above sea level.
The atmospheric pressure is measured with Barometer. There are different types of barometer used in measuring atmospheric pressure.
SIMPLE BAROMETER
This consist of an inverted burette (or long tube) initially filled with water or mercury and inverted in a beaker or trough of water or mercury. The column of water or mercury inside the burette is supported by the surrounding air pressure which acts through the water to the base. The height(h) of the column is a measure of the atmospheric pressure.
FORTIN BAROMETER
The Fortin barometer gives an accurate measurement of the atmospheric pressure, and it is used in the laboratory.
Fortin Barometer is used for more accurate measurement of atmospheric pressure than the simple or Torricellian barometer.
It is made up of a tube containing the mercury which is protected by enclosing it in a brass tube, the upper part of which is made of glass so that the mercury surface may be seen.
Readings are taken by a vernier moving over a millimetre scale of sufficient length to cover the full range of variation in barometric height.
https://youtu.be/3kUS5m1gb20
ANEROID BAROMETER.
The aneroid barometer is used to obtain an accurate measurement of the atmospheric pressure. It has the advantage of not containing a liquid. It is also used in weather forecast, in determining altitudes or heights above sea level as well as in airplanes as altimeters.
Aneroid barometer has no liquid and is widely used in home for showing weather changes. The essential part of an aneroid barometer is a flat cylindrical metal box or capsule, corrugated for strength, and hermetically sealed after having been partially exhausted of air. Increase in atmospheric pressure causes the box to case in slightly, while a decrease allows it to expand.
The movement of the box are magnified by a system of levers and transmitted to a fine chain wrapped round the spindle of a pointer.
https://youtu.be/0xkNFN5j6Wo
EVALUATION.
1. What is the main advantage of Aneroid barometer over types of barometers?
2. Explain the mode of operation of a simple barometer.
PRESSURE ON LIQUID.
The pressure exerted by a liquid have the following characteristics:
It increases in magnitude with depth. This is reflected in the speed and distance of water spouting out of the lowest jet.
Pressure at any point at the same level is same in all directions.
Pressure increases as density increases at the same depth
The pressure acting in a liquid act at right angle to the surface of the liquid.
If g=acceleration due to gravity.
ℓ=density of liquid.
h=liquid depth or height of liquid column supported by atmospheric pressure.
Then, exerted pressure P=h ℓ g.
https://youtu.be/Cvp6mLWbgaM
EVALUATION.
At what level below a liquid surface, will a tube containing a liquid of density 900kg/m3 have a pressure equal to 3600N//m[sup]2[/sup] ?
(g=10m/s[sup]2[/sup]).
A hydraulic press, has pistons of surface area 3.5x10[sup]-2[/sup]m[sup]2[/sup] and 6.5x10[sup]-2[/sup]m[sup]2[/sup], if the small piston is pushed down with a force 200N. What would the pressure exerted on the big piston be?.
ASSIGNMENT.
1 A force of 100N acts on an area of 10m2 .What is the pressure exerted on the surface? (A) 0.1Pa (B) 10Pa (C) 100 Pa (D) 0.01Pa.
2 The unit of pressure is? (A) N/M[sup]2[/sup] (B) NM[sup]2[/sup] (C) Joule (D) Watt.
3 The instrument used measure gas pressure is
(A) Bourdon gauge (B) Barometer (C) Clinical thermometer (D) Siphon
4 One of the following make use of air pressure in its function.
(A) Hydraulic press (B) Manometer (C) Aneroid barometer (D) Force pump
5 One of the following pumps has its piston head not totally tight to the barrel wall during the upstroke.
(A) Lift pump (B) Common pump (C) Force pump (D) Bicycle pump
THEORY
1. With the aid of a diagram explain the principle of the hydraulic press. Mention two uses of the hydraulic press.
A hydraulic press consists of two cylinders of cross-sectional area 0.2m[sup]2[/sup] and 5.0m[sup]2[/sup]. The piston in the smaller cylinder is pushed down with a force of 100N through a distance of 0.2m. Calculate
(a) the pressure transmitted by the fluid
(b) the force exerted by the piston in the large cylinder
2. Describe the construction and mode of operation of
(a) a lift pump
(b) a force pump.
READING ASSIGNMENT.
New school physics for SSS M.W Anyakoha page
GENERAL EVALUATION
1. State 3 laws of friction
2. State 3 applications of frictional force
MAIN TOPIC: PRESSURE
SPECIFIC TOPIC: DEFINITION AND PROPERTIES OF PRESSURE IN LIQUID
REFERENCE BOOK: Igcse Physics by Richard Woodside
OBJECTIVE: At the end of the lesson, the students should be able to:
(i) Define the word pressure
(ii) Identify the properties of pressure in liquid
CONTENT: PRESSURE
Pressure is defined as the force acting at right angle, normal or perpendicular per unit surface area in contact with a substance.
Thus the force applied on the surface area of a substance is called pressure. It is denoted by p, a scalar quantity and exists in solid, liquid and gases.
Thus,
Pressure = Force
Surface area
= F (N)
A (M[sup]2[/sup])
= Nm[sup]-2[/sup] or pascal.
The SI unit of pressure is Nm[sup]-2[/sup]
Example 1: Calculate the pressure on the surface of a rectangular box of weight 100N if the base of this box has an area of 2m[sup]2[/sup].
Solution
Weight (force) = 100N, surface-area(A) = 2m[sup]2[/sup]
P = F/A = [sup]100[/sup]/[sub]2[/sub] = 50N/m[sup]2[/sup]
Example 2: The weight on the narrow heel of a girl's shoes is 250N and the surface area of the heel in contact with the floor is 50mm[sup]2[/sup].Determine the pressure exerted on the heel.
Solution
F = 250N, A = 50mm[sup]2[/sup]= 50 x 10[sup]-6[/sup]m[sup]2[/sup]
Pressure = force/area = [sup]250[/sup]/[sub]50[/sub] x 10[sup]-6[/sup]
= 5 x 106N/m[sup]2[/sup]
PRESSURE IN LIQUID
Consider a beaker containing liquid at level h. The total force acting on the base at the beaker is called thrust. Thrust per unit area is called pressure of the liquid.
Pressure P = [sup]Total thrust[/sup]/[sub]Total area[/sub] = [sup]Ahpg[/sup]/[sub]A[/sub] = hpg
Example 1: A cylindrical jar of radius 7.0cm and height 25cm is filled with a liquid 0.8g/cm3. What is the pressure exerted at the bottom of the jar by the liquid? (take g as 10m/s2)
Solution
H = 25cm = 0.25m, PLiquid = 0.8g/cm3
P = hpg
= 0.25 x 800 10
= 2 x 103N/m2
Where h = height, A = Area, p = density and g = acceleration due to gravity.
Example 2: A tube whose area of cross-section is 5m2 is loaded with shat till its total weight is 32N. To what depth will it sink in water and what is the density of a liquid to which it sinks to a depth of 7.5m? (take g = 10m/s2)
Solution
Pressure to support tube = [sup]32[/sup]/[sub]5[/sub] = 6.4Nm[sup]-2[/sup]
P = hpg
h= [sup]p[/sup]/[sub]pg[/sub] = [sup]6.4[/sup]/[sub]1000[/sub] x 10 = [sup]6.4[/sup]/[sub]1[/sub] x 10[sup]4[/sup]
6.4 x 10[sup]-4[/sup]m
P = hpg, p = [sup]P[/sup]/[sub]hg[/sub], P = [sup]6.4[/sup]/[sub]10[/sub] x 7.5 = [sup]6.4[/sup]/[sub]75[/sub]
= 0.0853kg/m[sup]3[/sup]
View Solution below
https://drive.google.com/file/d/1nXiV3p ... view?pli=1
PROPERTIES OF PRESSURE IN LIQUID
(I) The pressure at any point in a liquid is exerted normally, perpendicularly or at a right angle in all directions.
(II) The pressure is the same at all points on the same horizontal plane in a liquid.
(III) The pressure at a point below the surface of a liquid is directly proportional to the depth.
(IV) Pressure is dependent on the surface area in contact.
(V) Pressure increases with density, i.e nature of liquid.
(VI) Pressure decreases with height.
(VII) Pressure is independent of the shape and volume.
https://youtu.be/hgjAgz1iswo
EVALUATION:
(i) List three characteristics of pressure in liquid
(ii) The horizontal door of a substance at a depth of 500m has an area of 0.4m2. Calculate the force exerted by the sea water on the door at this depth
(iii) State the application of pressure (atmospheric) to our everyday activities
(iv)The area of a substance is 500cm2 and the mass of the substance is 3kg, if gravity is acted upon where g=10m/s. Determine the pressure of the substance.
ASSIGNMENT:
A hydraulic braking system has a master cylinder of cross-area 6.00cm2 and four brake cylinders each of cross-area 2.00cm2. When a force of 500N is applied to the piston in the master cylinder, what is the pressure in the oil in the master cylinder? What is the pressure in the oil in each break cylinder? What force appears at the piston in each brake cylinder? (frictional forces may be neglected)
CONTENTS
Definition
Applications
Pressure and depth
Atmospheric pressure
DEFINTION:
Pressure is defined as the force acting normally or perpendicularly per unit area. The s.i. unit of pressure is N/m[sup]2[/sup] or Pascal (Pa).
P=F/A.
From the relation above, we can see that the smaller the area the larger is the pressure. For this reason, little or apparently no force is needed while using a sharp knife as against blunt one and also a pointed heel will exert a greater force than a flat heel due to the small area involved.
APPLICATIONS.
According to Pascal principle, pressure is transmitted equally throughout the length of a liquid in an enclosure. This principle of transmission of pressure in fluids has many important applications, e.g. hydraulic press, car brakes, pumps, syringe, siphon, etc.
The principle of hydraulic press is to transmit equal pressure to all parts of the liquid at the same level. Two cylinders are joined by a connecting tube. The base of one cylinder is smaller than the other. Tight piston is fitted into each cylinder with liquid between them. When a little force is applied on the small piston of a cross-sectional area; pressure is transmitted through the liquid to the second piston in a large cylinder.
Rubber Sucker: used for attaching notices to shop windows.
Siphon: The siphon is a bent tube used in transferring liquid from one vessel to another. It is fixed in a way to make the transferred liquid flow.
Pressure is also one of the factors that enables this process to work .Example of where siphon used is in a petrol or diesel tank. When petrol is to be taken out from the petrol tank to the outside keg, a siphon is used. The pressure at the end of the siphon inside the petrol tank must be greater than that outside , i.e the one inside the keg so that the liquid petrol will flow from the petrol tank to the available keg
It is a bent tube used in transferring liquid from a higher level to a lower level by taking advantage of the difference in pressure height.
Hydraulic Press: this operates by transmitting an equal pressure of a liquid equally in all directions when a small force is applied.
If cross sectional area of small piston=a.
Cross sectional area of large piston=A
and force applied at small area=f
Then, pressure transmitted P=F/A.
Force impacted on large piston due to pressure P=PA.
=FA/a.
Car Brake: Car break is an hydraulic device. The principle involved is the use of little force on a pedal to the breaking system. While the pedal is still pressed, a piston in a master cylinder is opened to perform the function. The fluid transmits pressure equally to the four wheels, making the breaking system balance.
It operates by transmitting equally in all directions same pressure when a force is applied at the brake pedal. Larger force on brake pads due to the pressure makes it possible for stoppage to occur at the same time.
Syringe: The syringe consists of a barrel and a tight fitted piston. When the nozzle is dipped into water, or if it is fitted with a needle and the needle is passed into a liquid (as used in hospitals), on drawing the piston upwards. The air pressure above liquid surface forces the liquid into the lower barrel, and when the piston is pushed down, the piston press tightly to the wall of the barrel and pushes the liquid out.
The syringe is commonly used in the hospital. It used the principle of bicycle pump where a piston is moved forward and backward in a very tight tube. When the piston at the holding point is drawn back, pressure pushes liquid into the tube and it is placed inside a bowl of wall before pulling. After pulling backward, liquid is seen inside the tube because, the atmospheric pressure has pushed the liquid inside the tube. If the piston is pushed forward, the pressure is moved out of the tube which makes the liquid inside the tube to move out through the applied pressure.
Common pump(or lift pump): The common pump or lift pump used for lifting water operates using the air pressure, it consists of a barrel with a valve that opens upward and a tight fitted piston with a valve that also opens upwards.
Pumps are devices generally used to compress or force fluids (liquid or gas) into a given space or to evacuate from it.
Examples of pumps are: bicycle pumps, water pumps, lift pumps, force pumps etc
EVALUATION.
1. Explain the term pressure. No force is required when using sharp knife as against blunt knife, explain.
2. State three applications of the principle of transmission of pressure in liquids.
3. Explain the action of simple mercury barometer and Fortin barometer.
ATMOSPHERIC PRESSURE.
Atmospheric pressure is the pressure exerted by the surrounding air. The atmospheric pressure at sea level is about 1.0x10[sup]5[/sup]
N/M[sup]2[/sup] (1 bar) or 760mmHg. It decreases in magnitude as the altitude increases above sea level.
The atmospheric pressure is measured with Barometer. There are different types of barometer used in measuring atmospheric pressure.
SIMPLE BAROMETER
This consist of an inverted burette (or long tube) initially filled with water or mercury and inverted in a beaker or trough of water or mercury. The column of water or mercury inside the burette is supported by the surrounding air pressure which acts through the water to the base. The height(h) of the column is a measure of the atmospheric pressure.
FORTIN BAROMETER
The Fortin barometer gives an accurate measurement of the atmospheric pressure, and it is used in the laboratory.
Fortin Barometer is used for more accurate measurement of atmospheric pressure than the simple or Torricellian barometer.
It is made up of a tube containing the mercury which is protected by enclosing it in a brass tube, the upper part of which is made of glass so that the mercury surface may be seen.
Readings are taken by a vernier moving over a millimetre scale of sufficient length to cover the full range of variation in barometric height.
https://youtu.be/3kUS5m1gb20
ANEROID BAROMETER.
The aneroid barometer is used to obtain an accurate measurement of the atmospheric pressure. It has the advantage of not containing a liquid. It is also used in weather forecast, in determining altitudes or heights above sea level as well as in airplanes as altimeters.
Aneroid barometer has no liquid and is widely used in home for showing weather changes. The essential part of an aneroid barometer is a flat cylindrical metal box or capsule, corrugated for strength, and hermetically sealed after having been partially exhausted of air. Increase in atmospheric pressure causes the box to case in slightly, while a decrease allows it to expand.
The movement of the box are magnified by a system of levers and transmitted to a fine chain wrapped round the spindle of a pointer.
https://youtu.be/0xkNFN5j6Wo
EVALUATION.
1. What is the main advantage of Aneroid barometer over types of barometers?
2. Explain the mode of operation of a simple barometer.
PRESSURE ON LIQUID.
The pressure exerted by a liquid have the following characteristics:
It increases in magnitude with depth. This is reflected in the speed and distance of water spouting out of the lowest jet.
Pressure at any point at the same level is same in all directions.
Pressure increases as density increases at the same depth
The pressure acting in a liquid act at right angle to the surface of the liquid.
If g=acceleration due to gravity.
ℓ=density of liquid.
h=liquid depth or height of liquid column supported by atmospheric pressure.
Then, exerted pressure P=h ℓ g.
https://youtu.be/Cvp6mLWbgaM
EVALUATION.
At what level below a liquid surface, will a tube containing a liquid of density 900kg/m3 have a pressure equal to 3600N//m[sup]2[/sup] ?
(g=10m/s[sup]2[/sup]).
A hydraulic press, has pistons of surface area 3.5x10[sup]-2[/sup]m[sup]2[/sup] and 6.5x10[sup]-2[/sup]m[sup]2[/sup], if the small piston is pushed down with a force 200N. What would the pressure exerted on the big piston be?.
ASSIGNMENT.
1 A force of 100N acts on an area of 10m2 .What is the pressure exerted on the surface? (A) 0.1Pa (B) 10Pa (C) 100 Pa (D) 0.01Pa.
2 The unit of pressure is? (A) N/M[sup]2[/sup] (B) NM[sup]2[/sup] (C) Joule (D) Watt.
3 The instrument used measure gas pressure is
(A) Bourdon gauge (B) Barometer (C) Clinical thermometer (D) Siphon
4 One of the following make use of air pressure in its function.
(A) Hydraulic press (B) Manometer (C) Aneroid barometer (D) Force pump
5 One of the following pumps has its piston head not totally tight to the barrel wall during the upstroke.
(A) Lift pump (B) Common pump (C) Force pump (D) Bicycle pump
THEORY
1. With the aid of a diagram explain the principle of the hydraulic press. Mention two uses of the hydraulic press.
A hydraulic press consists of two cylinders of cross-sectional area 0.2m[sup]2[/sup] and 5.0m[sup]2[/sup]. The piston in the smaller cylinder is pushed down with a force of 100N through a distance of 0.2m. Calculate
(a) the pressure transmitted by the fluid
(b) the force exerted by the piston in the large cylinder
2. Describe the construction and mode of operation of
(a) a lift pump
(b) a force pump.
READING ASSIGNMENT.
New school physics for SSS M.W Anyakoha page
GENERAL EVALUATION
1. State 3 laws of friction
2. State 3 applications of frictional force
MAIN TOPIC: PRESSURE
SPECIFIC TOPIC: DEFINITION AND PROPERTIES OF PRESSURE IN LIQUID
REFERENCE BOOK: Igcse Physics by Richard Woodside
OBJECTIVE: At the end of the lesson, the students should be able to:
(i) Define the word pressure
(ii) Identify the properties of pressure in liquid
CONTENT: PRESSURE
Pressure is defined as the force acting at right angle, normal or perpendicular per unit surface area in contact with a substance.
Thus the force applied on the surface area of a substance is called pressure. It is denoted by p, a scalar quantity and exists in solid, liquid and gases.
Thus,
Pressure = Force
Surface area
= F (N)
A (M[sup]2[/sup])
= Nm[sup]-2[/sup] or pascal.
The SI unit of pressure is Nm[sup]-2[/sup]
Example 1: Calculate the pressure on the surface of a rectangular box of weight 100N if the base of this box has an area of 2m[sup]2[/sup].
Solution
Weight (force) = 100N, surface-area(A) = 2m[sup]2[/sup]
P = F/A = [sup]100[/sup]/[sub]2[/sub] = 50N/m[sup]2[/sup]
Example 2: The weight on the narrow heel of a girl's shoes is 250N and the surface area of the heel in contact with the floor is 50mm[sup]2[/sup].Determine the pressure exerted on the heel.
Solution
F = 250N, A = 50mm[sup]2[/sup]= 50 x 10[sup]-6[/sup]m[sup]2[/sup]
Pressure = force/area = [sup]250[/sup]/[sub]50[/sub] x 10[sup]-6[/sup]
= 5 x 106N/m[sup]2[/sup]
PRESSURE IN LIQUID
Consider a beaker containing liquid at level h. The total force acting on the base at the beaker is called thrust. Thrust per unit area is called pressure of the liquid.
Pressure P = [sup]Total thrust[/sup]/[sub]Total area[/sub] = [sup]Ahpg[/sup]/[sub]A[/sub] = hpg
Example 1: A cylindrical jar of radius 7.0cm and height 25cm is filled with a liquid 0.8g/cm3. What is the pressure exerted at the bottom of the jar by the liquid? (take g as 10m/s2)
Solution
H = 25cm = 0.25m, PLiquid = 0.8g/cm3
P = hpg
= 0.25 x 800 10
= 2 x 103N/m2
Where h = height, A = Area, p = density and g = acceleration due to gravity.
Example 2: A tube whose area of cross-section is 5m2 is loaded with shat till its total weight is 32N. To what depth will it sink in water and what is the density of a liquid to which it sinks to a depth of 7.5m? (take g = 10m/s2)
Solution
Pressure to support tube = [sup]32[/sup]/[sub]5[/sub] = 6.4Nm[sup]-2[/sup]
P = hpg
h= [sup]p[/sup]/[sub]pg[/sub] = [sup]6.4[/sup]/[sub]1000[/sub] x 10 = [sup]6.4[/sup]/[sub]1[/sub] x 10[sup]4[/sup]
6.4 x 10[sup]-4[/sup]m
P = hpg, p = [sup]P[/sup]/[sub]hg[/sub], P = [sup]6.4[/sup]/[sub]10[/sub] x 7.5 = [sup]6.4[/sup]/[sub]75[/sub]
= 0.0853kg/m[sup]3[/sup]
View Solution below
https://drive.google.com/file/d/1nXiV3p ... view?pli=1
PROPERTIES OF PRESSURE IN LIQUID
(I) The pressure at any point in a liquid is exerted normally, perpendicularly or at a right angle in all directions.
(II) The pressure is the same at all points on the same horizontal plane in a liquid.
(III) The pressure at a point below the surface of a liquid is directly proportional to the depth.
(IV) Pressure is dependent on the surface area in contact.
(V) Pressure increases with density, i.e nature of liquid.
(VI) Pressure decreases with height.
(VII) Pressure is independent of the shape and volume.
https://youtu.be/hgjAgz1iswo
EVALUATION:
(i) List three characteristics of pressure in liquid
(ii) The horizontal door of a substance at a depth of 500m has an area of 0.4m2. Calculate the force exerted by the sea water on the door at this depth
(iii) State the application of pressure (atmospheric) to our everyday activities
(iv)The area of a substance is 500cm2 and the mass of the substance is 3kg, if gravity is acted upon where g=10m/s. Determine the pressure of the substance.
ASSIGNMENT:
A hydraulic braking system has a master cylinder of cross-area 6.00cm2 and four brake cylinders each of cross-area 2.00cm2. When a force of 500N is applied to the piston in the master cylinder, what is the pressure in the oil in the master cylinder? What is the pressure in the oil in each break cylinder? What force appears at the piston in each brake cylinder? (frictional forces may be neglected)
WEEK 6
EQUILIBRIUM OF FORCES.
CONTENTS.
Principles of moment
Conditions for equilibrium of a rigid body
Centre of gravity and stability
Couple
MOMENT
Moment is referred to as the turning effect of forces about a point or axis, e.g. opening of a door or a bottle-top, rotation of wheels of vehicles, etc.
Moment of a force: moment of a force about a point is the product of the force and the perpendicular distance of its line of action from the point. It is measured in Nm-Sl unit and is a vector quantity.
Consider a rigid body, having an axis of rotation passing through the centre and perpendicular to the plane and the force, acting on the body. The body rotates in the anti-clockwise direction around the axis. The effect of the turning of the force is called the moment of a force.
(a) Moment= f x h.
(b) Where F is the force and h is the perpendicular distance from the point 0 to the line of action of the force
(b)
In (b) above, moment is not f x h, but is given by: Moment= f x OP= F x h sin O (Sin OP is the perpendicular distance from O to the line of action of force F)
Thus, moment= force x perpendicular distance to the axis of rotation.
Couples: this is a pair of equal and parallel force acting on a body in opposite direction and whose lines of action do not coincide, which results in the rotation of the body about its axis, e.g. turning of water-tap, rotation of a corkscrew, etc.
Torque: Torque is the turning effect of a couple. Thus, it is defined as the product of the forces and the perpendicular distance between the line of action of the two forces. It is also measured in Nm and is a vector quantity.
The moment of a couple (torque)= F x 2r.
Resultants: resultant of two forces acting alone which would produce the same effect as the two forces put together. The effect of a force in a certain direction depends on the size of the forces and the angle between the force and the directions.
A body is said to be in equilibrium if under the action of several forces, it does accelerate uniformly, rotate with uniform angular velocity or remain at rest. Example a stone at rest, the earth rotating round the sun, a body moving along a path at uniform velocity.
EQUILIBRANT
Equilibrant is a single force which in equilibrium or in a balance state keeps all the forces acting together.
RESULTANT EQUILIBRANT
Resultant of two forces acting at a point - It keeps a single force in equilibrium
The result of two forces acting from a source - A balanced force keeps different force in a balance state.
CONDITIONS FOR EQUILIBRUM OF RIGID BODIES UNDER THE ACTION OF PARALLEL FORCE
The conditions of a body in equilibrium by action of parallel forces are:
(1) Total forces in one direction are equal to the total forces in opposite direction i.e total upward force is equal to the total downward force.
(2) The algebraic sum of the moment of all forces about any point should be zero i.e total clockwise(cw) moment of the forces about any point is equal to the total anti-clockwise (Acw) moment of the remaining forces about the same point.
(3) The total resultant force must be zero.
CONDITIONS FOR THE EQUILIBRUM OF RIGID BODIES UNDER THE ACTION OF NON-PARALLEL FORCES
(1) The line of action of the forces must meet at a point
(2) The resultant should be zero
(3) The algebraic sum of resolved components of the forces in any perpendicular direction is respectively zero.
CONDITIONS FOR THE EQUILIBRUM OF THREE NON-PARALLEL CO-PLANAR FORCE
(1) If any object is in equilibrium under the actions of three forces, the resultant of the two forces must be equal and opposite to the third force.
(2) The lines of actions of the forces must meet at a point.
(3) All the three forces must be concurrent i.e meet at a point.
(4) The line of action of the third force must pass through the point of intersection of the lines of action of other two forces.
(5) The algebraic sum of the forces at a point must be equal to zero.
(6) The forces can be represented in magnitude and direction by the sides of a triangle taken in order.
MOMENT OF A FORCE
The moment of a force is the product of the force and the perpendicular distance
https://youtu.be/izbxpjNg7Q8
CONDITIONS FOR EQUILIBRIUM
1.The sum of the upward forces must be equal to the sum of the downward forces.
2.The sum of the clockwise moment above a point must be equal to the sum of anticlockwise moment about the same point
Upward forces [ U[sub]F[/sub]] = F1 + F2 + F3
Downward forces [ D[sub]F[/sub]] = F4 + F5
For a body at equilibrium,
U[sub]F[/sub] ¬ = D[sub]F[/sub]
F[sub]1[/sub] + F[sub]2[/sub] + F[sub]3[/sub] = F[sub]4[/sub] + F[sub]5[/sub]
For a body at equilibrium,
Upward forces = downward force,
F[sub]1[/sub] + F[sub]2[/sub] = F[sub]3[/sub] + F[sub]4[/sub]
(F[sub]1[/sub]+F[sub]2[/sub]) – (F[sub]3[/sub]+F[sub]4[/sub])=0
Clockwise moment = F[sub]2[/sub] X[sub]2[/sub] + F[sub]4[/sub]X[sub]4[/sub]
Anticlockwise moment = F[sub]1[/sub]X[sub]1[/sub]+ F[sub]3[/sub]X[sub]3[/sub]
(F[sub]1[/sub]X[sub]1[/sub]+ F[sub]3[/sub]X[sub]3[/sub]) –(F[sub]2[/sub]X[sub]2[/sub] + F[sub]4[/sub]X[sub]4[/sub])= 0
sum of clockwise moment =sum of anticlockwise moment
COUPLE
A couple is a system of two parallel, equal and opposite forces acting along the same line. The effect of a couple is to rotate the body.
The moment of a couple is the product of one of the forces and the perpendicular distance between the lines of action of the two forces
In fig (i), M = f x 2r
In fig (ii), M = f x d
The distance between the two equal forces is called the arm of the couple, the moment of a couple is also called a torque
APPLICATION OF THE EFFECT OF COUPLES
1. It is easier to turn a tap on or off by applying couple
2. It is easier to turn a steering wheel of a vehicle by applying a couple with our two hands instead of a single force with one arm.
https://youtu.be/us7KNhguSuM
EXAMPLES 1 A light beam AB sits on two pivots C and D . A load of 1ON hangs at O,2m from the support at c. Find the value of the reaction forces P and Q at C and D.
P + Q = 10N
X 2 = Q (2 + 6 )
20 = 8Q
Q = 20/8 =2.5 N
Taking moment about D
P x8 = 10 x6
P = 60/8
=7.5N
Q = 10 -7.5
= 2.5 N
EXAMPLE 2 A pole AB of length 10m and weight 600N has its centre of gravity 4m from the end A, and lies on horizontal ground. Draw a diagram to show the the forces acting on the pole when the end B is lift this end. prove that this force applied at the end A will not be
Clockwise moment =600 x 4 =2400Nm
Anticlockwise moment =p x 10 = 10pNm
P =240Nm
If this force of 240Nm is applied at A, we have
P= 240Nm
Taking moment about B, we have
clockwise moment =240 x 10 =2400Nm
Anticlockwise moment =600 x 6 =3600 Nm
The anticlockwise moment is greater than the clockwise moment .
Therefore , the 240N force A will not be sufficient to lift the end A because the turning effect due to the 600N force far exceeds that due to the 240N force
Find the moment of the force of 20N in the diagram above about A and B
Taking moment about A
Cos 60 =d/3m
D= 3 cos 60
D = 1.5m
Moment about A =F x d
M = 20 x 1.5= 30 Nm
The Moment about B= 0
EXAMPLE 4. A uniform rod lm long weighing 100N is supported horizontally on two knife edges placed 10cm from its ends. What will be the reaction at the support when a 40N load is suspended 10cm from the mid point of the rod.
R1 + R2 = 140N
Taking moment about R1
R2 x 80 = (100 x 40 ) + (40 x 50 )
80R2 = 4000 + 2000
R2 = 6000/80
R2=75N
R1 = 140 – 75 =65N
EXAMPLE 5. A metre rule is found to balance horizontally at the 50cm mark, When a body of mass 60kg is suspended at the 6cm mark, the balance point is found to be at the 30cm mark, calculate.
the weight of the metre rule
the distances of the balance point to the 60kg mass if the mass is moved to the 13cm mark
600x(X)=720(37-X)
600x = 6640 – 720x
600x+ 720x = 6640
x = 6640/1320
x = 20. 18cm
https://youtu.be/l_Nck-X49qc
https://youtu.be/XXUW8PCaVXA
CENTRE OF GRAVITY.
The centre of gravity of a body is the point through which the line of action of the weight of the body always passes irrespective of the position of the body. It is also the point at which the entire weight of the body appears to be concentrated.
The centre of mass of a body is the point at which the total mass of the body appears to be concentrated. Sometimes, the center of mass may coincides with the centre of gravity for small objects.
Centre of gravity (C.G) is defined as the point in which the resultant weight of a body appears to act, when the entire weight seems to be concentrated in a point where the resultant gravitational force on a body appears to act i.e the point of application of the resultant due to the earth's attraction on it.
The centre of gravity of a body depends on the shape and size of the body. The position of the centre of gravity changes with the changes in the shape of the body. The C.G of a ring (circle) or a uniform disc (sphere) is at its centre.
For a uniform thin sheet or lamina such as square, rectangle, rhombus, or parallelogram i.e quadrilaterals, the C.G is at the intersection of its diagonals.
A triangular uniform lamina has its C.G where the medians intersect. A uniform rod or beam has its C.G at the mid-point or centre of the beam or rod.
Note: In a uniform rod or beam, the centre of mass and C.G coincides because the mass per unit length is the same along its whole length. But for a non-uniform beam, the centre of mass and C.G are different.
CENTRE OF MASS
Centre of mass of an object is defined as the point at which an applied force produces acceleration but no direction.
The centre of gravity (C.G ) of a non-uniform object can be determined by the following methods:
(i) Balancing method
(ii) Plumb line
TO LOCATE CENTRE OF GRAVITY BY A BALANCING METHOD
The centre of gravity of a long thin object such as a ruler or billiard may be found approximately by simple balancing it on a straight-edge.
Using a thin sheet or lamina of cardboard or metal, it is necessary to balance in two positions.
The thin sheet or lamina of cardboard or metal, is balanced on the edge of a straight ruler in two directions AB and CD and the lines of balance are marked with pencil lines. Since the C.G is situated on both lines, it must actually lie at their point of intersection.
The edge of the ruler should pass through the intersection of the previous lines.
TO LOCATE CENTRE OF GRAVITY BY MEANS OF PLUMB LINE
To locate C.G by mean of plumb line is one of the best means of finding the C.G of an irregularly shaped piece of cardboard.
Three small holes are made at well-spaced intervals round the edge of the cardboard. A stout pin is then put through one of the holes and held firmly by a clamp and stand so that the card can swing freely on it. The card will come to rest with its centre of gravity. A stout pin is then put through one of the holes and held firmly by a clamp and stand so that the card can swing freely on it. The card will come to rest with its centre of gravity vertically below the point of support by means of a plumb line.
N.B: A suitable plumb line is made from weight tied at the other end. This is hung from the pin and the position of the cotton marked on the card by two small pencil crosses. These crosses are joined by a pencil line.
https://youtu.be/R8wKV0UQtlo
EVALUATION.
(i) Explain what is meant by the moment of a force about a point
(ii) Distinguish between the resultant and the equilibrant of forces
(iii) State two conditions necessary for the equilibrium of three non-parallel co-planar forces
ASSIGNMENT:
1. Explain how the position of the centre of gravity of a body affects the equilibrium of the body.
2. With the aid of diagrams, explain how you can determine the centre of gravity of four named regular uniform bodies.
3. Describe an experiment to determine the centre of gravity of an irregular lamina.
STABILITY OF OBJECTS.
There are three types of equilibrium- stable equilibrium, unstable equilibrium, and neutral equilibrium.
1. Stable equilibrium: a body is said to be in stable equilibrium if it tends to return to its original position when slightly displaced. A low centre of gravity and wide base will put objects in stable equilibrium e.g. a cone resting on its base ; a racing car with low C.G and wide base; a ball or a sphere in the middle of a bowl.
2. Unstable equilibrium: a body is said to be in an unstable equilibrium if when slightly displaced it tends to move further away from its original position e.g. a cone or an egg resting on its apex. High C.G. and a narrow base usally causes unstable equilibrium.
3. Neutral equilibrium: a body is said to be in neutral equilibrium if when slightly displaced, it tends to come to rest in its new position e.g a cone or cylinder or an egg resting on its side.
https://youtu.be/4rG9u478X1Q
EVALUATION.
Students project.
Each student will make paper model of the three types of equilibrium.
ASSIGNMENT
1. The S.I unit of moment is (a) Jm (b)Wm (c)Nm
2. A uniform metre rule of mass 100g balances at the 40cm mark when a mass X is placed at the 10cm mark.What is the value of X? (a)33.33g (b)43.33g (C) 53.33g.
3. Two forces each of magnitude 10N act jn opposite directions at the end of a table.If the length of the table is 50cm.Find the moment of the couple on the tab le.(a)0.5Nm (b)5Nm (c) 50Nm.
4 A pole AB of length 5M and weigh 300N has its centre of gravity 2.0M from the end A,and lies on horizontal ground.Calculate the force required to begin to lift this end.(a) 60N (b)120N (c) 240N.
5 when a body is acted upon by several forces and it does not accelerates or rotates, the body is said to be in (a) space (b)equilibrium (C) motion.
THEORY.
1 State the conditions necessary for a body to be in equilibrium, mention the three types of equilibrium with atleast two examples each.
Use the diagram above to calculate the moment of the force of 10N about the point p.
READING ASSIGNMENT..
New Sch. Physics FOR SSS –M W ANYAKOHA PAGES 173-182.
GENERAL EVALUATION
1. A uniform beam AB of length 6m and mass 20kg rests on support P and Q placed 1m from each end of the beam. Masses of 10kg and 8kg are placed at A and B respectively. Calculate the reactions at P and Q (g = 9.8ms-2)
2. A box is pushed along a horizontal floor by a horizontal force of 60 N. There is a frictional force between the box and the floor of 50 N.What is the gain in kinetic energy of the box when it moves a distance of 4.0 m?
CONTENTS.
Principles of moment
Conditions for equilibrium of a rigid body
Centre of gravity and stability
Couple
MOMENT
Moment is referred to as the turning effect of forces about a point or axis, e.g. opening of a door or a bottle-top, rotation of wheels of vehicles, etc.
Moment of a force: moment of a force about a point is the product of the force and the perpendicular distance of its line of action from the point. It is measured in Nm-Sl unit and is a vector quantity.
Consider a rigid body, having an axis of rotation passing through the centre and perpendicular to the plane and the force, acting on the body. The body rotates in the anti-clockwise direction around the axis. The effect of the turning of the force is called the moment of a force.
(a) Moment= f x h.
(b) Where F is the force and h is the perpendicular distance from the point 0 to the line of action of the force
(b)
In (b) above, moment is not f x h, but is given by: Moment= f x OP= F x h sin O (Sin OP is the perpendicular distance from O to the line of action of force F)
Thus, moment= force x perpendicular distance to the axis of rotation.
Couples: this is a pair of equal and parallel force acting on a body in opposite direction and whose lines of action do not coincide, which results in the rotation of the body about its axis, e.g. turning of water-tap, rotation of a corkscrew, etc.
Torque: Torque is the turning effect of a couple. Thus, it is defined as the product of the forces and the perpendicular distance between the line of action of the two forces. It is also measured in Nm and is a vector quantity.
The moment of a couple (torque)= F x 2r.
Resultants: resultant of two forces acting alone which would produce the same effect as the two forces put together. The effect of a force in a certain direction depends on the size of the forces and the angle between the force and the directions.
A body is said to be in equilibrium if under the action of several forces, it does accelerate uniformly, rotate with uniform angular velocity or remain at rest. Example a stone at rest, the earth rotating round the sun, a body moving along a path at uniform velocity.
EQUILIBRANT
Equilibrant is a single force which in equilibrium or in a balance state keeps all the forces acting together.
RESULTANT EQUILIBRANT
Resultant of two forces acting at a point - It keeps a single force in equilibrium
The result of two forces acting from a source - A balanced force keeps different force in a balance state.
CONDITIONS FOR EQUILIBRUM OF RIGID BODIES UNDER THE ACTION OF PARALLEL FORCE
The conditions of a body in equilibrium by action of parallel forces are:
(1) Total forces in one direction are equal to the total forces in opposite direction i.e total upward force is equal to the total downward force.
(2) The algebraic sum of the moment of all forces about any point should be zero i.e total clockwise(cw) moment of the forces about any point is equal to the total anti-clockwise (Acw) moment of the remaining forces about the same point.
(3) The total resultant force must be zero.
CONDITIONS FOR THE EQUILIBRUM OF RIGID BODIES UNDER THE ACTION OF NON-PARALLEL FORCES
(1) The line of action of the forces must meet at a point
(2) The resultant should be zero
(3) The algebraic sum of resolved components of the forces in any perpendicular direction is respectively zero.
CONDITIONS FOR THE EQUILIBRUM OF THREE NON-PARALLEL CO-PLANAR FORCE
(1) If any object is in equilibrium under the actions of three forces, the resultant of the two forces must be equal and opposite to the third force.
(2) The lines of actions of the forces must meet at a point.
(3) All the three forces must be concurrent i.e meet at a point.
(4) The line of action of the third force must pass through the point of intersection of the lines of action of other two forces.
(5) The algebraic sum of the forces at a point must be equal to zero.
(6) The forces can be represented in magnitude and direction by the sides of a triangle taken in order.
MOMENT OF A FORCE
The moment of a force is the product of the force and the perpendicular distance
https://youtu.be/izbxpjNg7Q8
CONDITIONS FOR EQUILIBRIUM
1.The sum of the upward forces must be equal to the sum of the downward forces.
2.The sum of the clockwise moment above a point must be equal to the sum of anticlockwise moment about the same point
Upward forces [ U[sub]F[/sub]] = F1 + F2 + F3
Downward forces [ D[sub]F[/sub]] = F4 + F5
For a body at equilibrium,
U[sub]F[/sub] ¬ = D[sub]F[/sub]
F[sub]1[/sub] + F[sub]2[/sub] + F[sub]3[/sub] = F[sub]4[/sub] + F[sub]5[/sub]
For a body at equilibrium,
Upward forces = downward force,
F[sub]1[/sub] + F[sub]2[/sub] = F[sub]3[/sub] + F[sub]4[/sub]
(F[sub]1[/sub]+F[sub]2[/sub]) – (F[sub]3[/sub]+F[sub]4[/sub])=0
Clockwise moment = F[sub]2[/sub] X[sub]2[/sub] + F[sub]4[/sub]X[sub]4[/sub]
Anticlockwise moment = F[sub]1[/sub]X[sub]1[/sub]+ F[sub]3[/sub]X[sub]3[/sub]
(F[sub]1[/sub]X[sub]1[/sub]+ F[sub]3[/sub]X[sub]3[/sub]) –(F[sub]2[/sub]X[sub]2[/sub] + F[sub]4[/sub]X[sub]4[/sub])= 0
sum of clockwise moment =sum of anticlockwise moment
COUPLE
A couple is a system of two parallel, equal and opposite forces acting along the same line. The effect of a couple is to rotate the body.
The moment of a couple is the product of one of the forces and the perpendicular distance between the lines of action of the two forces
In fig (i), M = f x 2r
In fig (ii), M = f x d
The distance between the two equal forces is called the arm of the couple, the moment of a couple is also called a torque
APPLICATION OF THE EFFECT OF COUPLES
1. It is easier to turn a tap on or off by applying couple
2. It is easier to turn a steering wheel of a vehicle by applying a couple with our two hands instead of a single force with one arm.
https://youtu.be/us7KNhguSuM
EXAMPLES 1 A light beam AB sits on two pivots C and D . A load of 1ON hangs at O,2m from the support at c. Find the value of the reaction forces P and Q at C and D.
P + Q = 10N
X 2 = Q (2 + 6 )
20 = 8Q
Q = 20/8 =2.5 N
Taking moment about D
P x8 = 10 x6
P = 60/8
=7.5N
Q = 10 -7.5
= 2.5 N
EXAMPLE 2 A pole AB of length 10m and weight 600N has its centre of gravity 4m from the end A, and lies on horizontal ground. Draw a diagram to show the the forces acting on the pole when the end B is lift this end. prove that this force applied at the end A will not be
Clockwise moment =600 x 4 =2400Nm
Anticlockwise moment =p x 10 = 10pNm
P =240Nm
If this force of 240Nm is applied at A, we have
P= 240Nm
Taking moment about B, we have
clockwise moment =240 x 10 =2400Nm
Anticlockwise moment =600 x 6 =3600 Nm
The anticlockwise moment is greater than the clockwise moment .
Therefore , the 240N force A will not be sufficient to lift the end A because the turning effect due to the 600N force far exceeds that due to the 240N force
Find the moment of the force of 20N in the diagram above about A and B
Taking moment about A
Cos 60 =d/3m
D= 3 cos 60
D = 1.5m
Moment about A =F x d
M = 20 x 1.5= 30 Nm
The Moment about B= 0
EXAMPLE 4. A uniform rod lm long weighing 100N is supported horizontally on two knife edges placed 10cm from its ends. What will be the reaction at the support when a 40N load is suspended 10cm from the mid point of the rod.
R1 + R2 = 140N
Taking moment about R1
R2 x 80 = (100 x 40 ) + (40 x 50 )
80R2 = 4000 + 2000
R2 = 6000/80
R2=75N
R1 = 140 – 75 =65N
EXAMPLE 5. A metre rule is found to balance horizontally at the 50cm mark, When a body of mass 60kg is suspended at the 6cm mark, the balance point is found to be at the 30cm mark, calculate.
the weight of the metre rule
the distances of the balance point to the 60kg mass if the mass is moved to the 13cm mark
600x(X)=720(37-X)
600x = 6640 – 720x
600x+ 720x = 6640
x = 6640/1320
x = 20. 18cm
https://youtu.be/l_Nck-X49qc
https://youtu.be/XXUW8PCaVXA
CENTRE OF GRAVITY.
The centre of gravity of a body is the point through which the line of action of the weight of the body always passes irrespective of the position of the body. It is also the point at which the entire weight of the body appears to be concentrated.
The centre of mass of a body is the point at which the total mass of the body appears to be concentrated. Sometimes, the center of mass may coincides with the centre of gravity for small objects.
Centre of gravity (C.G) is defined as the point in which the resultant weight of a body appears to act, when the entire weight seems to be concentrated in a point where the resultant gravitational force on a body appears to act i.e the point of application of the resultant due to the earth's attraction on it.
The centre of gravity of a body depends on the shape and size of the body. The position of the centre of gravity changes with the changes in the shape of the body. The C.G of a ring (circle) or a uniform disc (sphere) is at its centre.
For a uniform thin sheet or lamina such as square, rectangle, rhombus, or parallelogram i.e quadrilaterals, the C.G is at the intersection of its diagonals.
A triangular uniform lamina has its C.G where the medians intersect. A uniform rod or beam has its C.G at the mid-point or centre of the beam or rod.
Note: In a uniform rod or beam, the centre of mass and C.G coincides because the mass per unit length is the same along its whole length. But for a non-uniform beam, the centre of mass and C.G are different.
CENTRE OF MASS
Centre of mass of an object is defined as the point at which an applied force produces acceleration but no direction.
The centre of gravity (C.G ) of a non-uniform object can be determined by the following methods:
(i) Balancing method
(ii) Plumb line
TO LOCATE CENTRE OF GRAVITY BY A BALANCING METHOD
The centre of gravity of a long thin object such as a ruler or billiard may be found approximately by simple balancing it on a straight-edge.
Using a thin sheet or lamina of cardboard or metal, it is necessary to balance in two positions.
The thin sheet or lamina of cardboard or metal, is balanced on the edge of a straight ruler in two directions AB and CD and the lines of balance are marked with pencil lines. Since the C.G is situated on both lines, it must actually lie at their point of intersection.
The edge of the ruler should pass through the intersection of the previous lines.
TO LOCATE CENTRE OF GRAVITY BY MEANS OF PLUMB LINE
To locate C.G by mean of plumb line is one of the best means of finding the C.G of an irregularly shaped piece of cardboard.
Three small holes are made at well-spaced intervals round the edge of the cardboard. A stout pin is then put through one of the holes and held firmly by a clamp and stand so that the card can swing freely on it. The card will come to rest with its centre of gravity. A stout pin is then put through one of the holes and held firmly by a clamp and stand so that the card can swing freely on it. The card will come to rest with its centre of gravity vertically below the point of support by means of a plumb line.
N.B: A suitable plumb line is made from weight tied at the other end. This is hung from the pin and the position of the cotton marked on the card by two small pencil crosses. These crosses are joined by a pencil line.
https://youtu.be/R8wKV0UQtlo
EVALUATION.
(i) Explain what is meant by the moment of a force about a point
(ii) Distinguish between the resultant and the equilibrant of forces
(iii) State two conditions necessary for the equilibrium of three non-parallel co-planar forces
ASSIGNMENT:
1. Explain how the position of the centre of gravity of a body affects the equilibrium of the body.
2. With the aid of diagrams, explain how you can determine the centre of gravity of four named regular uniform bodies.
3. Describe an experiment to determine the centre of gravity of an irregular lamina.
STABILITY OF OBJECTS.
There are three types of equilibrium- stable equilibrium, unstable equilibrium, and neutral equilibrium.
1. Stable equilibrium: a body is said to be in stable equilibrium if it tends to return to its original position when slightly displaced. A low centre of gravity and wide base will put objects in stable equilibrium e.g. a cone resting on its base ; a racing car with low C.G and wide base; a ball or a sphere in the middle of a bowl.
2. Unstable equilibrium: a body is said to be in an unstable equilibrium if when slightly displaced it tends to move further away from its original position e.g. a cone or an egg resting on its apex. High C.G. and a narrow base usally causes unstable equilibrium.
3. Neutral equilibrium: a body is said to be in neutral equilibrium if when slightly displaced, it tends to come to rest in its new position e.g a cone or cylinder or an egg resting on its side.
https://youtu.be/4rG9u478X1Q
EVALUATION.
Students project.
Each student will make paper model of the three types of equilibrium.
ASSIGNMENT
1. The S.I unit of moment is (a) Jm (b)Wm (c)Nm
2. A uniform metre rule of mass 100g balances at the 40cm mark when a mass X is placed at the 10cm mark.What is the value of X? (a)33.33g (b)43.33g (C) 53.33g.
3. Two forces each of magnitude 10N act jn opposite directions at the end of a table.If the length of the table is 50cm.Find the moment of the couple on the tab le.(a)0.5Nm (b)5Nm (c) 50Nm.
4 A pole AB of length 5M and weigh 300N has its centre of gravity 2.0M from the end A,and lies on horizontal ground.Calculate the force required to begin to lift this end.(a) 60N (b)120N (c) 240N.
5 when a body is acted upon by several forces and it does not accelerates or rotates, the body is said to be in (a) space (b)equilibrium (C) motion.
THEORY.
1 State the conditions necessary for a body to be in equilibrium, mention the three types of equilibrium with atleast two examples each.
Use the diagram above to calculate the moment of the force of 10N about the point p.
READING ASSIGNMENT..
New Sch. Physics FOR SSS –M W ANYAKOHA PAGES 173-182.
GENERAL EVALUATION
1. A uniform beam AB of length 6m and mass 20kg rests on support P and Q placed 1m from each end of the beam. Masses of 10kg and 8kg are placed at A and B respectively. Calculate the reactions at P and Q (g = 9.8ms-2)
2. A box is pushed along a horizontal floor by a horizontal force of 60 N. There is a frictional force between the box and the floor of 50 N.What is the gain in kinetic energy of the box when it moves a distance of 4.0 m?
WEEK 7
NEWTON’S LAWS OF MOTION.
CONTENTS.
Newton’s laws of motion
Conservation of linear momentum
CONTENT: NEWTON’S LAWS OF MOTION
LAW I.
Everybody perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon.
Projectiles persevere in their motions, so far as they are not retarded by the resistance of the air, or impelled downwards by the force of gravity. A top, whose parts by their cohesion are perpetually drawn aside from rectilinear motions, does not cease its rotation, otherwise than as it is retarded by the air. The greater bodies of the Planets and Comets, meeting with less resistance in more free spaces, preserve their motions both progressive and circular for a much longer time.
LAW II.
The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed
If any force generates a motion, a double force will generate double the motion, a triple force triple the motion, whether that force be impressed altogether and at once, or gradually and successively. And this motion (being always directed the same way with the generating force) if the body moved before, is added to or sub ducted from the former motion, according as they directly conspire with or are directly contrary to each other; or obliquely joined, when they are oblique, so as to produce a new motion compounded from the determination of both.
LAW III.
To every Action there is always opposed an equal Reaction: or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts
Whatever draws or presses another is as much drawn or pressed by that other. If you press a stone with your finger, the finger is also pressed by the stone. If a horse draws a stone tyed to a rope, the horse (if I may so say) will be equally drawn back towards the stone: For the distended rope, by the same endeavour to relax or unbend it self, will draw the horse as much towards the stone, as it does the stone towards the horse, and will obstruct the progress of the one as much as it advances that of the other.
https://youtu.be/ztt1mVYH6x4
EVALUATION:
Give two laws of motion and explain how it affects live
NEWTON’S FIRST LAW OF MOTION states “that everybody continues in its state of rest or of uniform motion in straight line unless it is acted upon by a force.”
This simply means that a body at rest will remain permanently there or a body moving with uniform velocity on a straight line will continue moving forever if it were possible for all the external opposing forces to be eliminated.
The tendency of a body to remain at rest or, if moving, to continue its motion in a straight line is called the inertia of the body. That is why Newton’s first law is otherwise referred to as the law of inertia
There are consequences of this law. For example, when a car had a head on collision with another car or the driver suddenly applies the brake, the passengers are likely to be injured when they hit the windscreen.
The reason is that an external force will only stop the car but not the passengers who tend to continue their linear motion. This necessitated, the use of safety precautions e.g seat belt
Also, if a stationary car is knocked forward from behind, the passengers may sustain neck injuries as their bodies tend to move forward in relation to the car while their neck move backward. Modern cars have rests to prevent this incident
https://youtu.be/5oi5j11FkQg
https://youtu.be/1XSyyjcEHo0
https://youtu.be/sabH4bJsxWA
NEWTON’S SECOND LAW OF MOTION states “that the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.”
This implies that when a heavy body and a light one are acted upon by the same force for the same time, the light body will build up a grater velocity than the heavy one. But they gain the same momentum
Momentum of a body is the product of the mass and velocity of the body.
The S.I unit of momentum is kgm/s.
IMPULSE
The impulse of a force and time. This impulse is also equal to the change in momentum and they therefore share the same unit (Ns)
F = [sup]m (v-u )[/sup]/[sub]t[/sub]
Ft = mv – mu ………………………….. 2
F x t = I (Ns) ……………………….. 3
mv- mu = change in momentum …………………… 4
https://youtu.be/5oi5j11FkQg
https://youtu.be/xzA6IBWUEDE
NEWTON’S THIRD LAW OF MOTION states that "to every action, there is an equal but opposite reaction". when a book is placed on a table, the downwards weight (force) of the book on the table is balanced by the upwards reaction of the table on the book
Another practical demonstration of this law can be observed when a bullet is fired from a gun, the person holding it experiences the backward recoil force of the gun (reaction) which is equal to the propulsive force (action ) acting on the bullet
According to Newton second law of motion, force is proportional to change in momentum ,therefore the momentum of the bullet is equal and opposite to the momentum of the gun i.e.
mass of bullet x muzzle velocity = mass of gun x recoil velocity,
hence, for a bullet of mass in and muzzle velocity , v the velocity v of the recoil of the gun is given by
M[sub]g[/sub]V[sub]g[/sub] = m[sub]b[/sub]v[sub]b[/sub] ……………………………………….. 5
V[sub]g[/sub] = [sup]m[sub]b[/sub]v[sub]b[/sub][/sup] /[sub]M[sub]g[/sub][/sub]
https://youtu.be/TVAxASr0iUY
https://youtu.be/y61_VPKH2B4
EVALUATION.
1. State Newtons laws of motion
2. Mention and explain the consequencies of each law stated above
CONSERVATION OF LINEAR MOMENTUM .
The principle of conservation of linear momentum states that when two or more bodies collide, their momentum remain constant provided there is no external force acting on the system.
This implies that in a closed or isolated system where there is no external forces , the total momentum after collision remains constant.
The principle is true for both elastic and inelastic collision.
https://youtu.be/M2xnGcaaAi4
TYPES OF COLLISION
There are two types of (a) collision- elastic and (b) inelastic.
IN ELASTIC COLLISION where the two bodies and then move with different velocities , both momentum and kinetic energy are conserved eg collision between gaseous particles , a ball which rebounds to its original height etc .If the two colliding bodies have masses m1and m2 initial velocities u1 and u2 and final velocities v1 and v2 , , the conservation principle can be mathematically expressed as
m[sub]1[/sub]u[sub]1[/sub] + m[sub]2[/sub]u[sub]2[/sub] = m[sub]1[/sub]v[sub]1[/sub] + m[sub]2[/sub]v[sub]2[/sub] ……………………………………. 6
IN AN INELASTIC COLLISION , the two bodies join together after the collision and with the same velocity momentum is conserved but kinetic energy is not conversed because part of it has been converted to heat or sound energy, leading to deformation.
Thus, the conversation principle can be re-written as
m[sub]1[/sub]U[sub]1[/sub] + m[sub]2[/sub]U[sub]2[/sub] = V[sub]12[/sub] (m[sub]1[/sub] +m[sub]2[/sub]) ……………………………….7
V[sub]12[/sub] = common velocity
Since momentum is a vector quantity, all the velocities must be measured in the same direction , assigning positive signs to the forward velocities and negative signs to the backward or opposite velocities .
TWO BODIES MOVING IN THE SAME DIRECTION BEFORE COLLISION
TWO BODIES TRAVELLING IN OPPOSITE DIRECTION
Worked example.
.
1 Two moving toys of masses 50kg and 30 kg are traveling on the same plane with speeds of 5 m/s and 3 m/s respectively in the same direction .If they collide and stick together , calculate their common velocity.
2. Two balls of masses 0.5 kg and 0.3kg move towards each other in the same line at speeds of 3 m/s and 4 m/s respectively .After the collision , the first balls has a speed of 1m/s in the opposite direction
What is the speed of the second ball after collision
3. A gun of mass 100kg fires a bullet of mass 20g at a speed of 400m/s .What is the recoil velocity of the gun ?
Momentum gun = momentum of bullet
M V = m v
10 x V = 0.002 x 400
V = [sup]0.002 x 400[/sup]/[sub]10[/sub]
V= 0.8 m/s.
https://youtu.be/2E9fY8H6O1g
EVALUATION.
1. State the principle of conservation of linear momentum.
2. Explain elastic and inelastic collision, and give two example of each
ASSIGNMENT
1. A ball of mass 0.5kg moving at 10m/s collides with another ball of equal mass at rest. If the two balls move off together after the impact, calculate their common velocity (A) 0,2m/s (B)0.5m/s (C) 5m/s (D) 10m/s.
2 A ball of mass 6kg moving with a velocity of 10m/s collides with a 2kg ball moving in the opposite direction with a velocity of 5m/s. After the collision the two balls coalesce
and move in the same direction. Calculate the velocity of the composite body.
(A) 5m/s (B) 6.25m/s (C) 8.75m/s (D) 12m/s
3 .A constant force of 5N acts for 5 seconds on a mass of 5kg initially at rest. Calculate the final momentum. (A) 125kgm/s (B) 25kgm/s (C) 15kgm/s (D) 5kgm/s.
4 When taking a penalty kick, a footballer applies a force of 30N for a periods of 0.05S. If the mass of the ball is 0.075kg, calculate the speed with which the ball moves off. (A) 4.5m/s (B) 11.25m/s (C) 20m/s (D) 45m/s.
5 A body of mass 40kg changes its velocity from 10m/s to 80m/s in 10 seconds. Calculate the force acting on the body. (A) 480N (B) 380N (C) 280N (D) 180N.
THEORY.
1. State the law of conservation of linear momentum. A 3kg rifle lays on a smooth table when it suddenly discharges, firing a bullet of 0.02kg with a speed of 500m/s. Calculate the recoil speed of the gun.
2. Distinguish between
(a) elastic and inelastic collisions (b) Inertial mass and weight
(c) Derive from Newtons law the relationship between Force, mass and acceleration.
READING ASSIGNMENT.
New Sch Physics FOR SSS-ANYAKOHA PAGES-161—165.
GENERAL EVALUATION
1. State Archimedes principle
2. A 15kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 200N and breaks as the elevator accelerates. What was the elevators minimum acceleration (g=10m/s2).
CONTENTS.
Newton’s laws of motion
Conservation of linear momentum
CONTENT: NEWTON’S LAWS OF MOTION
LAW I.
Everybody perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon.
Projectiles persevere in their motions, so far as they are not retarded by the resistance of the air, or impelled downwards by the force of gravity. A top, whose parts by their cohesion are perpetually drawn aside from rectilinear motions, does not cease its rotation, otherwise than as it is retarded by the air. The greater bodies of the Planets and Comets, meeting with less resistance in more free spaces, preserve their motions both progressive and circular for a much longer time.
LAW II.
The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed
If any force generates a motion, a double force will generate double the motion, a triple force triple the motion, whether that force be impressed altogether and at once, or gradually and successively. And this motion (being always directed the same way with the generating force) if the body moved before, is added to or sub ducted from the former motion, according as they directly conspire with or are directly contrary to each other; or obliquely joined, when they are oblique, so as to produce a new motion compounded from the determination of both.
LAW III.
To every Action there is always opposed an equal Reaction: or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts
Whatever draws or presses another is as much drawn or pressed by that other. If you press a stone with your finger, the finger is also pressed by the stone. If a horse draws a stone tyed to a rope, the horse (if I may so say) will be equally drawn back towards the stone: For the distended rope, by the same endeavour to relax or unbend it self, will draw the horse as much towards the stone, as it does the stone towards the horse, and will obstruct the progress of the one as much as it advances that of the other.
https://youtu.be/ztt1mVYH6x4
EVALUATION:
Give two laws of motion and explain how it affects live
NEWTON’S FIRST LAW OF MOTION states “that everybody continues in its state of rest or of uniform motion in straight line unless it is acted upon by a force.”
This simply means that a body at rest will remain permanently there or a body moving with uniform velocity on a straight line will continue moving forever if it were possible for all the external opposing forces to be eliminated.
The tendency of a body to remain at rest or, if moving, to continue its motion in a straight line is called the inertia of the body. That is why Newton’s first law is otherwise referred to as the law of inertia
There are consequences of this law. For example, when a car had a head on collision with another car or the driver suddenly applies the brake, the passengers are likely to be injured when they hit the windscreen.
The reason is that an external force will only stop the car but not the passengers who tend to continue their linear motion. This necessitated, the use of safety precautions e.g seat belt
Also, if a stationary car is knocked forward from behind, the passengers may sustain neck injuries as their bodies tend to move forward in relation to the car while their neck move backward. Modern cars have rests to prevent this incident
https://youtu.be/5oi5j11FkQg
https://youtu.be/1XSyyjcEHo0
https://youtu.be/sabH4bJsxWA
NEWTON’S SECOND LAW OF MOTION states “that the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.”
This implies that when a heavy body and a light one are acted upon by the same force for the same time, the light body will build up a grater velocity than the heavy one. But they gain the same momentum
Momentum of a body is the product of the mass and velocity of the body.
The S.I unit of momentum is kgm/s.
IMPULSE
The impulse of a force and time. This impulse is also equal to the change in momentum and they therefore share the same unit (Ns)
F = [sup]m (v-u )[/sup]/[sub]t[/sub]
Ft = mv – mu ………………………….. 2
F x t = I (Ns) ……………………….. 3
mv- mu = change in momentum …………………… 4
https://youtu.be/5oi5j11FkQg
https://youtu.be/xzA6IBWUEDE
NEWTON’S THIRD LAW OF MOTION states that "to every action, there is an equal but opposite reaction". when a book is placed on a table, the downwards weight (force) of the book on the table is balanced by the upwards reaction of the table on the book
Another practical demonstration of this law can be observed when a bullet is fired from a gun, the person holding it experiences the backward recoil force of the gun (reaction) which is equal to the propulsive force (action ) acting on the bullet
According to Newton second law of motion, force is proportional to change in momentum ,therefore the momentum of the bullet is equal and opposite to the momentum of the gun i.e.
mass of bullet x muzzle velocity = mass of gun x recoil velocity,
hence, for a bullet of mass in and muzzle velocity , v the velocity v of the recoil of the gun is given by
M[sub]g[/sub]V[sub]g[/sub] = m[sub]b[/sub]v[sub]b[/sub] ……………………………………….. 5
V[sub]g[/sub] = [sup]m[sub]b[/sub]v[sub]b[/sub][/sup] /[sub]M[sub]g[/sub][/sub]
https://youtu.be/TVAxASr0iUY
https://youtu.be/y61_VPKH2B4
EVALUATION.
1. State Newtons laws of motion
2. Mention and explain the consequencies of each law stated above
CONSERVATION OF LINEAR MOMENTUM .
The principle of conservation of linear momentum states that when two or more bodies collide, their momentum remain constant provided there is no external force acting on the system.
This implies that in a closed or isolated system where there is no external forces , the total momentum after collision remains constant.
The principle is true for both elastic and inelastic collision.
https://youtu.be/M2xnGcaaAi4
TYPES OF COLLISION
There are two types of (a) collision- elastic and (b) inelastic.
IN ELASTIC COLLISION where the two bodies and then move with different velocities , both momentum and kinetic energy are conserved eg collision between gaseous particles , a ball which rebounds to its original height etc .If the two colliding bodies have masses m1and m2 initial velocities u1 and u2 and final velocities v1 and v2 , , the conservation principle can be mathematically expressed as
m[sub]1[/sub]u[sub]1[/sub] + m[sub]2[/sub]u[sub]2[/sub] = m[sub]1[/sub]v[sub]1[/sub] + m[sub]2[/sub]v[sub]2[/sub] ……………………………………. 6
IN AN INELASTIC COLLISION , the two bodies join together after the collision and with the same velocity momentum is conserved but kinetic energy is not conversed because part of it has been converted to heat or sound energy, leading to deformation.
Thus, the conversation principle can be re-written as
m[sub]1[/sub]U[sub]1[/sub] + m[sub]2[/sub]U[sub]2[/sub] = V[sub]12[/sub] (m[sub]1[/sub] +m[sub]2[/sub]) ……………………………….7
V[sub]12[/sub] = common velocity
Since momentum is a vector quantity, all the velocities must be measured in the same direction , assigning positive signs to the forward velocities and negative signs to the backward or opposite velocities .
TWO BODIES MOVING IN THE SAME DIRECTION BEFORE COLLISION
TWO BODIES TRAVELLING IN OPPOSITE DIRECTION
Worked example.
.
1 Two moving toys of masses 50kg and 30 kg are traveling on the same plane with speeds of 5 m/s and 3 m/s respectively in the same direction .If they collide and stick together , calculate their common velocity.
2. Two balls of masses 0.5 kg and 0.3kg move towards each other in the same line at speeds of 3 m/s and 4 m/s respectively .After the collision , the first balls has a speed of 1m/s in the opposite direction
What is the speed of the second ball after collision
3. A gun of mass 100kg fires a bullet of mass 20g at a speed of 400m/s .What is the recoil velocity of the gun ?
Momentum gun = momentum of bullet
M V = m v
10 x V = 0.002 x 400
V = [sup]0.002 x 400[/sup]/[sub]10[/sub]
V= 0.8 m/s.
https://youtu.be/2E9fY8H6O1g
EVALUATION.
1. State the principle of conservation of linear momentum.
2. Explain elastic and inelastic collision, and give two example of each
ASSIGNMENT
1. A ball of mass 0.5kg moving at 10m/s collides with another ball of equal mass at rest. If the two balls move off together after the impact, calculate their common velocity (A) 0,2m/s (B)0.5m/s (C) 5m/s (D) 10m/s.
2 A ball of mass 6kg moving with a velocity of 10m/s collides with a 2kg ball moving in the opposite direction with a velocity of 5m/s. After the collision the two balls coalesce
and move in the same direction. Calculate the velocity of the composite body.
(A) 5m/s (B) 6.25m/s (C) 8.75m/s (D) 12m/s
3 .A constant force of 5N acts for 5 seconds on a mass of 5kg initially at rest. Calculate the final momentum. (A) 125kgm/s (B) 25kgm/s (C) 15kgm/s (D) 5kgm/s.
4 When taking a penalty kick, a footballer applies a force of 30N for a periods of 0.05S. If the mass of the ball is 0.075kg, calculate the speed with which the ball moves off. (A) 4.5m/s (B) 11.25m/s (C) 20m/s (D) 45m/s.
5 A body of mass 40kg changes its velocity from 10m/s to 80m/s in 10 seconds. Calculate the force acting on the body. (A) 480N (B) 380N (C) 280N (D) 180N.
THEORY.
1. State the law of conservation of linear momentum. A 3kg rifle lays on a smooth table when it suddenly discharges, firing a bullet of 0.02kg with a speed of 500m/s. Calculate the recoil speed of the gun.
2. Distinguish between
(a) elastic and inelastic collisions (b) Inertial mass and weight
(c) Derive from Newtons law the relationship between Force, mass and acceleration.
READING ASSIGNMENT.
New Sch Physics FOR SSS-ANYAKOHA PAGES-161—165.
GENERAL EVALUATION
1. State Archimedes principle
2. A 15kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 200N and breaks as the elevator accelerates. What was the elevators minimum acceleration (g=10m/s2).
WEEK 8
SIMPLE HARMONIC MOTION
CONTENT
Definition
Velocity, acceleration and energy
Forced vibration
DEFINITION
This is the periodic motion of a body or particle along a straight line such that the acceleration of the body is directed towards a fixed point .
A particle undergoing simple harmonic motion will move to and fro in a straight line under the influence of a force . This influential force is called a restoring force as it always directs the particle back to its equilibrium position.
Examples of simple harmonic motions are
i. loaded test tube in a liquid
ii Mass on a string
iii The simple pendulum
for a body performing simplr harmonic motion, the general equation is given as
y = A sin [ wt ± kx ]
where k = phase constant, w = angular velocity, t = time, A = amplitude,
As the particle P moves round the circle once, it sweeps through an angle θ = 3600 (or 2π radian) in the time T the period of motion. The rate of change of the angle θ with time (t) is known as the angular velocity ω
Angular velocity (ω) is defined by
This is similar to the relation distance = uniform velocity x time (s= =vt ) for motion in a straight line
A = r = radius of the circle
The linear velocity v at any point ,Q whose distance from C the central point is x is given by
V = ω √ A[sup]2[/sup] – X[sup]2[/sup] ………………………………………… 2
The minimum velocity ,Vm corresponds to the point at X = 0 that is the velocity at the central point or centre of motion .
Hence
Vm =ω A = ω r …………………………………………. 3
Thus the maximum velocity of the SHM occurs at the centre of the motion (X=0) while the minimum velocity occurs at the extreme position of motion (x=A ).
https://youtu.be/gZ_KnZHCn4M
EVALUATION.
1. A body of mass 0.2kg is executing simple harmonic motion with an amplitude of 20mm. The maximum force which acts upon it is 0.064N.Calculate (a) its maximum velocity (b) its period of oscillation.
2. A steel strip clamped at one end , vibrates with a frequency of 20HZ and an amplitude of 5mm at the free end , where a small mass of 2g is positioned. Find the velocity of the end when passing through the zero position.
RELATIONSHIP BETWEEN LINEAR ACCELERATION AND ANGULAR VELOCITY
X = A COS θ
Θ = ωt
X = A cos ω t
dx = -ωA sin ω t
dt
dv =-ω2 A cos ω t
dt
= - ω2X = - ω2A = - ω2r ………………………………………….. 4
The negative sign indicates that the acceleration is always inwards towards C while the displacement is measured outwards from C.
https://youtu.be/jNc2SflUl9U
ENERGY OF SIMPLE HARMONIC MOTION
Since force and displacement are involved, it follows that work and energy are involved in simple harmonic motion .
At any instant of the motion , the system may contain some energy as kinetic energy (KE ) or potential energy(PE) .The total energy (KE + PE ) for a body performing SHM is always conserved although it may change form between PE and KE .
When a mass is suspended from the end of a spring stretched vertically downwards and released , it oscillates in a simple harmonic motion .During this motion , the force tending to restore the spring to its elastic restoring force is simply the elastic restoring force which is given by
F= - ky …………………………………… 5
K is the force constant of the spring , but F = ma
The total work done in stretching the spring at distance y is given by
W = average force x displacement
W = ½ ky x y = ½ ky[sup]2[/sup] ………………………………… 6
Thus the maximum energy total energy stored in the spring is given by
W = ½ KA[sup]2[/sup] …………………………………. 7
A = amplitude ( maximum displacement fro equilibrium position ).
This maximum energy is conserved throughout the motion of the system .
At any stage of the oscillation , the total energy is
W = ½ KA[sup]2[/sup]
W= ½ mv[sup]2[/sup] + ½ ky[sup]2[/sup] ………………………………………….. 8
½ mv[sup]2[/sup] = ½ KA[sup]2[/sup] – ½ ky[sup]2 [/sup]
v[sup]2[/sup] = k/m (A[sup]2[/sup] –y[sup]2[/sup])
V = √k/m(A[sup]2[/sup]-y[sup]2[/sup])
The constant K is obtained from
Hooke’s law in which
F= mg = ke
Where e is the extension produced in the spring by a mass m
But V= ω√A[sup]2[/sup]-X[sup]2[/sup]
Therefore ω =√k/m
Hence the period T = 2π/ω
T = [sup]2π√m[/sup]/[sub]k[/sub]
EXAMPLE
A body of mass 20g is suspended from the end of a spiral spring whose force constant is 0.4Nm-1
The body is set into a simple harmonic motion with amplitude 0.2m. Calculate :
a. The period of the motion
b. The frequency of the motion
c. The angular speed
d. The total energy
e. The maximum velocity of the motion
f. The maximum acceleration
SOLUTION
a T = 2π √m/k
= 2π √ 0.02/0.4
= 0.447 π sec
= 1.41 sec
b. f=1/T = 1/1.41 = 0.71Hz
c. ω =2πf
= 2π x 0.71
= 4.46 rad. S-1
d. Total energy = ½ KA2
= ½ (0.4) (0.2)2
= 0.008 J
e. ½ mv2 = /12 KA2
Vm2 = 0.008 x 2
0.02
= 0.8
Vm= 0.89 m/s
Or V= ω A
= 4.462 x 0.2
= 3.98m/s2 .
EVALUATION.
A body of mass 0.5kg is attached to the end of a spring and the mass pulled down a distance 0.01m. Calculate (i) the period of oscillation (ii) the maximum kinetic energy of mass (iii) kinetic and potential energy of the spring when the body is 0.04m below its centre of oscillation.(k=50Nm)
FORCED VIBRATION AND RESONANCE
Vibrations resulting from the action of an external periodic force on an oscillating body are called forced vibrations. Every vibrating object possesses a natural frequency ((fo) of vibration. This is the frequency with which the object will oscillate when it is left undisturbed after being set into vibration. The principle of the sounding board of a piano or the diaphragm of a loudspeaker is based on the phenomenon of forced vibrations.
Whenever the frequency of a vibrating body acting on a system coincides with the natural frequency of the system, then the system is set into vibration with a relatively large amplitude. This phenomenon is called resonance.
https://youtu.be/L_7vxGgROfc
https://youtu.be/hHXEYdZja1o
EVALUATION.
Explain the terms forced vibrations, resonance. Give two examples of forced vibrations and two examples of resonance. Describe an experiment to demonstrate forced vibration and resonance..
ASSIGNMENT.
1. Which of the following correctly gives the relationship between linear speed v and angular speed w of a body moving uniformly in a circle of radius r?
(A) v=wr (B) v=w2r (C) v= wr2 (D) v=w/r.
2 The motion of a body is simple harmonic if the:
(A) acceleration is always directed towards a fixed point.
(B) path of motion is a straight line .
(c ) acceleration is directed towards a fixed point and proportional to its distance from the point.
(D) acceleration is proportional to the square of the distance from a fixed point.
3.The maximum kinetic energy of a simple pendulum occurs when the bob is at position.
(a) 1 (b) 2 (c) 3 (d) 4 (e) 5
4 The vibration resulting from the action of an external periodic force on the motion of a body is called:
(a) Forced vibration. (b) damped vibration. (c) natural vibration. (d) compound vibration.
5 The maximum potential energy of the swinging pendulum occurs positions
(A) 1and 5 (B) 2 and 4 (C) 3 only (D) 4 only (E) 5 and 3
THEORY
1. Define simple harmonic motion(SHM). A body moving with SHM has an amplitude of 10cm and a frequency of 100Hz. Find (a) the period of oscillation (b) the acceleration at the maximum displacement (c) the velocity at the centre of motion.
2 Define the following terms: frequency, period, amplitude of simple harmonic motion. What is the relation between period and frequency.
Reading assignment.
NEW SCH PHYSICS FOR SSS –ANYAKOHA pages 188-197
GENERAL EVALUATION
1. State the principle of floatation
2. A stone of mass 2.0Kg is thrown vertically upward with a velocity of 20.0m/s. Calculate the initial kinetic energy of the stone.
MAIN TOPIC: NEWTON'S LAWS OF MOTION (Review)
SPECIFIC TOPIC: CONSERVATION OF LINEAR MOMENTUM AND CONSERVATION OF ENERGY
REFERENCE BOOK: Igcse Physics by Richard Woodside
OBJECTIVE: At the end of the lesson, the students should be able to:
Explain the conservation of linear momentum and conservation of energy
CONTENT: CONSERVATION OF LINEAR MOMENTUM AND CONSERVATION OF ENERGY
The study of the effect of forces on bodies that in motion is a branch of science called dynamics. The study of motion involves also the study of forces producing the motion. We define
Force as any agent that changes or tends to change the state of rest or of uniform motion of a body in a straight line.
NEWTON'S FIRST LAW AND ITS APPLICATION
A body will continue in its present state of rest or, if it is in motion, will continue to move with uniform speed in a straight line unless it is acted upon by a force.
This tendency of a body to remain in its state of rest or uniform motion is called the inertia of the body. For this reason, Newton's first law is sometimes called "the law of inertia".
As an example of inertia, think of the effect on a car driver if his car is standing still and another car crashes into his car from behind. His car is suddenly knocked forward. His body is pushed forward by the seat, but his head stays still, in its state of rest,and is jerked back in relation to his body. For this reason, neck injuries are common in accidents where cars are hit from behind and many modern cars have headrests to protect drivers and passengers from injury.
https://youtu.be/1XSyyjcEHo0
https://youtu.be/sabH4bJsxWA
MOMENTUM
Momentum is an important property of a moving object. It explains its tendency to continue moving in a straight line
The momentum of a body is defined as the product of its mass and its velocity.
Momentum = mv
The unit of momentum corresponds to kgms-1. Thus a bullet having a small mass 0.01kg, moving with a high velocity of 1000 ms-1 and a heavy ball of mass 100kg moving with a small speed of 1ms-1 have the same momentum. We also note that the greater the force it will exert on the body it hits.
More powerful brakes are required to stop a heavy lorry than a light car moving with the same speed.
NEWTON'S SECOND LAW AND ITS APPLICATIONS
The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.
Force α change in momentum/time taken
Suppose a force F acts on a body of mass m for a time t and causes it to change its velocity from u to v, then Newton's second law can be stated as
F α [sup]mv - mu[/sup]/[sub]t[/sub]
So F α [sup]m(v-u)[/sup]/[sub]t[/sub]
but [sup] v - u[/sup]/[sub]t[/sub] = a(acceleration)
Therefore, F α ma
F = kma
If we take m = 1kg and a = 1ms-1, the unit of force is chosen to make F = 1 when k = 1. The SI unit of force is called the Newton (N),and it is the force which produces an acceleration of 1ms-2 when it acts on a mass of 1kg. Thus when F is in newton, m in kg, a in ms-2 we have
F = ma
This equation is recognised as a fundamental equation of dynamics and is one of the most important equations of physics. It must be noted that, when using the above equation, the force F must be the resultant force acting on the body.
Example 1: a body of mass 2kg undergoes a constant horizontal acceleration of 5ms-2.Calculate the resultant horizontal force acting on the body. What will be the resultant force on the body when it moves with a uniform velocity of 10ms-1?
Solution
Given that a = 5ms-2, m= 2kg, then Newton's fundamental equation
F = ma becomes
F = 5 x 2
F = 10N
Which is the resultant force on the body.If the object moves with uniform velocity, it means that it not accelerating, i.e. a = 0
F = ma
= m x 0
= 0
The resultant force on the body is zero.
DEDUCTIONS FROM NEWTON'S SECOND LAW
IMPULSE of a force
Now F = [sup]m(v - u)[/sup]/[sub]t[/sub]
or Ft = mv - mu
the quantity Ft is called the IMPULSE I of the force.
Impulse is the product of a large force and a very short time during which it acts.
The unit of impulse is the newton second(Ns). Since this quantity is equal to change in momentum it means that momentum can also be expressed in Ns.
Therefore, I = Ft = change in momentum
Example 1: A force of 10N acts for 20s. What is the change in momentum of the body?
Solution
Impulse = Ft = change in momentum
Therefore, change in momentum
10 x 20
200Ns
Example 2: A body of mass 5kg moving with a speed of 30ms-1 is suddenly hit by another body moving in the same direction, thereby changing the speed of the former body to 60ms-2. What is the impulse received by the first body?
Solution
Impulse = change in momentum
mv = mu
= m(v - u)
= 5(60 - 30)
= 150Ns
https://youtu.be/xzA6IBWUEDE
NEWTON'S THIRD LAW AND ITS APPLICATION
To every action there is an equal and opposite reaction
When we place an object on a table the reaction of the table on the object(the vertical force exerted on the object by the table) is equal and opposite to the action of the object on the table (the weight of the object bearing down on the table)
Again, if a moving car A hits a stationary car B, the force exerted on B by A (the reaction) will be the same as the action of B on A. This is why both cars are damaged.
Application
When a bullet is shot out of a gun the person firing it experiences the backwards recoil force of the gun(a reaction) and this is equal to the propulsive force(action) acting on the bullet.
Thus for a bullet of mass, m , and velocity, v, the velocity V of recoil of the gun is given as MV = -mv
V = -mv
M
Where M is the mass of the gun
Newton's third law has a very useful application in the operation of jet aeroplanes and rockets. The application is based on the fact that a large mass of very hot gases issues from the nozzle behind the jet or rocket. The velocity and mass per second of the gases are so high that considerable momentum is imparted to the stream of gas. An equal and opposite momentum is imparted to the rocket or aeroplane which undergoes a forward thrust.
https://youtu.be/y61_VPKH2B4
LAW OF CONSERVATION OF MOMENTUM
Newton's second and third laws enable us to formulate an important conservation law known as the law of conservation of momentum.
In a system of colliding objects the total momentum is conserved, provided there is no net external force acting on the system.
It can also be stated as follows:
The total momentum of an isolated or closed system of colliding bodies remains constant.
Thus if two or more bodies collide in a closed system, the total momentum before collision is equal to the total momentum after collision.
https://youtu.be/8HXISImmGuQ
EVALUATION:
(i) Explain what is meant by the moment of a force about a point.
(ii) Distinguish between the resultant and the equilibrant of forces
ASSIGNMENT:
(i) State two conditions necessary for the equilibrium of three non-parallel co-planar forces.
CONTENT
Definition
Velocity, acceleration and energy
Forced vibration
DEFINITION
This is the periodic motion of a body or particle along a straight line such that the acceleration of the body is directed towards a fixed point .
A particle undergoing simple harmonic motion will move to and fro in a straight line under the influence of a force . This influential force is called a restoring force as it always directs the particle back to its equilibrium position.
Examples of simple harmonic motions are
i. loaded test tube in a liquid
ii Mass on a string
iii The simple pendulum
for a body performing simplr harmonic motion, the general equation is given as
y = A sin [ wt ± kx ]
where k = phase constant, w = angular velocity, t = time, A = amplitude,
As the particle P moves round the circle once, it sweeps through an angle θ = 3600 (or 2π radian) in the time T the period of motion. The rate of change of the angle θ with time (t) is known as the angular velocity ω
Angular velocity (ω) is defined by
This is similar to the relation distance = uniform velocity x time (s= =vt ) for motion in a straight line
A = r = radius of the circle
The linear velocity v at any point ,Q whose distance from C the central point is x is given by
V = ω √ A[sup]2[/sup] – X[sup]2[/sup] ………………………………………… 2
The minimum velocity ,Vm corresponds to the point at X = 0 that is the velocity at the central point or centre of motion .
Hence
Vm =ω A = ω r …………………………………………. 3
Thus the maximum velocity of the SHM occurs at the centre of the motion (X=0) while the minimum velocity occurs at the extreme position of motion (x=A ).
https://youtu.be/gZ_KnZHCn4M
EVALUATION.
1. A body of mass 0.2kg is executing simple harmonic motion with an amplitude of 20mm. The maximum force which acts upon it is 0.064N.Calculate (a) its maximum velocity (b) its period of oscillation.
2. A steel strip clamped at one end , vibrates with a frequency of 20HZ and an amplitude of 5mm at the free end , where a small mass of 2g is positioned. Find the velocity of the end when passing through the zero position.
RELATIONSHIP BETWEEN LINEAR ACCELERATION AND ANGULAR VELOCITY
X = A COS θ
Θ = ωt
X = A cos ω t
dx = -ωA sin ω t
dt
dv =-ω2 A cos ω t
dt
= - ω2X = - ω2A = - ω2r ………………………………………….. 4
The negative sign indicates that the acceleration is always inwards towards C while the displacement is measured outwards from C.
https://youtu.be/jNc2SflUl9U
ENERGY OF SIMPLE HARMONIC MOTION
Since force and displacement are involved, it follows that work and energy are involved in simple harmonic motion .
At any instant of the motion , the system may contain some energy as kinetic energy (KE ) or potential energy(PE) .The total energy (KE + PE ) for a body performing SHM is always conserved although it may change form between PE and KE .
When a mass is suspended from the end of a spring stretched vertically downwards and released , it oscillates in a simple harmonic motion .During this motion , the force tending to restore the spring to its elastic restoring force is simply the elastic restoring force which is given by
F= - ky …………………………………… 5
K is the force constant of the spring , but F = ma
The total work done in stretching the spring at distance y is given by
W = average force x displacement
W = ½ ky x y = ½ ky[sup]2[/sup] ………………………………… 6
Thus the maximum energy total energy stored in the spring is given by
W = ½ KA[sup]2[/sup] …………………………………. 7
A = amplitude ( maximum displacement fro equilibrium position ).
This maximum energy is conserved throughout the motion of the system .
At any stage of the oscillation , the total energy is
W = ½ KA[sup]2[/sup]
W= ½ mv[sup]2[/sup] + ½ ky[sup]2[/sup] ………………………………………….. 8
½ mv[sup]2[/sup] = ½ KA[sup]2[/sup] – ½ ky[sup]2 [/sup]
v[sup]2[/sup] = k/m (A[sup]2[/sup] –y[sup]2[/sup])
V = √k/m(A[sup]2[/sup]-y[sup]2[/sup])
The constant K is obtained from
Hooke’s law in which
F= mg = ke
Where e is the extension produced in the spring by a mass m
But V= ω√A[sup]2[/sup]-X[sup]2[/sup]
Therefore ω =√k/m
Hence the period T = 2π/ω
T = [sup]2π√m[/sup]/[sub]k[/sub]
EXAMPLE
A body of mass 20g is suspended from the end of a spiral spring whose force constant is 0.4Nm-1
The body is set into a simple harmonic motion with amplitude 0.2m. Calculate :
a. The period of the motion
b. The frequency of the motion
c. The angular speed
d. The total energy
e. The maximum velocity of the motion
f. The maximum acceleration
SOLUTION
a T = 2π √m/k
= 2π √ 0.02/0.4
= 0.447 π sec
= 1.41 sec
b. f=1/T = 1/1.41 = 0.71Hz
c. ω =2πf
= 2π x 0.71
= 4.46 rad. S-1
d. Total energy = ½ KA2
= ½ (0.4) (0.2)2
= 0.008 J
e. ½ mv2 = /12 KA2
Vm2 = 0.008 x 2
0.02
= 0.8
Vm= 0.89 m/s
Or V= ω A
= 4.462 x 0.2
= 3.98m/s2 .
EVALUATION.
A body of mass 0.5kg is attached to the end of a spring and the mass pulled down a distance 0.01m. Calculate (i) the period of oscillation (ii) the maximum kinetic energy of mass (iii) kinetic and potential energy of the spring when the body is 0.04m below its centre of oscillation.(k=50Nm)
FORCED VIBRATION AND RESONANCE
Vibrations resulting from the action of an external periodic force on an oscillating body are called forced vibrations. Every vibrating object possesses a natural frequency ((fo) of vibration. This is the frequency with which the object will oscillate when it is left undisturbed after being set into vibration. The principle of the sounding board of a piano or the diaphragm of a loudspeaker is based on the phenomenon of forced vibrations.
Whenever the frequency of a vibrating body acting on a system coincides with the natural frequency of the system, then the system is set into vibration with a relatively large amplitude. This phenomenon is called resonance.
https://youtu.be/L_7vxGgROfc
https://youtu.be/hHXEYdZja1o
EVALUATION.
Explain the terms forced vibrations, resonance. Give two examples of forced vibrations and two examples of resonance. Describe an experiment to demonstrate forced vibration and resonance..
ASSIGNMENT.
1. Which of the following correctly gives the relationship between linear speed v and angular speed w of a body moving uniformly in a circle of radius r?
(A) v=wr (B) v=w2r (C) v= wr2 (D) v=w/r.
2 The motion of a body is simple harmonic if the:
(A) acceleration is always directed towards a fixed point.
(B) path of motion is a straight line .
(c ) acceleration is directed towards a fixed point and proportional to its distance from the point.
(D) acceleration is proportional to the square of the distance from a fixed point.
3.The maximum kinetic energy of a simple pendulum occurs when the bob is at position.
(a) 1 (b) 2 (c) 3 (d) 4 (e) 5
4 The vibration resulting from the action of an external periodic force on the motion of a body is called:
(a) Forced vibration. (b) damped vibration. (c) natural vibration. (d) compound vibration.
5 The maximum potential energy of the swinging pendulum occurs positions
(A) 1and 5 (B) 2 and 4 (C) 3 only (D) 4 only (E) 5 and 3
THEORY
1. Define simple harmonic motion(SHM). A body moving with SHM has an amplitude of 10cm and a frequency of 100Hz. Find (a) the period of oscillation (b) the acceleration at the maximum displacement (c) the velocity at the centre of motion.
2 Define the following terms: frequency, period, amplitude of simple harmonic motion. What is the relation between period and frequency.
Reading assignment.
NEW SCH PHYSICS FOR SSS –ANYAKOHA pages 188-197
GENERAL EVALUATION
1. State the principle of floatation
2. A stone of mass 2.0Kg is thrown vertically upward with a velocity of 20.0m/s. Calculate the initial kinetic energy of the stone.
MAIN TOPIC: NEWTON'S LAWS OF MOTION (Review)
SPECIFIC TOPIC: CONSERVATION OF LINEAR MOMENTUM AND CONSERVATION OF ENERGY
REFERENCE BOOK: Igcse Physics by Richard Woodside
OBJECTIVE: At the end of the lesson, the students should be able to:
Explain the conservation of linear momentum and conservation of energy
CONTENT: CONSERVATION OF LINEAR MOMENTUM AND CONSERVATION OF ENERGY
The study of the effect of forces on bodies that in motion is a branch of science called dynamics. The study of motion involves also the study of forces producing the motion. We define
Force as any agent that changes or tends to change the state of rest or of uniform motion of a body in a straight line.
NEWTON'S FIRST LAW AND ITS APPLICATION
A body will continue in its present state of rest or, if it is in motion, will continue to move with uniform speed in a straight line unless it is acted upon by a force.
This tendency of a body to remain in its state of rest or uniform motion is called the inertia of the body. For this reason, Newton's first law is sometimes called "the law of inertia".
As an example of inertia, think of the effect on a car driver if his car is standing still and another car crashes into his car from behind. His car is suddenly knocked forward. His body is pushed forward by the seat, but his head stays still, in its state of rest,and is jerked back in relation to his body. For this reason, neck injuries are common in accidents where cars are hit from behind and many modern cars have headrests to protect drivers and passengers from injury.
https://youtu.be/1XSyyjcEHo0
https://youtu.be/sabH4bJsxWA
MOMENTUM
Momentum is an important property of a moving object. It explains its tendency to continue moving in a straight line
The momentum of a body is defined as the product of its mass and its velocity.
Momentum = mv
The unit of momentum corresponds to kgms-1. Thus a bullet having a small mass 0.01kg, moving with a high velocity of 1000 ms-1 and a heavy ball of mass 100kg moving with a small speed of 1ms-1 have the same momentum. We also note that the greater the force it will exert on the body it hits.
More powerful brakes are required to stop a heavy lorry than a light car moving with the same speed.
NEWTON'S SECOND LAW AND ITS APPLICATIONS
The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.
Force α change in momentum/time taken
Suppose a force F acts on a body of mass m for a time t and causes it to change its velocity from u to v, then Newton's second law can be stated as
F α [sup]mv - mu[/sup]/[sub]t[/sub]
So F α [sup]m(v-u)[/sup]/[sub]t[/sub]
but [sup] v - u[/sup]/[sub]t[/sub] = a(acceleration)
Therefore, F α ma
F = kma
If we take m = 1kg and a = 1ms-1, the unit of force is chosen to make F = 1 when k = 1. The SI unit of force is called the Newton (N),and it is the force which produces an acceleration of 1ms-2 when it acts on a mass of 1kg. Thus when F is in newton, m in kg, a in ms-2 we have
F = ma
This equation is recognised as a fundamental equation of dynamics and is one of the most important equations of physics. It must be noted that, when using the above equation, the force F must be the resultant force acting on the body.
Example 1: a body of mass 2kg undergoes a constant horizontal acceleration of 5ms-2.Calculate the resultant horizontal force acting on the body. What will be the resultant force on the body when it moves with a uniform velocity of 10ms-1?
Solution
Given that a = 5ms-2, m= 2kg, then Newton's fundamental equation
F = ma becomes
F = 5 x 2
F = 10N
Which is the resultant force on the body.If the object moves with uniform velocity, it means that it not accelerating, i.e. a = 0
F = ma
= m x 0
= 0
The resultant force on the body is zero.
DEDUCTIONS FROM NEWTON'S SECOND LAW
IMPULSE of a force
Now F = [sup]m(v - u)[/sup]/[sub]t[/sub]
or Ft = mv - mu
the quantity Ft is called the IMPULSE I of the force.
Impulse is the product of a large force and a very short time during which it acts.
The unit of impulse is the newton second(Ns). Since this quantity is equal to change in momentum it means that momentum can also be expressed in Ns.
Therefore, I = Ft = change in momentum
Example 1: A force of 10N acts for 20s. What is the change in momentum of the body?
Solution
Impulse = Ft = change in momentum
Therefore, change in momentum
10 x 20
200Ns
Example 2: A body of mass 5kg moving with a speed of 30ms-1 is suddenly hit by another body moving in the same direction, thereby changing the speed of the former body to 60ms-2. What is the impulse received by the first body?
Solution
Impulse = change in momentum
mv = mu
= m(v - u)
= 5(60 - 30)
= 150Ns
https://youtu.be/xzA6IBWUEDE
NEWTON'S THIRD LAW AND ITS APPLICATION
To every action there is an equal and opposite reaction
When we place an object on a table the reaction of the table on the object(the vertical force exerted on the object by the table) is equal and opposite to the action of the object on the table (the weight of the object bearing down on the table)
Again, if a moving car A hits a stationary car B, the force exerted on B by A (the reaction) will be the same as the action of B on A. This is why both cars are damaged.
Application
When a bullet is shot out of a gun the person firing it experiences the backwards recoil force of the gun(a reaction) and this is equal to the propulsive force(action) acting on the bullet.
Thus for a bullet of mass, m , and velocity, v, the velocity V of recoil of the gun is given as MV = -mv
V = -mv
M
Where M is the mass of the gun
Newton's third law has a very useful application in the operation of jet aeroplanes and rockets. The application is based on the fact that a large mass of very hot gases issues from the nozzle behind the jet or rocket. The velocity and mass per second of the gases are so high that considerable momentum is imparted to the stream of gas. An equal and opposite momentum is imparted to the rocket or aeroplane which undergoes a forward thrust.
https://youtu.be/y61_VPKH2B4
LAW OF CONSERVATION OF MOMENTUM
Newton's second and third laws enable us to formulate an important conservation law known as the law of conservation of momentum.
In a system of colliding objects the total momentum is conserved, provided there is no net external force acting on the system.
It can also be stated as follows:
The total momentum of an isolated or closed system of colliding bodies remains constant.
Thus if two or more bodies collide in a closed system, the total momentum before collision is equal to the total momentum after collision.
https://youtu.be/8HXISImmGuQ
EVALUATION:
(i) Explain what is meant by the moment of a force about a point.
(ii) Distinguish between the resultant and the equilibrant of forces
ASSIGNMENT:
(i) State two conditions necessary for the equilibrium of three non-parallel co-planar forces.
WEEK 9
MAIN TOPIC: SIMPLE HARMONIC MOTION
SPECIFIC TOPIC: DEFINITION OF TERMS
REFERENCE BOOK: Igcse Physics by Richard Woodside
OBJECTIVE: At the end of the lesson, the students should be able to:
(i) Explain simple harmonic motion.
(ii) Define amplitude, frequency and period
CONTENT: SIMPLE HARMONIC MOTION
Simple Harmonic Motion is cause by restoring forces and is defined ass the motion of an object whose acceleration is proportional to the distance from a fixed point and is always directed towards that point, e.g the prongs of a tuning fork , mass hanging from a spiral spring, the swinging of a pendulum bob, loaded test-tube floating and oscillating vertically in a liquid.
https://youtu.be/8bTvHbrJ8Cg
https://youtu.be/uM2HpLBVAkA
https://youtu.be/JSBw-JyFgZk
https://youtu.be/LRmZSz0t_HU
SPEED AND ACCELERATION OF A S.H.M
Velocity during S.H.M : It is observed that when a small object moves at a constant angular speed round a circle, the motion of the object which is perpendicular to a fixed diameter of the circle is a simple harmonic motion.
Let us consider a point, C, moving round a circle of radius, r, with uniform speed, w. As C moves round the circle once, the foot of the perpendicular D moves from B through O to A and back to B. The velocity of D along charges continuously as the point moves along AB. It has its greatest velocity at O, the centre motion of d but a zero velocity at A and B. This to and fro motion along the fixed diameter is defined as simple harmonic motion.
SIMPLE HARMONIC MOTION FORMULA
Acceleration = rw2
The acceleration towards the centre =rw2 sinθ where r sinθ = y; using resolution into component where acceleration = -w2y.
DISPLACEMENT
From r sin θ = y
Since θ = wt
Y =r sin wt. (This shows that the graph g against θ or y against wt is a sine curve
VELOCITY
V = +-r cos wt = +-w r2 - y2
Velocity max = wr where y = r
Since a = w2y (The acceleration is maximum where y = r)
So amax = -w2y ............................(i)
Acceleration = a = -w2y. Period = T = 2∏/w
Velocity = +- w r2 - y2 ...........................(ii)
https://youtu.be/xwp8EqGceWE
https://youtu.be/WkLO7tQYfDU
PERIOD, FREQUENCY AND AMPLITUDE OF S.H.M
The movement of a simple pendulum is an example of a simple harmonic motion. The bob of a simple harmonic moving through a small angle θ has a simple harmonic motion.
AMPLITUDE: The amplitude of a simple pendulum is the maximum displacement of the moving bob.
CYCLE: One complete oscillation is called cycle.
PERIOD: Period, T, of the motion is the time taken to complete one cycle.
FREQUENCY: Frequency, F, is the number of circles per period.
https://youtu.be/vxdmK3LTwug
EVALUATION:
(1) Explain the following (i) simple harmonic motion (ii) frequency (iii) period
(2) Find the relationship between frequency and period
MAIN TOPIC: SIMPLE HARMONIC MOTION
SPECIFIC TOPIC: MATHEMATICAL REPRESENTATION OF FREQUENCY AND PERIOD
REFERENCE BOOK: SENIOR SECONDARY PHYSICS,BY OKEKE.
OBJECTIVE: At the end of the lesson, the students should be able to:
Use the derived formula to solve some problems
CONTENT: MATHEMATICAL REPRESENTATION OF FREQUENCY AND PERIOD OF S.H.M (view below)
https://drive.google.com/file/d/1Yz3kK5 ... BzfcT/view
EVALUATION:
A particle makes 240 revolutions per minute on a circle of radius 2m. Find (i) its period (ii) angular velocity (iii) linear velocity and its acceleration.
ASSIGNMENT:
The amplitude of a particle executing simple harmonic motion is 5cm, while its angular frequency is 10radian-1. Calculate the maximum acceleration of the particle.
SPECIFIC TOPIC: DEFINITION OF TERMS
REFERENCE BOOK: Igcse Physics by Richard Woodside
OBJECTIVE: At the end of the lesson, the students should be able to:
(i) Explain simple harmonic motion.
(ii) Define amplitude, frequency and period
CONTENT: SIMPLE HARMONIC MOTION
Simple Harmonic Motion is cause by restoring forces and is defined ass the motion of an object whose acceleration is proportional to the distance from a fixed point and is always directed towards that point, e.g the prongs of a tuning fork , mass hanging from a spiral spring, the swinging of a pendulum bob, loaded test-tube floating and oscillating vertically in a liquid.
https://youtu.be/8bTvHbrJ8Cg
https://youtu.be/uM2HpLBVAkA
https://youtu.be/JSBw-JyFgZk
https://youtu.be/LRmZSz0t_HU
SPEED AND ACCELERATION OF A S.H.M
Velocity during S.H.M : It is observed that when a small object moves at a constant angular speed round a circle, the motion of the object which is perpendicular to a fixed diameter of the circle is a simple harmonic motion.
Let us consider a point, C, moving round a circle of radius, r, with uniform speed, w. As C moves round the circle once, the foot of the perpendicular D moves from B through O to A and back to B. The velocity of D along charges continuously as the point moves along AB. It has its greatest velocity at O, the centre motion of d but a zero velocity at A and B. This to and fro motion along the fixed diameter is defined as simple harmonic motion.
SIMPLE HARMONIC MOTION FORMULA
Acceleration = rw2
The acceleration towards the centre =rw2 sinθ where r sinθ = y; using resolution into component where acceleration = -w2y.
DISPLACEMENT
From r sin θ = y
Since θ = wt
Y =r sin wt. (This shows that the graph g against θ or y against wt is a sine curve
VELOCITY
V = +-r cos wt = +-w r2 - y2
Velocity max = wr where y = r
Since a = w2y (The acceleration is maximum where y = r)
So amax = -w2y ............................(i)
Acceleration = a = -w2y. Period = T = 2∏/w
Velocity = +- w r2 - y2 ...........................(ii)
https://youtu.be/xwp8EqGceWE
https://youtu.be/WkLO7tQYfDU
PERIOD, FREQUENCY AND AMPLITUDE OF S.H.M
The movement of a simple pendulum is an example of a simple harmonic motion. The bob of a simple harmonic moving through a small angle θ has a simple harmonic motion.
AMPLITUDE: The amplitude of a simple pendulum is the maximum displacement of the moving bob.
CYCLE: One complete oscillation is called cycle.
PERIOD: Period, T, of the motion is the time taken to complete one cycle.
FREQUENCY: Frequency, F, is the number of circles per period.
https://youtu.be/vxdmK3LTwug
EVALUATION:
(1) Explain the following (i) simple harmonic motion (ii) frequency (iii) period
(2) Find the relationship between frequency and period
MAIN TOPIC: SIMPLE HARMONIC MOTION
SPECIFIC TOPIC: MATHEMATICAL REPRESENTATION OF FREQUENCY AND PERIOD
REFERENCE BOOK: SENIOR SECONDARY PHYSICS,BY OKEKE.
OBJECTIVE: At the end of the lesson, the students should be able to:
Use the derived formula to solve some problems
CONTENT: MATHEMATICAL REPRESENTATION OF FREQUENCY AND PERIOD OF S.H.M (view below)
https://drive.google.com/file/d/1Yz3kK5 ... BzfcT/view
EVALUATION:
A particle makes 240 revolutions per minute on a circle of radius 2m. Find (i) its period (ii) angular velocity (iii) linear velocity and its acceleration.
ASSIGNMENT:
The amplitude of a particle executing simple harmonic motion is 5cm, while its angular frequency is 10radian-1. Calculate the maximum acceleration of the particle.