PHYSICS
CHAPTER 2
CHAPTER 2: Kinematics of linear motion
1
PHYSICS
CHAPTER 2
Kinematics of linear motion 2.1 2.2 2.3 2.4
Linear Motion Uniformly Accelerated Motion Free Falling Body Projectile Motion
2
PHYSICS CHAPTER 2 Learning Outcome: 2.1 Linear Motion At the end of this chapter, students should be able to: Define and distinguish between i. distance and displacement ii. speed and velocity iii. instantaneous velocity, average velocity and uniform velocity. iv. instantaneous acceleration, average acceleration and uniform acceleration. Sketch graphs of displacement-time, velocity-time and acceleration-time. Determine the distance travelled, displacement, velocity and acceleration from appropriate graphs. 3
PHYSICS
CHAPTER 2
2.1. Linear motion (1-D) 2.1.1. Distance,
d
scalar quantity. is defined as the length of actual path between two points. points For example : Q
P
The length of the path from P to Q is 25 cm.
4
PHYSICS 2.1.2
CHAPTER 2
Displacement,s
vector quantity is defined as the distance between initial point and final point in a straight line. line The S.I. unit of displacement is metre (m).
Example 1: An object P moves 20 m to the east after that 10 m to the south and finally moves 30 m to west. Determine the displacement of P N relative to the original position. Solution : O
W
20 m
θ P
θ 10 m
S
E 10 m
20 m 5
PHYSICS
CHAPTER 2
The magnitude of the displacement is given by
and its direction is
2.1.3 Speed, v
is defined the rate of change of distance. distance scalar quantity. Equation:
change of distance speed = time interval
Δd v= Δt 6
PHYSICS 2.1.4
CHAPTER 2
Velocity, v
is a vector quantity. The S.I. unit for velocity is m s-1.
Average velocity, vav
is defined as the rate of change of displacement. displacement Equation: change of displacement
vav =
time interval
s2 − s1 vav = t 2 − t1 Δs vav = Δt
Its direction is in the same direction of the change in displacement. displacement
7
PHYSICS
CHAPTER 2
Instantaneous velocity, v is defined as the instantaneous rate of change of displacement. displacement Equation:
limit ∆ s v= ∆t→ 0∆t
ds v= dt
An object is moving in uniform velocity if
ds = constant dt 8
PHYSICS
CHAPTER 2
s
s1
0
Q
The gradient of the tangent to the curve at point Q = the instantaneous velocity at time, t = t1
t
t1 Therefore
Gradient of s-t graph = velocity 9
PHYSICS 2.1.5
CHAPTER 2
Acceleration, a
vector quantity The S.I. unit for acceleration is m s-2.
Average acceleration, aav
is defined as the rate of change of velocity. velocity Equation: a = change of velocity av
time interval
v2 − v1 aav = t 2 − t1
Δv aav = Δt
Its direction is in the same direction of motion. motion The acceleration of an object is uniform when the magnitude of velocity changes at a constant rate and along fixed 10
PHYSICS
CHAPTER 2
Instantaneous acceleration, a is defined as the instantaneous rate of change of velocity. velocity Equation: limit
∆v a= ∆t→ 0∆t 2
dv d s a= = 2 dt dt
An object is moving in uniform acceleration if
dv = constant dt 11
PHYSICS
v
CHAPTER 2
Deceleration, a is a negative acceleration. acceleration The object is slowing down meaning the speed of the object decreases with time. time
Q
v1
The gradient of the tangent to the curve at point Q = the instantaneous acceleration at time, t = t1
0
Therefore
t1
t
Gradient of v-t graph = acceleration 12
PHYSICS 2.1.6
CHAPTER 2 Graphical methods
Displacement against time graph (s-t)
s
s
Gradient increases with time
Gradient = constant 0
s (a) Uniform velocity
t
t
0
(b) The velocity increases with time
Q
(c) P
R
Gradient at point R is negative.
Gradient at point Q is zero.
0
t
The direction of velocity is changing.
The velocity is zero.
13
PHYSICS
CHAPTER 2
Velocity versus time graph (v-t)
v
v Uniform velocity
v
Uniform acceleration
B
C
A 0
t1 (a) t2
t
0
t1
(b) t2
t
0
t1
t
t2 (c)
Area under the v-t graph = displacement
The gradient at point A is positive – a > 0(speeding up) The gradient at point B is zero – a= 0 The gradient at point C is negative – a < 0(slowing down) 14
PHYSICS
CHAPTER 2 From the equation of instantaneous velocity,
ds v= dt
∫ ds = ∫ vdt Therefore
s=
∫
t2 t1
vdt
s = shaded area under the v − t graph Simulation 2.1
Simulation 2.2
Simulation 2.3 15
PHYSICS
CHAPTER 2
Example 2 : A toy train moves slowly along a straight track according to the displacement, s against time, t graph in figure 2.1.
s (cm)
10 8 6 4 2 Figure 2.1 0
2
4
6
8
10 12 14
t (s)
a. Explain qualitatively the motion of the toy train. b. Sketch a velocity (cm s-1) against time (s) graph. c. Determine the average velocity for the whole journey. d. Calculate the instantaneous velocity at t = 12 s.
16
PHYSICS
CHAPTER 2
Solution : 0 to 6 s
:
6 to 10 s : 10 to 14 s : b.
v (cm s−1) 1.50
0.68
0
2
4
6
8
10 12 14
t (s) 17
PHYSICS
CHAPTER 2
Solution : c. v
d.
av
s2 − s1 = t 2 − t1
v = average velocity from 10 s to 14 s s2 − s1 v= t 2 − t1
18
PHYSICS
CHAPTER 2
Example 3 : A velocity-time (v-t) graph in figure 2.2 shows the motion of a lift.
v (m s −1) 4 2 0 -2 Figure 2.2
5
10 15
20 25 30 35 40 45
50
t (s)
-4
a. Describe qualitatively the motion of the lift. b. Sketch a graph of acceleration (m s-1) against time (s). c. Determine the total distance travelled by the lift and its displacement. d. Calculate the average acceleration between 20 s to 40 s. 19
PHYSICS
CHAPTER 2
Solution : a. 0 to 5 s 5 to 15 s 15 to 20 s 20 to 25 s 25 to 30 s 30 to 35 s
: Lift moves upward from rest with acceleration of 0.4 m s−2. : The velocity of the lift from 2 m s−1 to 4 m s−1 but the acceleration to 0.2 m s−2. : Lift : Lift : Lift : Lift moves
35 to 40 s :
Lift moving
40 to 50 s :
20
PHYSICS
CHAPTER 2
Solution : −2 b. a (m s ) 0.8 0.6 0.4 0.2 0 -0.2
5
10 15
20 25 30 35 40 45
50
t (s)
-0.4 -0.6 -0.8
21
PHYSICS
CHAPTER 2
Solution : −1 v (m s ) c. i. 4 2 0 -2
A1 5
A2 10 15
A3 20 25 30 A35 40 45 4 A5
50
t (s)
-4
Total distance = area under the graph of v-t = A1 + A 2 + A 3 + A 4 + A 5
1 1 1 1 1 Total distance = ( 2 )( 5) + ( 2 + 4 )(10 ) + ( 5 + 10 )( 4 ) + ( 5)( 4 ) + (15 + 5)( 4 ) 2 2 2 2 2
22
PHYSICS
CHAPTER 2
Solution : c. ii. Displacement
= area under the graph of v-t = A1 + A 2 + A 3 + A 4 + A 5
1 1 1 1 1 Displacement = ( 2 )( 5) + ( 2 + 4 )(10 ) + ( 5 + 10 )( 4 ) + ( 5)( − 4 ) + (15 + 5)( − 4 ) 2 2 2 2 2
d.
v2 − v1 aav = t 2 − t1
23
PHYSICS
CHAPTER 2
Exercise 2.1 : 1. Figure 2.3 shows a velocity versus time graph for an object constrained to move along a line. The positive direction is to the right.
Figure 2.3
a. Describe the motion of the object in 10 s. b. Sketch a graph of acceleration (m s-2) against time (s) for the whole journey. c. Calculate the displacement of the object in 10 s. 24 ANS. : 6 m
PHYSICS
CHAPTER 2
Exercise 2.1 : 2. A train pulls out of a station and accelerates steadily for 20 s until its velocity reaches 8 m s−1. It then travels at a constant velocity for 100 s, then it decelerates steadily to rest in a further time of 30 s. a. Sketch a velocity-time graph for the journey. b. Calculate the acceleration and the distance travelled in each part of the journey. c. Calculate the average velocity for the journey. Physics For Advanced Level, 4th edition, Jim Breithaupt, Nelson Thornes, pg.15, no. 1.11
ANS. : 0.4 m s−2,0 m s−2,-0.267 m s−2, 80 m, 800 m, 120 m; 6.67 m s−1.
25
PHYSICS CHAPTER 2 Learning Outcome: 2.2 Uniformly accelerated motion At the end of this chapter, students should be able to: Derive and apply equations of motion with uniform acceleration:
v = u + at 1 2 s = ut + at 2 2 2 v = u + 2as
26
PHYSICS
CHAPTER 2
2.2. Uniformly accelerated motion
From the definition of average acceleration, uniform (constant) constant acceleration is given by
v− u a= t
v = u + at where
v u a t
(1)
: final velocity : initial velocity : uniform (constant) acceleration : time
27
PHYSICS
CHAPTER 2 From equation (1), the velocity-time graph is shown in figure velocity 2.4:
v
u Figure 2.4
t
0
time
From the graph,
The displacement after time, s = shaded area under the graph = the area of trapezium Hence,
1 s = ( u + v)t 2
(2) 28
PHYSICS
CHAPTER 2 By substituting eq. (1) into eq. (2) thus
1 s = [ u + ( u + at ) ]t 2
1 2 s = ut + at 2
From eq. (1), From eq. (2),
( v − u ) = at 2s (v + u) =
( v + u )( v − u ) =
(3)
multiply
t 2s ( at ) t
v 2 = u 2 + 2as
(4) 29
PHYSICS
CHAPTER 2 Notes: equations (1) – (4) can be used if the motion in a straight line with constant acceleration.
For a body moving at constant velocity, ( a = 0) the equations (1) and (4) become
v= u
Therefore the equations (2) and (3) can be written as
s = vt
constant velocity
30
PHYSICS
CHAPTER 2
Example 4 : A plane on a runway takes 16.2 s over a distance of 1200 m to take off from rest. Assuming constant acceleration during take off, calculate a. the speed on leaving the ground, b. the acceleration during take off. Solution : a= ?
u= 0
a. Use
v= ?
s = 1200 m t = 16.2 s 1 s = ( u + v )t 2 31
PHYSICS
CHAPTER 2
Solution : b. By using the equation of linear motion,
v 2 = u 2 + 2as
OR
1 2 s = ut + at 2
32
PHYSICS
CHAPTER 2
Example 5 : A bus travelling steadily at 30 m s−1 along a straight road passes a stationary car which, 5 s later, begins to move with a uniform acceleration of 2 m s−2 in the same direction as the bus. Determine a. the time taken for the car to acquire the same velocity as the bus, b. the distance travelled by the car when it is level with the bus. −1 −2 Solution : vb = 30 m s = constant ; u c = 0; ac = 2 ms −1 v = v = 30 m s a. Given c b Use vc = u c + ac t c
33
PHYSICS
CHAPTER 2
b.
b c
vb = 30 m s − 1
b
b
ac = 2 m s − 2
uc = 0
tb = 0 s
vb
c
tb = 5 s
tb = t
s c = sb
From the diagram,
tb = t ; t c = t − 5 s c = sb
1 2 uc tc + ac tc = vbtb 2
vb
Therefore
sc = vb t
34
PHYSICS
CHAPTER 2
Example 6 : A particle moves along horizontal line according to the equation
s = 3t 3 − 4t 2 + 2t
Where s is displacement in meters and t is time in seconds. At time, t =2.00 s, determine a. the displacement of the particle, b. Its velocity, and c. Its acceleration. Solution : a. t =2.00 s ;
3
2
s = 3t − 4t + 2t
35
PHYSICS
CHAPTER 2
Solution : b. Instantaneous velocity at t = 2.00 s, Use
ds v= dt
(
d 3 v= 3t − 4t 2 + 2t dt Thus
)
v = 9( 2.00 ) − 8( 2.00 ) + 2 2
36
PHYSICS
CHAPTER 2
Solution : c. Instantaneous acceleration at t = 2.00 s, Use
dv a= dt
Hence
a = 18( 2.00 ) − 8
37
PHYSICS
CHAPTER 2
Exercise 2.2 : 1. A speedboat moving at 30.0 m s-1 approaches stationary buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.50 m s-2 by reducing the throttle. a. How long does it take the boat to reach the buoy? b. What is the velocity of the boat when it reaches the buoy? No. 23,pg. 51,Physics for scientists and engineers with modern physics, Serway & Jewett,6th edition. ANS. : 4.53 s; 14.1 m s−1
2. An unmarked police car travelling a constant 95 km h -1 is passed by a speeder traveling 140 km h-1. Precisely 1.00 s after the speeder passes, the policemen steps on the accelerator; if the police car’s acceleration is 2.00 m s -2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)? No. 44, pg. 41,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd edition. 38 ANS. : 14.4 s
PHYSICS
CHAPTER 2
Exercise 2.2 : 3. A car traveling 90 km h-1 is 100 m behind a truck traveling 75 km h-1. Assuming both vehicles moving at constant velocity, calculate the time taken for the car to reach the truck. No. 15, pg. 39,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd edition. ANS. : 24 s
4. A car driver, travelling in his car at a constant velocity of 8 m s-1, sees a dog walking across the road 30 m ahead. The driver’s reaction time is 0.2 s, and the brakes are capable of producing a deceleration of 1.2 m s-2. Calculate the distance from where the car stops to where the dog is crossing, assuming the driver reacts and brakes as quickly as possible. ANS. : 1.73 m 39
PHYSICS CHAPTER 2 Learning Outcome: 2.3 Free falling body At the end of this chapter, students should be able to: Describe and use equations for free falling body. For upward and downward motion, use
a = −g = −9.81 m s−2
40
PHYSICS
CHAPTER 2
2.3. Free falling body
is defined as the vertical motion of a body at constant acceleration, g under gravitational field without air resistance. In the earth’s gravitational field, the constant acceleration known as acceleration due to gravity or free-fall acceleration or gravitational acceleration. acceleration the value is g = 9.81 m s−2 the direction is towards the centre of the earth (downward). Note: In solving any problem involves freely falling bodies or free fall motion, the assumption made is ignore the air resistance. resistance
41
PHYSICS
CHAPTER 2 Sign convention:
+ From the sign convention thus, +
-
a= −g
Table 2.1 shows the equations of linear motion and freely falling bodies. Linear motion
v = u + at 2
Table 2.1
2
v = u + 2 as 1 2 s = ut + at 2
Freely falling bodies
v = u − gt
v 2 = u 2 − 2 gs 1 2 s = ut − gt 2 42
PHYSICS
CHAPTER 2 An example of freely falling body is the motion of a ball thrown vertically upwards with initial velocity, u as shown in figure 2.5. velocity = 0
v= u
H
Figure 2.5
u v
Assuming air resistance is negligible, the acceleration of the ball, a = −g when the ball moves upward and its velocity decreases to zero when the ball reaches the maximum height, H. 43
PHYSICS
CHAPTER 2
The graphs in figure 2.6 show the motion of the ball moves up and down. Derivation of equations At the maximum height or displacement, H where t = t1, its velocity,
v= 0 hence
v = u − gt
0 = u − gt1
therefore the time taken for the ball reaches H, Simulation 2.4
u t1 = g
Figure 2.6
s
v =0
H
0 v u
t1
2t1
0
t1
2t1
t1
2t1
t
t
−u a 0
−g
t 44
PHYSICS
CHAPTER 2 To calculate the maximum height or displacement, H: 1 2 use either
s = ut1 −
2
gt1
Where s
OR 2
=H
2
v = u − 2 gs
0 = u 2 − 2 gH
maximum height,
2
u H= 2g
Another form of freely falling bodies expressions are
v = u − gt v 2 = u 2 − 2 gs
1 2 s = ut − gt 2
v y = u y − gt 2 2 v y = u y − 2 gs y 1 2 s y = u y t − gt 2
45
PHYSICS
CHAPTER 2
Example 7 : A ball is thrown from the top of a building is given an initial velocity of 10.0 m s−1 straight upward. The building is 30.0 m high and the ball just misses the edge of the roof on its way down, as shown in B figure 2.7. Calculate a. the maximum height of the stone from point A. b. the time taken from point A to C. u =10.0 m s−1 c. the time taken from point A to D. A d. the velocity of the stone when it reaches point D.
C
(Given g = 9.81 m s−2)
30.0 m
Figure 2.7
46
D
PHYSICS
CHAPTER 2
Solution : a. At the maximum height, H, vy = 0 and u
B
v 2y = u 2y − 2 gs y
= uy = 10.0 m s−1 thus
u C
A
b. From point A to C, the vertical displacement,
1 2 s y = u y t − gt 2
sy= 0 m thus
30.0 m
D 47
PHYSICS
CHAPTER 2
Solution : c. From point A to D, the vertical displacement, sy= −30.0 m thus
B
1 2 s y = u y t − gt 2
u C
A
a 30.0 m
By using
b
t=
− b±
c 2
b − 4ac 2a OR
D
Time don’t have negative value. 48
PHYSICS
CHAPTER 2
Solution : d. Time taken from A to D is t = 3.69 s thus
B
v y = u y − gt v y = (10.0 ) − ( 9.81)( 3.69 )
u C
A
OR From A to D, sy = −30.0 m 2
30.0 m
D
2
v y = u y − 2 gs y
v y = (10.0 ) − 2( 9.81)( − 30.0 ) 2
2
Therefore the ball’s velocity at D is 49
PHYSICS
CHAPTER 2
Example 8 : A book is dropped 150 m from the ground. Determine a. the time taken for the book reaches the ground. b. the velocity of the book when it reaches the ground. (given g = 9.81 m s-2) Solution :
uy = 0 m s
−1
a. The vertical displacement is
sy = −150 m Hence
s y = − 150 m
150 m
1 2 s y = u y t − gt 2
50
PHYSICS
CHAPTER 2
Solution : b. The book’s velocity is given by
uy = 0
v y = u y − gt
OR
s y = − 150 m
2
2
v y = u y − 2 gs y vy = ?
Therefore the book’s velocity is
51
PHYSICS
CHAPTER 2
Exercise 2.3 : 1. A ball is thrown directly downward, with an initial speed of 8.00 m s−1, from a height of 30.0 m. Calculate a. the time taken for the ball to strike the ground, b. the ball’s speed when it reaches the ground. ANS. : 1.79 s; 25.6 m s−1
2. A falling stone takes 0.30 s to travel past a window 2.2 m tall as shown in figure 2.8.
2.2 m
to travel this distance took 0.30 s
Figure 2.8
From what height above the top of the windows did the stone fall? ANS. : 1.75 m
52
PHYSICS CHAPTER 2 Learning Outcome: 2.4 Projectile motion At the end of this chapter, students should be able to: Describe and use equations for projectile,
ux = uy = ax = ay =
u cos θ u sin θ 0 −g
Calculate time of flight, maximum height, range, maximum range, instantaneous position and velocity.
53
PHYSICS CHAPTER 2 2.4. Projectile motion
A projectile motion consists of two components: vertical component (y-comp.)
horizontal component (x-comp.)
motion under constant acceleration, ay= −g motion with constant velocity thus ax= 0
The path followed by a projectile is called trajectory is shown in y figure 2.9.
v1y P
Simulation 2.5 Figure 2.9
uy A
v1
θ1 v1x
u
θ ux
B
v Q
sy=H
v2y
v2x θ2 v2 C
t1
sx= R
t2
x
54
PHYSICS
CHAPTER 2 From figure 2.9, The x-component of velocity along AC (horizontal) at any point is constant,
u x = u cos θ
The y-component (vertical) of velocity varies from one point to another point along AC. but the y-component of the initial velocity is given by
u y = u sin θ
55
PHYSICS
CHAPTER 2 Table 2.2 shows the x and y-components, magnitude and direction of velocities at points P and Q.
Velocity
Point P
x-comp.
v1 x = u x = u cos θ
v2 x = u x = u cos θ
y-comp.
v1 y = u y − gt1
v2 y = u y − gt 2
magnitude
( v1x )
direction
v1 =
2
Point Q
( )
+ v1 y
2
− 1
v1 y θ1 = tan v1 x
v2 =
( v2 x )
2
( )
+ v2 y
2
− 1
v2 y θ 2 = tan v2 x
Table 2.2
56
PHYSICS
CHAPTER 2
2.4.1 Maximum height, H
The ball reaches the highest point at point B at velocity, v where x-component of the velocity, v = v = u = u cos θ x x y-component of the velocity, v = 0 y y-component of the displacement, s = H y
Use
v y2 = u y2 − 2 gs y
0 = ( u sin θ
)
2
− 2 gH
u sin θ H= 2g 2
2
57
PHYSICS
CHAPTER 2
2.4.2 Time taken to reach maximum height, ∆t’ At maximum height, H
Time, t
Use
= ∆t’ and vy= 0
v y = u y − gt
0 = ( u sin θ ) − g∆ t '
u sin θ ∆ t' = g
2.4.3 Flight time, ∆t (from point A to point C)
∆ t = 2∆ t ' 2u sin θ ∆t= g 58
PHYSICS
CHAPTER 2
2.4.4 Horizontal range, R and value of R maximum
Since the x-component for velocity along AC is constant hence
From the displacement formula with uniform velocity, thus the x-component of displacement along AC is
u x = v x = u cos θ
s x = u x t and s x = R R = ( u cos θ )( ∆ t ) 2u sin θ R = ( u cos θ ) g 2 u ( 2 sin θ cos θ ) R= g
59
PHYSICS
CHAPTER 2 From the trigonometry identity, thus
sin 2θ = 2 sin θ cos θ 2
u R= sin 2θ g
The value of R maximum when θ = 45° and sin 2θ = 1 therefore 2
Rmax
u = g
Simulation 2.6 60
PHYSICS
CHAPTER 2
2.4.5 Horizontal projectile
Figure 2.10 shows a ball bearing rolling off the end of a table with an initial velocity, u in the horizontal direction.
u
u
vx vy
h Figure 2.10
A
Horizontal component along path AB.
Vertical component along path AB.
v
B
x
velocity, u x = u = v x = constant displacement, s x = x initial velocity, u y = 0
displacement, s y = − h
Simulation 2.7 61
PHYSICS
CHAPTER 2
Time taken for the ball to reach the floor (point B), t By using the equation of freely falling bodies,
1 2 s y = u y t − gt 2 1 2 − h = 0 − gt 2
t=
2h g
Horizontal displacement, x Use condition below : The time taken for the ball free fall to point A
Figure 2.11
=
The time taken for the ball to reach point B
(Refer to figure 2.11) 62
PHYSICS
CHAPTER 2 Since the x-component of velocity along AB is constant, thus the horizontal displacement, x
sx = u xt
and
x = u
2h g
sx = x
Note : In solving any calculation problem about projectile motion, the air resistance is negligible. negligible
63
PHYSICS
CHAPTER 2
Example 9 : y
u Figure 2.12 O
H
θ = 60.0°
P
R
v1y Figure 2.12 shows a ball thrown by superman with an initial speed, u = 200 m s-1 and makes an angle, θ = 60.0° to the horizontal. Determine a. the position of the ball, and the magnitude and direction of its velocity, when t = 2.0 s.
v1x v1 Q
v2y
x v2x v2 64
PHYSICS
CHAPTER 2
b. the time taken for the ball reaches the maximum height,
H and
calculate the value of H. c. the horizontal range, R d. the magnitude and direction of its velocity when the ball reaches the ground (point P). e. the position of the ball, and the magnitude and direction of its velocity at point Q if the ball was hit from a flat-topped hill with the time at point Q is 45.0 s. (given g = 9.81 m s-2) Solution : The component of Initial velocity :
65
PHYSICS
CHAPTER 2
Solution : a. i. position of the ball when t = 2.0 s , Horizontal component :
sx = u xt
Vertical component :
1 2 s y = u y t − gt 2
therefore the position of the ball is 66
PHYSICS
CHAPTER 2
Solution : a. ii. magnitude and direction of ball’s velocity at t = 2.0 s , Horizontal component :
v x = u x = 100 m s − 1
Vertical component :
v y = u y − gt
Magnitude,
v=
v x2 + v 2y =
(100) 2 + (153) 2
vy − 1 153 θ = tan = tan 100 vx
Direction,
− 1
67
PHYSICS
CHAPTER 2
Solution : b. i. At the maximum height, H :
vy = 0
Thus the time taken to reach maximum height is given by
v y = u y − gt
ii. Apply
1 s y = u y t − gt 2
68
PHYSICS
CHAPTER 2
Solution : c.
Flight time = 2×(the time taken to reach the maximum height)
t = 2(17.6 )
Hence the horizontal range, R is
s x = u xt
d.
When the ball reaches point P thus s y = 0 The velocity of the ball at point P, −1 Horizontal component: v1 x = u x = 100 m s Vertical component: v1 y = u y − gt
69
PHYSICS
CHAPTER 2
Solution : Magnitude, v1
Direction,
=
v +v = 2 1x
2 1y
(100)
2
+ ( − 172)
v1 y − 1 − 172 = tan θ = tan 100 v1x −1
therefore the direction of ball’s velocity is θ = 300 from positive x-axis anticlockwise e. The time taken from point O to Q is 45.0 s. i. position of the ball when t = 45.0 s, Horizontal component :
sx = u xt
70
2
PHYSICS
CHAPTER 2
Solution : Vertical component :
1 2 s y = u y t − gt 2
therefore the position of the ball is (4500 m, −2148 m) e. ii. magnitude and direction of ball’s velocity at t = 45.0 s , Horizontal component :
v2 x = u x = 100 m s
−1
Vertical component :
v2 y = u y − gt 71
PHYSICS
CHAPTER 2
Solution : Magnitude, v2
=
v2 =
v +v 2 2x
(100)
2
2 2y
+ ( − 269 )
2
v2 y θ = tan v2 x
Direction,
−1
therefore the direction of ball’s velocity is
72
PHYSICS
CHAPTER 2
Example 10 : A transport plane travelling at a constant velocity of 50 m s −1 at an altitude of 300 m releases a parcel when directly above a point X on level ground. Calculate a. the flight time of the parcel, b. the velocity of impact of the parcel, c. the distance from X to the point of impact. (given g = 9.81 m s-2) Solution :
u = 50 m s − 1 300 m X
d
73
PHYSICS
CHAPTER 2
Solution : The parcel’s velocity = plane’s velocity thus a.
u = 50 m s − 1 u x = u = 50 m s − 1 and u y = 0 m s − 1
The vertical displacement is given by Thus the flight time of the parcel is
1 2 s y = u y t − gt 2
74
PHYSICS
CHAPTER 2
Solution : b. The components of velocity of impact of the parcel: −1 Horizontal component: v x = u x = 50 m s Vertical component: v y = u y − gt
v y = 0 − ( 9.81)( 7.82 )
Magnitude, v
=
v +v = 2 x
2 y
( 50)
2
+ ( − 76.7 )
2
vy − 1 − 76.7 θ = tan = tan 50 vx
Direction,
−1
therefore the direction of parcel’s velocity is 75
PHYSICS
CHAPTER 2
Solution : c.
Let the distance from X to the point of impact is d. Thus the distance, d is given by
sx = u xt
76
PHYSICS
CHAPTER 2
Exercise 2.4 : Use gravitational acceleration, g = 9.81 m s−2 1. A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket, as in figure 2.13. If he shoots the ball at a 40.0° angle above the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m.
Figure 2.13 ANS. : 10.7 m s−1
77
PHYSICS
CHAPTER 2
Exercise 2.4 : 2. An apple is thrown at an angle of 30° above the horizontal from the top of a building 20 m high. Its initial speed is 40 m s−1. Calculate a. the time taken for the apple to strikes the ground, b. the distance from the foot of the building will it strikes the ground, c. the maximum height reached by the apple from the ground. ANS. : 4.90 s; 170 m; 40.4 m
3. A stone is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m s −1 at 40° above the horizontal. How far above or below its original level will the stone strike the opposite wall? ANS. : 10.3 m below the original level.
78
PHYSICS
CHAPTER 2
THE END… Next Chapter…
CHAPTER 3 : Force, Momentum and Impulse
79
PHYSICS
CHAPTER 3
CHAPTER 3 MOMENTUM AND IMPULSE
1
PHYSICS 3.0
CHAPTER 3
MOMENTUM AND IMPULSE
3.1 Momentum and impulse
3.2
Conservation of linear momentum
2
PHYSICS
CHAPTER 3
Learning Outcome: 3.1 Momentum and impulse At the end of this chapter, students should be able to:
Define momentum.
Define impulse determine impulse
Use
and use F-t graph to
3
PHYSICS 3.1.1 Linear momentum,
p
CHAPTER 3
is defined as the product between mass and velocity. velocity is a vector quantity. Equation :
p = mv
The S.I. unit of linear momentum is kg m s-1. The direction of the momentum is the same as the direction of the velocity. velocity It can be resolve into vertical (y) component and horizontal (x) component.
p
py θ
p x = p cos θ = mv cos θ
p y = p sin θ = mv sin θ
px
4
PHYSICS
3.1.2 Impulse, J
CHAPTER 3
Let a single constant force, F acts on an object in a short time interval (collision), thus the Newton’s 2nd law can be written as
∑ where
dp F= F= = constant dt
J = Fdt = dp = p2 − p1
p2 : final momentum p1 : initial momentum
F : impulsive force
is defined as the product of a force, F and the time, t OR the change of momentum. momentum is a vector quantity whose direction is the same as the constant force on the object. 5
PHYSICS
CHAPTER 3
The S.I. unit of impulse is N s or kg m s−1. If the force acts on the object is not constant then
J=
∫
t2 t1
Fdt = Fav dt
Fav : average impulsive force
where Since impulse and momentum are both vector quantities, then it is often easiest to use them in component form :
consider 2-D collision only
J x = ( Fav ) x dt = p2 x − p1 x = m( v x − u x )
(
J y = ( Fav ) y dt = p2 y − p1 y = m v y − u y
)
J z = ( Fav ) z dt = p2 z − p1 z = m( v z − u z ) 6
PHYSICS
CHAPTER 3
When two objects in collision, the impulsive force, F against time, t graph is given by the Figure 3.20.
F
Figure 3.20
0 t1
t2
t
Shaded area under the F−t graph = impulse Picture 3.1
Picture 3.3
Picture 3.2 7
PHYSICS
CHAPTER 3
Example 3.1 : A 0.20 kg tennis ball strikes the wall horizontally with a speed of 100 m s−1 and it bounces off with a speed of 70 m s−1 in the opposite direction. a. Calculate the magnitude of impulse delivered to the ball by the wall, b. If the ball is in contact with the wall for 10 ms, determine the magnitude of average force exerted by the wall on the ball. Solution : m1 = 0.20 kg
u1 = 100 m s − 1 1 Wall (2)
v1 = 70 m s
−1
1
v2 = u 2 = 0 8
PHYSICS
CHAPTER 3
Solution : a. From the equation of impulse that the force is constant,
J = dp = p2 − p1
J = m1 ( v1 − u1 )
Therefore the magnitude of the impulse is 34 N s. s b. Given the contact time,
J = Fav dt
9
PHYSICS
CHAPTER 3
Example 3.2 : F ( kN )
18
0 0.2
1.0
1.8
t ( ms )
Figure 3.21
An estimated force-time curve for a tennis ball of mass 60.0 g struck by a racket is shown in Figure 3.21. Determine a. the impulse delivered to the ball, b. the speed of the ball after being struck, assuming the ball is being served so it is nearly at rest initially. 10
PHYSICS
CHAPTER 3 −3
Solution : m = 60.0 × 10 kg a. From the force-time graph,
J = area under the F − t graph
= 0 J = dp = m( v − u )
b. Given the ball’s initial speed, u
11
PHYSICS
CHAPTER 3
Exercise 3.1 : 1. A steel ball with mass 40.0 g is dropped from a height of 2.00 m onto a horizontal steel slab. The ball rebounds to a height of 1.60 m. a. Calculate the impulse delivered to the ball during impact. b. If the ball is in contact with the slab for 2.00 ms, determine the average force on the ball during impact. ANS. : 0.47 N s; 237. 1 N 2. A golf ball (m = 46.0 g) is struck with a force that makes an angle of 45° with the horizontal. The ball lands 200 m away on a flat fairway. If the golf club and ball are in contact for 7.00 ms, calculate the average force of impact. (neglect the air resistance.) ANS. : 293 N 12
PHYSICS
CHAPTER 3
Exercise 3.1 : 3.
Figure 3.22
A tennis ball of mass, m = 0.060 kg and a speed, v = 28 m s−1 strikes a wall at a 45° angle and rebounds with the same speed at 45° as shown in Figure 3.22. Calculate the impulse given by the wall. ANS. : 2.4 N s to the left or −2.4 N s 13
PHYSICS CHAPTER 3 3.2 Conservation of linear momentum 3.2.1 Principle of conservation of linear momentum
states “In an isolated (closed) system, the total momentum of that system is constant.” constant OR “When the net external force on a system is zero, the total momentum of that system is constant.” constant In a Closed system,
∑
F= 0
From the Newton’s second law, thus
∑
dp F= = 0 dt
dp = 0 14
PHYSICS Therefore
CHAPTER 3 p = constant
∑ ∑
then
p x = constant p y = constant
According to the principle of conservation of linear momentum, we obtain The total of initial momentum = the total of final momentum OR
∑
pi =
∑
pf
15
PHYSICS
CHAPTER 3
Linear momentum in one dimension collision
Example 3.3 :
u A = 6 m s− 1
uB = 3 m s− 1 B
A
Figure 3.14
Figure 3.14 shows an object A of mass 200 g collides head-on with object B of mass 100 g. After the collision, B moves at a speed .of 2 m s-1 to the left. Determine the velocity of A after Collision :Solution m = 0.200 kg; m = 0.100 kg; u = − 6 m s − 1 A
B
−1
u B = 3 m s ; vB = − 2 m s
∑
pi =
∑
−1
A
pf
16
PHYSICS
CHAPTER 3
Linear momentum in two dimension collision
Example 3.4 : m1
u1
m2
50 m1
Before collision
v1
After collision
Figure 3.15
A tennis ball of mass m1 moving with initial velocity u1 collides with a soccer ball of mass m2 initially at rest. After the collision, the tennis ball is deflected 50° from its initial direction with a velocity v1 as shown in figure 3.15. Suppose that m1 = 250 g, m2 = 900 g, u1 = 20 m s−1 and v1 = 4 m s−1. Calculate the magnitude and direction of soccer ball after the collision. Simulation 3.3 17
PHYSICS
CHAPTER 3 m1 = 0.250 kg; m2 = 0.900 kg; u1 = 20 m s − 1 ;
Solution :
u 2 = 0; v1 = 4 m s − 1 ; θ1 = 50
From the principle of conservation of linear momentum,
∑
pi =
∑
pf
The x-component of linear momentum,
pix = p fx m1u1 x + m2 u 2 x = m1v1 x + m2 v2 x
∑
∑
18
PHYSICS
CHAPTER 3
Solution : The y-component oflinear momentum,
∑
piy =
∑
p fy
0 = m1v1 y + m2 v2 y
Magnitude of the soccer ball,
v2 =
( v2 x ) 2 +
(v )
2
2y
v2 y − 1 0.851 = tan θ 2 = tan 4.84 v2 x
Direction of the soccer ball,
− 1
19
PHYSICS
CHAPTER 3
Exercise 3.2 : 1. An object P of mass 4 kg moving with a velocity 4 m s−1 collides elastically with another object Q of mass 2 kg moving with a velocity 3 m s−1 towards it. a. Determine the total momentum before collision. b. If P immediately stop after the collision, calculate the final velocity of Q. c. If the two objects stick together after the collision, calculate the final velocity of both objects. ANS. : 10 kg m s−1; 5 m s−1 to the right; 1.7 m s−1 to the right 2. A marksman holds a rifle of mass mr = 3.00 kg loosely in his hands, so as to let it recoil freely when fired. He fires a bullet of mass mb = 5.00 g horizontally with a velocity 300 m s-1. Determine a. the recoil velocity of the rifle, b. the final momentum of the system. ANS. : −0.5 m s−1; U think.
20
PHYSICS
CHAPTER 3
3. 1.20 kg
1.80 kg
Before 0.630 m s-1
After
1.40 m s-1
Figure 3.16
In Figure 3.16 show a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless tabletop. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second block, with mass 1.80 kg. Speeds of 0.630 m s−1 and 1.40 m s-1, respectively, are thereby given to the blocks. Neglecting the mass removed from the first block by the bullet, determine a. the speed of the bullet immediately after it emerges from the first block and .b. the initial speed of the bullet ANS. : 721 m s−1; 937.4 m s−1
21
PHYSICS
CHAPTER 3
Exercise 3.2 : 4. A ball moving with a speed of 17 m s−1 strikes an identical ball that is initially at rest. After the collision, the incoming ball has been deviated by 45° from its original direction, and the struck ball moves off at 30° from the original direction as shown in Figure 3.17. Calculate the speed of each ball after the collision.
Figure 3.17
ANS. : 8.80 m s− 1; 12.4 m s−1 22
PHYSICS
CHAPTER 3
3.2.2 Collision
is defined as an isolated event in which two or more bodies (the colliding bodies) exert relatively strong forces on each other for a relatively short time. time
Two types of collisions :
Elastic collision
Inelastic (non-elastic) collision
23
PHYSICS
CHAPTER 3
Elastic collision is defined as one in which the total kinetic energy (as well as total momentum) of the system is the same before and after the collision. collision Figure 3.18 shows the head-on collision of two billiard balls. 1
Before collision
At collision
After collision
m1u1 m2 u 2
1
m1v1
2
1
2
Simulation 3.4
2 Figure 3.18
m2 v 2 24
PHYSICS
CHAPTER 3
The properties of elastic collision are a. The total momentum is conserved. conserved
∑
pi =
∑
pf
b. The total kinetic energy is conserved. conserved
∑
Ki =
∑
Kf
OR
1 1 1 1 2 2 2 m1u1 + m2 u 2 = m1v1 + m2 v22 2 2 2 2 25
PHYSICS
CHAPTER 3
Inelastic (non-elastic) collision is defined as one in which the total kinetic energy of the system is not the same before and after the collision (even though the total momentum of the system is conserved). conserved) Figure 3.19 shows the model of a completely inelastic collision of two billiard balls. u = 0 Before collision
At collision
After collision (stick together)
1
m1u1
2
2
m2 1
1
Simulation 3.5
2
2
v
Figure 3.19
26
PHYSICS
CHAPTER 3
Caution: Not all the inelastic collision is stick together. together In fact, inelastic collisions include many situations in which the bodies do not stick
The properties of inelastic collision are a. The total momentum is conserved. conserved
∑
pi =
∑
pf
b. The total kinetic energy is not conserved because some of the energy is converted to internal energy and some of it is transferred away by means of sound or heat. heat But the total energy is conserved. conserved
∑
Ei =
∑
Ef OR
∑
Ki =
∑
K f + losses energy 27
PHYSICS
CHAPTER 3
Example 3.5 : Ball A of mass 400 g and velocity 4 m s-1 collides with ball B of mass 600 g and velocity 10 m s-1. After collision, A and B will move together. Determine the final velocity of both balls if A and B moves in the opposite direction initially Solution : mA = 0.4 kg, uA = 4 m s-1 , mB = 0.6 kg, uB = -10 m s-1, inelastic collision Before collision
A
A B
After collision
∑
uA
pi =
∑
uB
B
v= ?
pf
By using the principle of conservation of linear momentum, thus A A B B A B
m u + m u = ( m + m )v
28
PHYSICS
CHAPTER 3
Solution :
m Au A + mB u B v= m A + mB
Final velocity of both balls is - 4.4 m s
-1
29
PHYSICS
CHAPTER 3
Example 3.6 : A ball A of mass 1 kg moving at a velocity of 4 m s-1 collides with ball B of mass 2 kg which at rest. Calculate the velocity of both balls after collision if the collision is an elastic collision. Solution : Given mA = 1 kg, uA = 4 m s-1, mB = 2 kg, uB = 0 m s-1, elastic collision Before collision
After collision
A
u A u B = 0ms − 1
vA = ?
B
vB = ? A
B
30
PHYSICS
CHAPTER 3
Solution : Apply principle of conservation of momentum,
m Au A + m B u B = m A v A + m B v B 1(4) + 2(0) = 1(v A ) + 2(vB ) vA = 4 - 2 v B m s − 1 ……..(1) Apply principle of conservation of kinetic energy,
1 1 1 2 2 2 m A (u A ) = m A (v A ) + mB (vB ) 2 2 2 m A (u A ) = m A (v A ) + mB (vB ) 2
2
2
31
PHYSICS
CHAPTER 3
Solution :
1(4) 2 = 1(v A ) 2 + 2(vB ) 2 2
16 = v A + vB
2
………..(2)
Substitute equation (1) into equation (2) 2
16 = (4 − 2vB ) + 2vB 2 2 16 = 4(2 − vB ) + 2vB 2
8 = 2(2 − vB ) + vB 2
2
32
PHYSICS
CHAPTER 3
Solution :
8vB = 3vB
2
Substitute vB = 2.67 ms
−1
into equation (1),
vA = 4 - 2 (2.67)
33
PHYSICS
CHAPTER 3
THE END… Next Chapter… CHAPTER 4 :
Forces
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PHYSICS
CHAPTER 4
CHAPTER 4: FORCES
1
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PHYSICS
CHAPTER 4
4. FORCES 4.1 Basic of Forces and Free Body Diagram 4.2 Newton’s Laws of Motion
2
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PHYSICS CHAPTER 4 Learning Outcome: 4.1 Basic of Forces and Free Body Diagram At the end of this chapter, students should be able to: Identify the forces acting on a body in different situations. Weight Tension Normal force Friction Determine weight, static friction and kinetic friction Draw free body diagram Determine the resultant force
3
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PHYSICS
CHAPTER 4
4.1 Basic of Forces and Free Body Diagram Weight, is defined as the force exerted on a body under gravitational field. It is a vector quantity.
It is dependant on where it is measured, because the value of g varies at different localities on the earth’s surface. It always directed toward the centre of the earth or in the same direction of acceleration due to gravity, g. The S.I. unit is kg m s-2 or Newton (N). Equation:
W mg
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PHYSICS
CHAPTER 4
Tension, T The tension force is the force that is transmitted through a string, rope, cable or wire when it is pulled tight by forces acting from opposite ends. The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire.
Figure 4.1
5
PHYSICS
CHAPTER 4
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Normal (reaction) force,
Figure 4.2
N or R
is defined as a reaction force that exerted by the surface to an object interact with it and the direction always perpendicular to the surface. An object lies at rest on a flat horizontal surface as shown in Figure 4.2. Action: weight of an object is exerted on the N horizontal surface Reaction: surface is exerted a force, N on the object
W mg
F
y
N mg 0
Therefore
A free body diagram is defined as a diagram showing the chosen body by itself, with vectors drawn to show the magnitude and directions of all the forces applied to the body by the other bodies that interact with it.
N mg 6
PHYSICS
CHAPTER 4
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Friction
is defined as a force that resists the motion of one surface relative to another with which it is in contact. is independent of the area of contact between the two surfaces.. is directly proportional to the reaction force. OR
f N
f N
where
f : frictional force
μ : coefficient of friction
N : reaction force
Coefficient of friction, is defined as the ratio between frictional force to reaction force. OR
f N
is dimensionless and depends on the nature of the surfaces. 7
PHYSICS www.kms.matrik.edu.my/physics
CHAPTER 4
There are three types of frictional force :
Static, fs (frictional force act on the object before its move)
Kinetic, fk (frictional force act on the object when its move)
Rolling, fr (frictional force act on the object when its rolling)
f s s N
f k k N
f r r N
Can be ignored where thus
fr fk fs r k s
Caution: The direction of the frictional force exerted by a surface on an object is always in the opposite direction of the motion. The frictional and the reaction forces are always perpendicular. Simulation 4.1 8
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PHYSICS
CHAPTER 4
Example 4.1: A mass is resting on a flat surface which has a normal force of 98N, with a coefficient of static friction of 0.35. What force would it take to move the object? Solution:
N = 98N, μs = 0.35
f s s N
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PHYSICS
CHAPTER 4
Example 4.2: A 15 kg piece of wood is placed on top of another piece of wood. There is 35N of static friction measured between them. Determine the coefficient of static friction between the two pieces of wood. Solution: N = mg = 15(9.81) = 147.15 N, Fs = 35 N
fs s N
10
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PHYSICS
CHAPTER 4
Example 4.3 A dock worker loading crates on a ship finds that a 15 kg crate, initially at rest on a horizontal surface, requires a 50 N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 30 N is required to keep it moving with a constant speed. The acceleration of gravity is 9.8 ms-2. Find the coefficient of kinetic friction. Solution: Mass of crate = m = 15 kg Force required to set the crate in motion = F1 = 50 N Force required to keep the crate in moving at constant speed = fk = 30 N Acceleration of gravity = g = 9.81 ms-2 Normal force, N = mg = =
11
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PHYSICS Solution: From
CHAPTER 4 fk k N
12
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PHYSICS Resultant force
CHAPTER 4
Is defined as a single force that represents the combined effect of two or more forcesy
Example 4.4:
F1 (10 N) 30o
x
O 30o
F3 (40 N)
F2 (30 N)
The figure above shows three forces F1, F2 and F3 acted on a particle O. Calculate the magnitude and direction of the 13 resultant force on particle O.
PHYSICS
CHAPTER 4 y
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Solution :
F2
F3 x
F2 x
30o 60o
30o
x O
F3 y
F3
Fr F F1 F2 F3 Fr Fx Fy Fx F1x F2 x F3x Fy F1 y F2 y F3 y
F2 y F1
14
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PHYSICS Vector
CHAPTER 4 x-component
y-component
F1
F1x 0 N
F1 y F1 F1 y 10 N
F2
F2 x 30 cos 60 F2 x 15 N
F3
F3 x 40 cos 30 F3 x 34.6 N
F2 y 30 sin 60 F2 y 26 N
F3 y 40 sin 30 F3 y 20 N
Vector sum
15
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PHYSICS
CHAPTER 4
Solution : The magnitude of the resultant force is
Fr
F F 2
2
x
y
y
1 θ tan
and
Fy Fx
Fr
Fy
162 18
Fx
O
x
Its direction is 162 from positive x-axis OR 18 above negative xaxis. 16
PHYSICS
CHAPTER 4
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Exercise 4.1: 1. Given three vectors P, Q and R as shown in Figure 4.3.
Q 24 m s 2
R 10 m s 2
P 35 m s 2
y
50 0
x
Figure 4.3 Calculate the resultant vector of P, Q and R. ANS. : 49.4 m s2; 70.1 above + x-axis
17
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PHYSICS CHAPTER 4 Learning Outcome: 4.2 Newton’s laws of motion At the end of this chapter, students should be able to: State Newton’s First Law Define mass as a measure of inertia. Define the equilibrium of a particle. Apply Newton’s First Law in equilibrium of forces State and apply Newton’s Second Law
dp d dv dm F mv v m dt dt dt dt
State and apply Newton’s Third Law.
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PHYSICS
CHAPTER 4
4.2 Newton’s laws of motion 4.2.1 Newton’s first law of motion
states “an object at rest will remain at rest, or continues to move with uniform velocity in a straight line unless it is acted upon by a external forces” OR
Fnett
F 0
The first law gives the idea of inertia.
19
PHYSICS
CHAPTER 4
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4.2.2 Inertia Inertia is defined as the tendency of an object to resist any change in its state of rest or motion. is a scalar quantity.
Mass, m is defined as a measure of a body’s inertia. is a scalar quantity. The S.I. unit of mass is kilogram (kg). The value of mass is independent of location. If the mass of a body increases then its inertia will increase.
20
PHYSICS www.kms.matrik.edu.my/physics
CHAPTER 4
Figures 4.4a and 4.4b show the examples of real experience of inertia.
Figure 4.4
21
PHYSICS
CHAPTER 4
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4.2.3 Equilibrium of a particle
is defined as the vector sum of all forces acting on a particle (point) must be zero. The equilibrium of a particle ensures the body in translational equilibrium and its condition is given by
F Fnett 0
Newton’s first law of motion This is equivalent to the three independent scalar equations along the direction of the coordinate axes,
F
x
0,
F
y
0,
F
z
0
There are two types of equilibrium of a particle. It is
Static equilibrium (v=0) body remains at rest (stationary).
Dynamic equilibrium (a=0) body moving at a uniform (constant) velocity. 22
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PHYSICS
CHAPTER 4
Problem solving strategies for equilibrium of particle
a
The following procedure is recommended when dealing with problems involving the equilibrium of a particle: Sketch a simple diagram of the system to help conceptualize the problem. Sketch a separate free body diagram for each body. Choose a convenient coordinate axes for each body and construct a table to resolve the forces into their components. Apply the condition for equilibrium of a particle in component form : Fx 0 and Fy 0
Solve the component equations for the unknowns.
23
PHYSICS
CHAPTER 4
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Example 4.5: A load of 250 kg is hung by a crane’s cable. The load is pulled by a horizontal force such that the cable makes a 30 angle to the vertical plane. If the load is in the equilibrium, calculate a. the magnitude of the tension in the cable, b. the magnitude of the horizontal force. (Given g =9.81 m s2) Solution : m 250 kg Free body diagram of the load :
30
T Ty 30 F
60
F
Tx
mg 24
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PHYSICS
CHAPTER 4
Solution : m 250 kg 1st method : a. Force x-component (N)
mg F T
y-component (N)
mg 250 9.81 2453 0 T sin 60
0
F T cos 60
Since the load is in the equilibrium, then Thus
F
F 0
Fx 0 y
0
F T cos 60 0
(1)
T sin 60 2453 0 (2)
b. By substituting eq. (2) into eq. (1), therefore
F 2833 cos 60 0
25
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PHYSICS
CHAPTER 4
Solution : m 250 kg 2nd method : a. Since the load is in the equilibrium, then a closed triangle of forces can be sketched as shown below.
30 mg
T
F b.
From the closed triangle of forces, hence
mg cos30 T
F sin 30 T
F sin 30 2833
26
PHYSICS
CHAPTER 4
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Example 4.6:
F1 12 N
F2 20 N
30.0 45.0
55.0
A
F3 30 N Figure 4.5 Calculate the magnitude and direction of a force that balance the three forces acted at point A as shown in Figure 4.5.
27
PHYSICS
CHAPTER 4 F1 12 N; F2 20 N; F3 30 N
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Solution :
Force
x-component (N)
y-component (N)
F1 F2 F3 F
12 cos 55.0 6.88 20 cos 30.0 17.3 30 cos 45.0 21.2 Fx
12 sin 55.0 9.83 20 sin 30.0 10.0 30 sin 45.0 21.2 Fy
To find a force to balance the three forces means the system must be in equilibrium hence
F
x
0
28
PHYSICS
CHAPTER 4
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Solution :
F
0 9.83 10.0 21.2 Fy 0 y
The magnitude of the force,
F and its direction,
Fx 2 Fy 2
31.6 2 1.37 2
1
Fy θ tan Fx 1 1.37
θ tan 31.6 29
PHYSICS
CHAPTER 4
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Example 4.7:
F
50.0
Figure 4.6 A window washer pushes his scrub brush up a vertical window at constant speed by applying a force F as shown in Figure 4.6. The brush weighs 10.0 N and the coefficient of kinetic friction is k= 0.125. Calculate
a. the magnitude of the force F , b. the normal force exerted by the window on the brush.
30 30
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Solution : W 10.0 N; μ 0.125 k a. The free body diagram of the brush :
F
constant speed
50.0
N
fk
W
Force
F W N fk
x-component (N)
y-component (N)
F cos 50.0
F sin 50.0
0
10.0
N
0 μk N 0.125 N
0
The brush moves up atconstant speed (a=0) so that
Thus
F
Fx 0 y
0
F ma 0
N F cos 50.0
(1)
F sin 50.0 0.125 N 10.0
(2)
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Solution : a. By substituting eq. (1) into eq. (2), thus
F sin 50.0 0.125 F cos 50.0 10.0 b. Therefore the normal force exerted by the window on the brush is given by
N F cos 50.0
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Exercise 4.2: Use gravitational acceleration, g = 9.81 m s2 1.
Figure 4.7 The system in Figure 5.8 is in equilibrium, with the string at the centre exactly horizontal. Calculate
a. the tensions T1, T2 and T3. b. the angle . ANS. : 49 N, 28 N, 57 N; 29 33
PHYSICS
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Exercise 4.2: 2.
Figure 4.8 A 20 kg ball is supported from the ceiling by a rope A. Rope B pulls downward and to the side on the ball. If the angle of A to the vertical is 20 and if B makes an angle of 50 to the vertical as shown in Figure 4.8, Determine the tension in ropes A and B. ANS. : 134 N; 300 N 34
PHYSICS
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Exercise 4.2: 3.
Figure 4.9 A block of mass 3.00 kg is pushed up against a wall by a force P that makes a 50.0 angle with the horizontal as show in Figure 4.9. The coefficient of static friction between the block and the wall is 0.250. Determine the possible values for the magnitude of P that allow the block to remain stationary. ANS. : 31.8 N; 48.6 N 35
PHYSICS
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Newton’s second law of motion
states “the rate of change of linear momentum of a moving body is proportional to the resultant force and is in the same direction as the force acting on it” OR its can be represented by
dp F dt
where
F : resultant force
dp : change in linear momentum
dt : time interval
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CHAPTER 4
From the Newton’s 2nd law of motion, it also can be written as
dp an p mv F dt d dm d mv dv F v dt m dt F dt
Case 1: Object at rest or in motion with constant velocity but with changing mass. For example : Rocket
dm dv F v m dt dt dm F v dt
and
dv 0 dt
37
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CHAPTER 4
Case 2: Object at rest or in motion with constant velocity and constant mass.
dm dv dv dm 0 F v m where 0 and dt dt dt dt st law of motion Newton’s 1 F 0 dp Thus F 0 dt
p constant
38
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CHAPTER 4
Case 3: Object with constant mass but changing velocity.
dm dv dm an F v m 0 dt dt dt d dv dv and a F m dt dt
F ma
where F : resultant force
m : mass of an object a : acceleration
The direction of the resultant force always in the same direction of the motion or acceleration.
39
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CHAPTER 4
Newton’s 2nd law of motion restates that “The acceleration of an object is directly proportional to the nett force acting on it and inversely proportional to its mass”. OR F
a
m
One newton(1 N) is defined as the amount of nett force that gives an acceleration of one metre per second squared to a body with a mass of one kilogramme. OR 1 N = 1 kg m s-2 Notes: F is a nett force or effective force or resultant force. The force which causes the motion of an object. If the forces act on an object and the object moving at uniform acceleration (not at rest or not in the equilibrium) hence
Fnett
F ma
40
PHYSICS
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Newton’s third law of motion
states “every action force has a reaction force that is equal in magnitude but opposite in direction”. For example : When the student push on the wall it will push back with the same force. (refer to Figure 4.10) B (wall)
FBA
A (hand)
FAB
FAB FBA
Figure 4.10
Where
FAB is a force by the hand on the wall (action) FBA is a force by the wall on the hand (reaction) 41
PHYSICS
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When a book is placed on the table. (refer to Figure 4.11) Force by the table on the book (reaction)
Figure 4.11
Force by the book on the table (action)
If a car is accelerating forward, it is because its tyres are pushing backward on the road and the road is pushing forward on the tyres. A rocket moves forward as a result of the push exerted on it by the exhaust gases which the rocket has pushed out.
In all cases when two bodies interact, the action and reaction forces act on different bodies. 42
PHYSICS
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Applications of Newton’s 2nd law of motion
From the Newton’s second law of motion, we arrived at equation
F F ma There are five steps in applying the equation above to solve nett
problems in mechanics: Identify the object whose motion is considered. Determine the forces exerted on the object. Draw a free body diagram for each object. is defined as a diagram showing the chosen body by itself, with vectors drawn to show the magnitude and directions of all the forces applied to the body by the other bodies that interact with it. Choose a system of coordinates so that calculations may be simplified. Apply the equation above, Along x-axis: Fx ma
Along y-axis: F
y
x
may
43
PHYSICS
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Example 4.8: Three wooden blocks connected by a rope of negligible mass are being dragged by a horizontal force, F in Figure 4.12.
F
m1
T1
m2
T2
m3
Figure 4.12
Suppose that F = 1000 N, m1 = 3 kg, m2 = 15 kg and m3 = 30 kg. Determine a. the acceleration of blocks system. b. the tension of the rope, T1 and T2. Neglect the friction between the floor and the wooden blocks.
44
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CHAPTER 4
Solution : a. For the block, m1 = 3 kg
F
a
m1
T1
F F
x
x
F T1 m1a 1000 T1 3a (1)
For the block, m2 = 15 kg
T1
a
m2
T2
F F
x
T1 T2 m2 a
x
T1 T2 15a (2)
For the block, m3 = 30 kg
T2
a
F
x
T2 m3 a
m3
(3) 45
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CHAPTER 4
Solution : a. By substituting eq. (3) into eq. (2) thus
T1 45a 0
(4)
Eq. (1)(4) :
b. By substituting the value of acceleration into equations (4) and (3), therefore
46
PHYSICS
CHAPTER 4
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Example 4.9: Two objects of masses m1 = 10 kg and m2 = 15 kg are connected by a light string which passes over a smooth pulley as shown in Figure 4.13. Calculate a. the acceleration of the object of mass 10 kg. b. the tension in the each string. (Given g = 9.81 m s2) Solution : a. For the object m1= 10 kg,
T1
a
Simulation 4.2
F
T1 m1 g m1a
where
T1 T2 T
y
T 10 g 10a W1 m1 g
m1 m2 Figure 4.13
(1) 47
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CHAPTER 4
Solution : a. For the object m2= 15 kg,
T2
a
F F
m2 g T2 m2 a y 15 g T 15a T 15 g 15a y
(2)
Eq. (1) + (2) :
W2 m2 g b. Substitute the value of acceleration into equation (1) thus
T 109.81 101.96
Therefore 48
PHYSICS
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Example 4.10: Two blocks, A of mass 10 kg and B of mass 30 kg, are side by side and in contact with each another. They are pushed along a smooth floor under the action of a constant force F of magnitude 200 N applied to A as shown in Figure 4.14. Determine a. the acceleration of the blocks, B A F b. the force exerted by A on B. Simulation 4.3 Solution :
Figure 4.14
mA 10 kg; mB 30 kg; F 200 N
a. Let the acceleration of the blocks is a. Therefore
F m
mB a F mA mB a x
A
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CHAPTER 4
Solution : b. For the object A,
a
F
A
FBA
F
x
F FBA mAa
200 FBA 105.0
From the Newton’s 3rd law, thus OR For the object B,
a
F
x
FAB
FAB mB a
B 50
PHYSICS
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Exercise 4.3: 1. A block is dragged by forces, F1 and F2 of the magnitude 20 N and 30 N respectively as shown in Figure 4.15. The frictional force f exerted on the block is 5 N. If the weight of the block is 200 N and it is move horizontally, determine the acceleration of the block. (Given g = 9.81 m
s2)
F1
a
50
f
20
F2
Figure 4.15
ANS. : 1.77 m s2
51
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Exercise 4.3: 2. One 3.5 kg paint bucket is hanging by a massless cord from another 3.5 kg paint bucket, also hanging by a massless cord as shown in Figure 4.16. If the two buckets are pulled upward with an acceleration of 1.60 m s2 by the upper cord, calculate the tension in each cord. (Given g = 9.81 m s2)
ANS. : 39.9 N; 79.8 N
Figure 4.16
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CHAPTER 4
THE END… Next Chapter… CHAPTER 5 : Work, Energy and Power
53
CHAPTER 5
WORK, ENERGY AND POWER
CHAPTER 5: Work, Energy and Power (3 Hours)
1
CHAPTER 5
WORK, ENERGY AND POWER
Learning Outcome: 5.1 Work (1 hour)
At the end of this chapter, students should be able to: (a) Define and use work done by a force.
W F s (b) Determine work done from the forcedisplacement graph. 2
CHAPTER 5 5.1
WORK, ENERGY AND POWER
Work, W
Work done by a constant force
is defined as the product of the component of the force parallel to the displacement times the displacement of a body. OR is defined as the scalar (dot) product between force and displacement of a body. 3
CHAPTER 5 Mathematically :
WORK, ENERGY AND POWER
W F s W F cos θ s Fs cos θ
Where,
F : magnitude of force
s : displaceme nt of the body
θ : the angle between F and s
4
CHAPTER 5
WORK, ENERGY AND POWER
It is a scalar quantity.
Dimension :
W F s W ML2T 2
The S.I. unit of work is kg m2 s2 or joule (J).
The joule (1 J) is defined as the work done by a force of 1 N which results in a displacement of 1 m in the direction of the force.
1 J 1 N m 1 kg m 2 s 2 5
CHAPTER 5
WORK, ENERGY AND POWER
Work done by a variable force Figure 5.1 shows a force, F whose magnitude changes with the displacement, s. For a small displacement, s1 the force remains almost constant at F1 and work done therefore becomes W1=F1 s1 .
6
CHAPTER 5
WORK, ENERGY AND POWER F/N FN
F4 F1 0
s1
Figure 5.1
W1 s1
s4
s2
ssN
To find the total work done by a variable force, W when the displacement changes from s=s1 to s=s2, we can divide the displacement into N small successive displacements :
s1 , s2 , s3 , …, sN Thus
W F1s1 F2 s2 ... FN s N
7
CHAPTER 5
WORK, ENERGY AND POWER
When N , s 0, therefore s2
W Fds s1
W the area under the force - displaceme nt graph F/N
Work = Area
0 s1
s2 s/m
8
CHAPTER 5
WORK, ENERGY AND POWER
Applications of work’s equation Case 1 :
Work done by a horizontal force, F on an object (Figure 4.2).
F
Figure 5.2
W Fs cos θ and θ 0 W Fs s
Case 2 :
Work done by a vertical force, F on an object (Figure 4.3).
F
Figure 5.3
W Fs cos θ and θ 90 W 0J s 9
CHAPTER 5
WORK, ENERGY AND POWER
Case 3 :
Work done by a horizontal forces, F1 and F2 on an object (Figure 5.4). F
W F s cos 0 1 1 1 W F s cos 0 2 2 s W W1 W2 F1s F2 s W F1 F2 s and Fnett F1 F2
F2
Figure 5.4
W W
nett
Case 4 :
Fnett s
Work done by a force, F and frictional force, f on an object (Figure 5.5). F
f
Figure 5.5
s
Wnett Fnett s and Fnett F cos θ f ma Wnett F cos f s OR Wnett mas
10
CHAPTER 5
WORK, ENERGY AND POWER
Caution :
Work done on an object is zero when F = 0 or s = 0 and = 90.
11
CHAPTER 5
Sign for work.
WORK, ENERGY AND POWER
W Fs cos
If 0< <90 (acute angle) then cos > 0 (positive value) therefore
W > 0 (positive) work done on the system ( by the external force) where energy is transferred to the system.
If 90< <180 (obtuse angle) then value) therefore
cos <0 (negative
W < 0 (negative) work done by the system where energy is transferred from the system.
12
CHAPTER 5
WORK, ENERGY AND POWER
Example 5.1 : You push your physics reference book 1.50 m along a horizontal table with a horizontal force of 5.00 N. The frictional force is 1.60 N. Calculate a. the work done by the 5.00 N force, b. the work done by the frictional force, c. the total work done on the book. Solution : F 5.00 N
f 1.60 N
s 1.50 m a. Use work’s equation of constant force,
WF Fs cosθ and θ 0 13
CHAPTER 5
WORK, ENERGY AND POWER
Solution : b. W f
fs cos θ and θ 180 W f 1.60 1.50 cos180
c.
W W W W 7.50 2.40 F
f
OR
W F
nett
s
W F f s
W 5.00 1.601.50 14
CHAPTER 5
WORK, ENERGY AND POWER
Example 5.2 : A box of mass 20 kg moves up a rough plane which is inclined to the horizontal at 25.0. It is pulled by a horizontal force F of magnitude 250 N. The coefficient of kinetic friction between the box and the plane is 0.300. a. If the box travels 3.80 m along the plane, determine i. the work done on the box by the force F, ii. the work done on the box by the gravitational force, iii. the work done on the box by the reaction force, iv. the work done on the box by the frictional force, v. the total work done on the box. b. If the speed of the box is zero at the bottom of the plane, calculate its speed when it is travelled 3.80 m. (Given g = 9.81 m s2) 15
CHAPTER 5
WORK, ENERGY AND POWER
Solution : m 20 kg; F
250 N; μk 0.300; s 3.80 m
a
N
y
mg sin 25 x fk 25
Fx
Fy
25
F
25
mg cos 25 W mg
s
a. Consider the work done along inclined plane, thus i. and
WF Fx s cos θ
θ 0
WF 250 cos 25 3.80 cos 0 16
CHAPTER 5
WORK, ENERGY AND POWER
Solution : a. ii. Wg
mg sin 25 s cos θ and θ 180
Wg 20 9.81sin 25 3.80 cos180
iii.
WN Ns cos θ and θ 90
iv.
Wf Wf Wf Wf
f k s cos θ and θ 180 μk N s cos180 μk F sin 25 mg cos 25 s 0.300 250 sin 25 209.81cos 25 3.80
17
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WORK, ENERGY AND POWER
Solution : a. v. W
W W W W W 861 315 0 323 F
g
N
f
b. Given u 0 By using equation of work for nett force,
W mas
223 20 a3.80
Hence by using the equation of linear motion,
v 2 u 2 2as v 2 0 22.933.80
18
CHAPTER 5
WORK, ENERGY AND POWER
Example 5.3 : F (N)
5 0
4
3
5
6
7
s(m)
Figure 5.6
A horizontal force F is applied to a 2.0 kg radio-controlled car as it moves along a straight track. The force varies with the displacement of the car as shown in Figure 5.6. Calculate the work done by the force F when the car moves from 0 to 7 m. Solution :
W area under the F s graph 1 1 W 6 5 35 7 6 4 2 2
19
CHAPTER 5 Exercise 5.1 :
WORK, ENERGY AND POWER
1. A block of mass 2.50 kg is pushed 2.20 m along a frictionless horizontal table by a constant 16.0 N force directed 25.0 below the horizontal. Determine the work done on the block by a. the applied force, b. the normal force exerted by the table, and c. the gravitational force. d. Determine the total work on the block. (Given g = 9.81 m s2) ANS. : 31.9 J; (b) & (c) U think; 31.9 J 2. A trolley is rolling across a parking lot of a supermarket. You apply a constant force F 30 ˆi 40ˆj N to the trolley as it undergoes a displacement s 9.0ˆi 3.0ˆj m . Calculate
a. the work done on the trolley by the force F, b. the angle between the force and the displacement of the trolley. 20 ANS. : 150 J; 108
CHAPTER 5 Exercise 5.1 :
WORK, ENERGY AND POWER y
F3
3.
F1
35 x
F2
50 Figure 5.7
Figure 5.7 shows an overhead view of three horizontal forces acting on a cargo that was initially stationary but that now moves across a frictionless floor. The force magnitudes are F1 = 3.00 N, F2 = 4.00 N and F3 = 10.0 N. Determine the total work done on the cargo by the three forces during the first 4.00 m of displacement. ANS. : 15.3 J 21
CHAPTER 5
WORK, ENERGY AND POWER
Learning Outcome: 5.2
Energy And Conservation Of Energy
At the end of this chapter, students should be able to: (a) Define and use kinetic energy,
1 K mv 2 2 (b) Define and use potential energy: i. gravitational potential energy,
U mgh ii. elastic potential energy for spring,
1 2 U kx 2 (c) State and use the principle of conservation of energy. (d) Explain the work-energy theorem and use the related equation. 22
CHAPTER 5
WORK, ENERGY AND POWER
Energy is defined as the system’s ability to do work. The S.I. unit for energy is same to the unit of work (joule, J). The dimension of energy 2 2
Energy Work ML T
is a scalar quantity. Table 5.1 summarises some common types of energy.
Forms of Energy
Description
Chemical
Energy released when chemical bonds between atoms and molecules are broken.
Electrical
Energy that is associated with the flow of electrical charge.
Heat
Energy that flows from one place to another as a result of a temperature difference.
Internal
Total of kinetic and potential energy of atoms or molecules within a body. 23
CHAPTER 5 Forms of Energy Nuclear
Mass
WORK, ENERGY AND POWER Description Energy released by the splitting of heavy nuclei. Energy released when there is a loss of small amount of mass in a nuclear process. The amount of energy can be calculated from Einstein’s mass-energy equation, E = mc2
Radiant Heat Energy associated with infra-red radiation. Sound Mechanical a. Kinetic b. Gravitational potential c. Elastic potential
Energy transmitted through the propagation of a series of compression and rarefaction in solid, liquid or gas. Energy associated with the motion of a body. Energy associated with the position of a body in a gravitational field. Energy stored in a compressed or stretched spring. Table 5.1
24
CHAPTER 5
WORK, ENERGY AND POWER
Conservation of energy 5.2.1 Kinetic energy, K
is defined as the energy of a body due to its motion. Equation : where K : kinetic energy of a body
1 2 K mv 2
m : mass of a body v : speed of a body
Work-kinetic energy theorem
Consider a block with mass, m moving along the horizontal surface (frictionless) under the action of a constant nett force, Fnett undergoes a displacement, s in Figure 4.8.
Fnett
Figure 5.8
F F
nett
ma
m
s (1)
25
CHAPTER 5
WORK, ENERGY AND POWER
By using an equation of linear motion:
v 2 u 2 2as v2 u 2 a 2s
(2)
By substituting equation (2) into (1), we arrive
v2 u 2 Fnett m 2s 1 2 1 2 Fnett s mv mu K f K i 2 2 Therefore
Wnett K
states “the work done by the nett force on a body equals the change in the body’s kinetic energy”. 26
CHAPTER 5
WORK, ENERGY AND POWER
Example 5.4 : A stationary object of mass 3.0 kg is pulled upwards by a constant force of magnitude 50 N. Determine the speed of the object when it is travelled upwards through 4.0 m. (Given g = 9.81 m s2) Solution : m 3.0 kg ; F 50 N; s 4.0 m; u 0 The nett force acting on the object is given by
F
s F
Fnett F mg 50 3.09.81
mg
mg
By applying the work-kinetic energy theorem, thus
Wnett K f K i 1 2 Fnett s mv 0 2 1 20.64.0 3.0v 2 2
27
CHAPTER 5
WORK, ENERGY AND POWER
Example 5.5 : A block of mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 below the horizontal. If the block starts from rest, calculate its final speed. You can ignore the friction. (Given g = 9.81 m s2) Solution : m 2.00 kg ; s 0.750 m; u 0
N
a
y
mg sin 36.9 mg cos 36.9
36.9
mg
s
x 36.9
28
CHAPTER 5
WORK, ENERGY AND POWER
Solution : m 2.00 kg ; s 0.750 m; u 0 Since the motion of the block along the incline surface thus nett force is given by
Fnett mg sin 36.9 Fnett 2.00 9.81sin 36.9 By using the work-kinetic energy theorem, thus
Wnett K f K i 1 2 Fnett s mv 0 2 1 11.80.750 2.00v 2 2 29
CHAPTER 5
WORK, ENERGY AND POWER
Example 5.6 : F (N) 10
0
4
6
5
7
10
s(m)
Figure 5.9
An object of mass 2.0 kg moves along the x-axis and is acted on by a force F. Figure 5.9 shows how F varies with distance travelled, s. The speed of the object at s = 0 is 10 m s1. Determine a. the speed of the object at s = 10 m, b. the kinetic energy of the object at s = 6.0 m. 30
CHAPTER 5
WORK, ENERGY AND POWER
Solution : m 2.0 kg; u 10 m s 1 a. W area under the F s graph from 0 m to 10 m
1 1 W 6 410 10 6 10 7 5 2 2 By using the work-kinetic energy theorem, thus
W K f Ki 1 2 1 W mv mu 2 2 2 1 1 2 2 32.5 2.0v 2.010 2 2
31
CHAPTER 5
WORK, ENERGY AND POWER
Solution : b. W area under
the F s graph from 0 m to 6 m
1 W 6 410 2 By using the work-kinetic energy theorem, thus
W K f Ki 1 W K f mu 2 2 1 2 50 K f 2.010 2
32
CHAPTER 5 Exercise 5.2 :
WORK, ENERGY AND POWER
Use gravitational acceleration, g = 9.81 m s2 1. A bullet of mass 15 g moves horizontally at velocity of 250 m s1.It strikes a wooden block of mass 400 g placed at rest on a floor. After striking the block, the bullet is embedded in the block. The block then moves through 15 m and stops. Calculate the coefficient of kinetic friction between the block and the floor. ANS. : 0.278 2. A parcel is launched at an initial speed of 3.0 m s1 up a rough plane inclined at an angle of 35 above the horizontal. The coefficient of kinetic friction between the parcel and the plane is 0.30. Determine a. the maximum distance travelled by the parcel up the plane, b. the speed of the parcel when it slides back to the starting point. ANS. : 0.560 m; 1.90 m s1 33
CHAPTER 5
WORK, ENERGY AND POWER
5.2.2 Potential Energy
is defined as the energy stored in a body or system because of its position, shape and state.
Gravitational potential energy, U is defined as the energy stored in a body or system because of its position. Equation :
U mgh where U : gravitatio nal potential energy
m : mass of a body g : acceleration due to gravity h : height of a body from the initial position
The gravitational potential energy depends only on the height of the object above the surface of the Earth. 34
CHAPTER 5
WORK, ENERGY AND POWER
Work-gravitational potential energy theorem
Consider a book with mass, m is dropped from height, h1 to height, h2 as shown in the Figure 5.10.
s
mg
h2
mg
h1
Figure 5.10
The work done by the gravitational force (weight) is
Wg mgs mg h1 h2 Wg mgh1 mgh2 U i U f Wg U f U i U
Therefore in general,
W U
states “ the change in gravitational potential energy as the negative of the work done by the gravitational force”. 35
CHAPTER 5
WORK, ENERGY AND POWER
Negative sign in the equation indicates that
When the body moves down, h decreases, the gravitational force does positive work because U
<0.
When the body moves up, h increases, the work done by gravitational force is negative because U >0. For calculation, use
W U U f U i where
U f : final gravitatio nal potential energy U i : initial gravitatio nal potential energy W : work done by a gravitatio nal force
36
CHAPTER 5
WORK, ENERGY AND POWER
Example 5.7 :
F
20.0 m
Figure 5.11
In a smooth pulley system, a force F is required to bring an object of mass 5.00 kg to the height of 20.0 m at a constant speed of 3.00 m s1 as shown in Figure 5.11. Determine a. the force, F
b. the work done by the force, F. (Given g = 9.81 m s-2) 37
CHAPTER 5
WORK, ENERGY AND POWER
Solution : m 5.00 kg; s h 20.0 m; v constant 3.00 m s 1 a. Since the object moves at the constant speed, thus F
Constant speed
F
Fnett 0 F mg
mg s
mg
b. From the equation of work,
W Fs cos θ and θ 0
OR
W Fs cos θ and θ 0 W U mgh 38
CHAPTER 5
WORK, ENERGY AND POWER
Elastic potential energy, Us is defined as the energy stored in in elastic materials as the result of their stretching or compressing. Springs are a special instance of device which can store elastic potential energy due to its compression or stretching.
Hooke’s Law states “the restoring force, Fs of spring is directly proportional to the amount of stretch or compression (extension or elongation), x if the limit of proportionality is not exceeded” OR
Fs x
Fs kx where
Fs : the restoring force of spring
k : the spring constant or force constant x : the amount of stretch or compressio n ( x f -xi )
39
CHAPTER 5
WORK, ENERGY AND POWER
Negative sign in the equation indicates that the direction of Fs is always opposite to the direction of the amount of stretch or compression (extension), x. Case 1: The spring is hung vertically and its is stretched by a suspended object with mass, m as shown in Figure 5.12.
Figure 5.12
Initial position
Fs
x Final position
The spring is in equilibrium, thus
Fs W mg
W mg
40
CHAPTER 5
WORK, ENERGY AND POWER
Case 2: The spring is attached to an object and it is stretched and compre5sed by a force, F as shown in Figure 5.13.
Fs is negative F s x is positive
F The spring is in equilibrium, hence
x x0
Fs 0 x0
Fs F
(Equilibrium position)
F
x 0
Fs Fs is positive
x is negative
x Figure 5.13
41
CHAPTER 5
WORK, ENERGY AND POWER
Caution: For calculation, use :
Fs kx F
Dimension of spring constant, k :
The unit of k is kg s2 or N m1
where
F : applied force
Fs k MT 2 x
From the Hooke’s law (without “” sign), a restoring force, Fs against extension of the spring, x graph is shown in Figure 5.14.
Fs F
0
Figure 5.14
W area under the Fs x graph 1 1 W Fx1 W kx1 x1 2 2 1 2 W kx1 U s 2 x x 1
42
CHAPTER 5
WORK, ENERGY AND POWER
The equation of elastic potential energy, Us for compressing or stretching a spring is
1 2 1 U s kx Fs x 2 2
The work-elastic potential energy theorem,
W U s
OR
1 2 1 2 W U sf U si kx f kxi 2 2
Notes : Work-energy theorem states the work done by the nett force on a body equals the change in the body’s total energy” OR
Wnett E
E E f
i 43
CHAPTER 5
WORK, ENERGY AND POWER
Example 5.8 : A force of magnitude 800 N caused an extension of 20 cm on a spring. Determine the elastic potential energy of the spring when a. the extension of the spring is 30 cm. b. a mass of 60 kg is suspended vertically from the spring. (Given g = 9.81 m s-2) Solution : F 800 N; x 0.200 m From the Hooke’s law,
Fs F kx 800 k 0.20
a. Given x=0.300 m,
1 2 U s kx 2 1 2 U s 4 10 3 0.300 2
44
CHAPTER 5
WORK, ENERGY AND POWER
Solution : b. Given m=60 kg. When the spring in equilibrium, thus
Fs
x
Fnett 0 Fs mg kx mg 4 103 x 60 9.81
Therefore
W mg
1 2 U s kx 2 1 2 3 U s 4 10 0.147 2
45
CHAPTER 5
WORK, ENERGY AND POWER
5.2.3 Principle of conservation of energy
states “in an isolated (closed) system, the total energy of that system is constant”. According to the principle of conservation of energy, we get The initial of total energy = the final of total energy OR
E E i
f
Conservation of mechanical energy In an isolated system, the mechanical energy of a system is the sum of its potential energy, U and the kinetic energy, K of the objects are constant.
E K U constant OR
Ki U i K f U f
46
CHAPTER 5
WORK, ENERGY AND POWER
Example 5.9 : A 1.5 kg sphere is dropped from a height of 30 cm onto a spring of spring constant, k = 2000 N m1 . After the block hits the spring, the spring experiences maximum compression, x as shown in Figure 5.15. a. Describe the energy conversion occurred after the sphere is dropped onto the spring until the spring experiences maximum compression, x. b. Calculate the speed of the sphere just before strikes the spring.
c. Determine the maximum compression, x.
30 cm
x
Before
After
Figure 5.15
(Given g = 9.81 m s-2) 47
CHAPTER 5
WORK, ENERGY AND POWER
Solution : a.
h 30 cm
v
h0
x h1 h2 (1) The spring is not stretched hence Us = 0. The sphere is at height h0 above ground therefore U = mgh0 and it is stationary hence K = 0.
E
1
mgh0
(3) (2) The spring is not stretched The sphere is at height h2 hence Us = 0. The sphere is above the ground after compressing the spring by x. at height h1 above ground The speed of the sphere at with speed, v just before this moment is zero. Hence strikes the spring. Therefore
1 E2 mgh1 mv 2 2
1 2 E3 mgh2 kx 2 48
CHAPTER 5
WORK, ENERGY AND POWER
Solution : m 1.5 kg; h 0.30 m; k 2000 N m 1 b. Applying the principle of conservation of energy involving the situation (1) and (2),
E E 1
2
1 2 mgh0 mgh1 mv 2 1 2 mg h0 h1 mv and h h0 h1 2 v 2 gh
v 29.810.30
49
CHAPTER 5
WORK, ENERGY AND POWER
Solution : m 1.5 kg; h 0.30 m; k 2000 N m 1 c. Applying the principle of conservation of energy involving the situation (2) and (3),
E E 2
3
1 2 1 2 mgh1 mv mgh2 kx 2 2
1 2 1 2 mg h1 h2 mv kx and x h1 h2 2 2 1 1 2 1.59.81x 1.52.43 2000 x 2 2 2
50
CHAPTER 5
WORK, ENERGY AND POWER
Example 5.10 :
m1
m1 m2
u1
m2
h
Figure 5.16
A bullet of mass, m1=5.00 g is fired into a wooden block of mass, m2=1.00 kg suspended from some light wires as shown in Figure 5.16. The block, initially at rest. The bullet embeds in the block, and together swing through a height, h=5.50 cm. Calculate a. the initial speed of the bullet. b. the amount of energy lost to the surrounding. (Given g = 9.81 m s2)
51
CHAPTER 5
WORK, ENERGY AND POWER
Solution : m1 a.
5.00 10 3 kg; m2 1.00 kg; h 5.50 10 2 m
v12 0 u2 0 m1
u1
m2
m1 m2
(1)
(2)
m1 m2
u12
h (3)
Applying the principle of conservation of energy involving the E2 E3 situation (2) and (3),
K U 1 m1 m2 u12 2 m1 m2 gh 2 u12 2 gh 29.81 5.50 10 2
52
CHAPTER 5
WORK, ENERGY AND POWER
Solution : m1 5.00 10 3 kg; m2 1.00 kg; h 5.50 10 2 Applying the principle of conservation of linear momentum involving the situation (1) and (2),
p1
m
p2
m1u1 m1 m2 u12 5.00 10 3 u1 5.00 10 3 1.00 1.04
b. The energy lost to the surrounding, Q is given by
Q
E E 1
2
1 1 2 2 Q m1 u1 m1 m 2 u12 2 2 1 1 2 2 3 Q 5.00 10 209 5.00 10 3 1.00 1.04 2 2
53
CHAPTER 5
WORK, ENERGY AND POWER
Example 5.11 : Smooth pulley
Q P
2m
Figure 5.17
Objects P and Q of masses 2.0 kg and 4.0 kg respectively are connected by a light string and suspended as shown in Figure 5.17. Object Q is released from rest. Calculate the speed of Q at the instant just before it strikes the floor. (Given g = 9.81 m s2) 54
CHAPTER 5
WORK, ENERGY AND POWER
Solution :
mP 2.0 kg; mQ 4.0 kg; h 2 m; u 0 Smooth pulley
v
Q P
Smooth pulley
2m Initial
P
2m Q
v
Final
Applying the principle of conservation of mechanical energy,
E E i
U Q U P K P KQ 1 1 2 mQ gh mP gh mP v mQ v 2 2 2 1 1 2 4.09.812 2.09.812 2.0v 4.0v 2 2 2 f
55
CHAPTER 5 Exercise 5.3 :
WORK, ENERGY AND POWER
Use gravitational acceleration, g = 9.81 m s2 1. If it takes 4.00 J of work to stretch a spring 10.0 cm from its initial length, determine the extra work required to stretch it an additional 10.0 cm. ANS. : 12.0 J 2. A book of mass 0.250 kg is placed on top of a light vertical spring of force constant 5000 N m1 that is compressed by 10.0 cm. If the spring is released, calculate the height of the book rise from its initial position. ANS. : 10.2 m 3. A 60 kg bungee jumper jumps from a bridge. She is tied to a bungee cord that is 12 m long when unstretched and falls a total distance of 31 m. Calculate a. the spring constant of the bungee cord. b. the maximum acceleration experienced by the jumper. ANS. : 100 N m1; 22 m s2 56
CHAPTER 5 Exercise 5.3 :
WORK, ENERGY AND POWER
4.
Figure 5.18
A 2.00 kg block is pushed against a light spring of the force constant, k = 400 N m-1, compressing it x =0.220 m. When the block is released, it moves along a frictionless horizontal surface and then up a frictionless incline plane with slope =37.0 as shown in Figure 5.18. Calculate a. the speed of the block as it slides along the horizontal surface after leaves the spring. b. the distance travelled by the block up the incline plane before it slides back down. ANS. : 3.11 m s1; 0.81 m 57
CHAPTER 5 Exercise 5.3 :
WORK, ENERGY AND POWER
C
u
5.
A 10 m
B
D
Figure 5.19
A ball of mass 0.50 kg is at point A with initial speed, u =4 m s1 at a height of 10 m as shown in Figure 5.19 (Ignore the frictional force). Determine a. the total energy at point A, b. the speed of the ball at point B where the height is 3 m, c. the speed of the ball at point D, d. the maximum height of point C so that the ball can pass over it. ANS. : 53.1 J; 12.4 m s1; 14.6 m s1; 10.8 m 58
CHAPTER 5
WORK, ENERGY AND POWER
Learning Outcome: 5.3 Power and mechanical efficiency (1 hour) At the end of this chapter, students should be able to: (a) Define and use power: W Average power,
Pav
(b) (c)
Instantaneous Power,
t
dW P dt
P F v
Derive and apply the formulae Define and use mechanical efficiency,
η
Poutput Pinput
100%
and the consequences of heat dissipation. 59
CHAPTER 5
WORK, ENERGY AND POWER
5.3 Power and mechanical efficiency 5.3.1 Power, P
is defined as the rate at which work is done. OR the rate at which energy is transferred. If an amount of work, W is done in an amount of time t by a force, the average power, Pav due to force during that time interval is
W E Pav t t
The instantaneous power, P is defined as the instantaneous rate of doing work, which can be write as
W dW P limit t 0 t dt 60
CHAPTER 5
WORK, ENERGY AND POWER
is a scalar quantity. The dimension of the power is
W ML2T 2 P t T
ML2T 3
The S.I. unit of the power is kg m2 s3 or J s1 or watt (W). Unit conversion of watt (W), horsepower (hp) and foot pounds per second (ft. lb s1)
1 hp 746 W 550 ft. lb s 1
Consider an object that is moving at a constant velocity v along a frictionless horizontal surface and is acted by a constant force, F directed at angle above the horizontal as shown in Figure 5.20. The object undergoes a displacement of ds.
F
Figure 5.20
ds
61
CHAPTER 5
WORK, ENERGY AND POWER
Therefore the instantaneous power, P is given by
dW and dW F cos θ ds P dt ds F cos θ ds and v P dt dt P Fv cos θ OR
P F v where
F : magnitude of force v : magnitude of velocity θ : the angle between F and v 62
CHAPTER 5
WORK, ENERGY AND POWER
Example 5.12 : An elevator has a mass of 1.5 Mg and is carrying 15 passengers through a height of 20 m from the ground. If the time taken to lift the elevator to that height is 55 s. Calculate the average power required by the motor if no energy is lost. (Use g = 9.81 m s2 and the average mass per passenger is 55 kg) Solution : h 20 m; Δt 55 s
M = mass of the elevator + mass of the 15 passengers M = 1500 + (5515) = 2325 kg According to the definition of average power,
E Pav t
Mgh Pav t
63
CHAPTER 5
WORK, ENERGY AND POWER
Example 5.13 : An object of mass 2.0 kg moves at a constant speed of 5.0 m s1 up a plane inclined at 30 to the horizontal. The constant frictional force acting on the object is 4.0 N. Determine a. the rate of work done against the gravitational force, b. the rate of work done against the frictional force, c. the power supplied to the object. (Given g = 9.81 m s2 ) 1 Solution : m 2.0 kg; v 5.0 m s constant; f 4.0 N
N
v
s
y
mg sin 30 x f 30
30
mg cos 30 W mg
64
CHAPTER 5
WORK, ENERGY AND POWER
Solution : m 2.0 kg; v 5.0 m s 1 constant; f 4.0 N a. the rate of work done against the gravitational force is given by
Wg t Wg
mg sin 30 s cos θ and θ 180
t
s s and v mg sin 30 t t t Wg mg sin 30 v t Wg 2.09.81sin 30 5.0 t Wg OR Fg v cos θ t Wg mg sin 30 v cos180 t
65
CHAPTER 5
WORK, ENERGY AND POWER
Solution : m 2.0 kg; v 5.0 m s 1 constant; f b. The rate of work done against the frictional force is
W f t
fv cos θ
and θ
4.0 N
180
c. The power supplied to the object, Psupplied = the power lost against gravitational and frictional forces, Plost
Psupplied
Wg t
W f t
66
CHAPTER 5
WORK, ENERGY AND POWER
5.3.2 Mechanical efficiency,
Efficiency is a measure of the performance of a machines, engine and etc... The efficiency of a machine is defined as the ratio of the useful (output) work done to the energy input. is a dimensionless quantity (no unit). Equations:
Wout η 100% Ein OR
Pout η 100% Pin where Pout : power produced by the system Pin : power supplied to a system
67
CHAPTER 5
WORK, ENERGY AND POWER
Notes :
In practice, Pout< Pin hence < 100%. The system loses energy to its surrounding because it may have encountered resistances such as surface friction or air resistance. The energy which is dissipated to the surroundings, may be in the form of heat or sound.
Example 5.14 : A 1.0 kW motor is used to lift an object of mass 10 kg vertically upwards at a constant speed. The efficiency of the motor is 75 %. Determine a. the rate of heat dissipated to the surrounding. b. the vertical distance travelled by the object in 5.0 s. (Given g = 9.81 m s2 ) 68
CHAPTER 5
WORK, ENERGY AND POWER
Solution : m 10.0 kg; η 75%; Pin 1000 a. The output power of the motor is given by
W
Pout η 100 % Pin Pout 75 100 1000
Therefore the rate of heat dissipated to the surrounding is
Rate of heat dissipated Pin Pout 1000 750
b.
Pout Fv cos θ where θ 0 and F mg Pout mgv cos 0 Since the speed is constant hence the vertical distance in 5.0 s h is v
t
69
CHAPTER 5 Exercise 5.4 :
WORK, ENERGY AND POWER
Use gravitational acceleration, g = 9.81 m s2 1. A person of mass 50 kg runs 200 m up a straight road inclined at an angle of 20 in 50 s. Neglect friction and air resistance. Determine a. the work done, b. the average power of the person. ANS. : 3.36104 J; 672 W 2. Electrical power of 2.0 kW is delivered to a motor, which has an efficiency of 85 %. The motor is used to lift a block of mass 80 kg. Calculate a. the power produced by the motor. b. the constant speed at which the block being lifted vertically upwards by the force produced by the motor. (neglect air resistance) ANS. : 1.7 kW; 2.17 m s1 70
CHAPTER 5 Exercise 5.4 :
WORK, ENERGY AND POWER
3.
10
1
Figure 5.21
A car of mass 1500 kg moves at a constant speed v up a road with an inclination of 1 in 10 as shown in Figure 5.21. All resistances against the motion of the car can be neglected. If the engine car supplies a power of 12.5 kW, calculate the speed v. ANS. : 8.50 m s1
71
PHYSICS
CHAPTER 6
CHAPTER 6: Circular motion (3 Hours)
1
PHYSICS CHAPTER 6 Learning Outcome: 6.1 Uniform circular motion (1 hour) At the end of this chapter, students should be able to: Describe graphically the uniform circular motion. In terms of velocity with constant magnitude (only the direction of the velocity changes).
2
PHYSICS
CHAPTER 6
6.1 Uniform circular motion
is defined as a motion in a circle (circular arc) at a constant speed. Consider an object which does move with uniform circular motion as shown in Figure 6.1.
r θ O
s
The length of a circular arc, s is given by
s rθ
where
θ : angle which the arc subtends to the centre of the circle in radian r : radius of the circular path Figure 6.1
3
PHYSICS 6.1.1
CHAPTER 6
Linear (tangential) velocity , v
It is directed tangentially to the circular path and always perpendicular to the radius of the circular path as shown in Figure 6.2.
v
v
r
r O
r Figure 6.2
v
In uniform circular motion, the magnitude of the linear velocity (speed) of an object is constant but the direction is continually changing. The unit of the tangential (linear) velocity is m s1. 4
PHYSICS
CHAPTER 6
The linear velocity, v is difficult to measure but we can measure the period, T of an object in circular motion.
Period, T is defined as the time taken for one complete revolution (cycle/rotation). The unit of the period is second (s).
Frequency, f is defined as the number of revolutions (cycles/rotations) completed in one second. The unit of the frequency is hertz (Hz) or s1. Equation : 1
f
T
Let the object makes one complete revolution in circular motion, thus the distance travelled is 2r (circumference of the circle),
the time interval is one period, T. 5
PHYSICS
CHAPTER 6
From the definition of speed,
changeof distance v time interval 2r v T If
2 ω 2f T
OR
v 2rf
therefore
v rω
where
ω : angular ve locity (angular frequency) r : radius of the circular path
Note: The unit of angular velocity (angular frequency) is rad s1 (radian per second).
Unit conversion of angle, :
rad 180 2 rad 360 6
PHYSICS CHAPTER 6 Learning Outcome: 6.2 Centripetal force (2 hours) At the end of this chapter, students should be able to: Define and use centripetal acceleration and centripetal acceleration,
use
2
v ac r
Define and solve problem on centripetal force,
mv 2 Fc r 7
PHYSICS 6.2.1
CHAPTER 6
Centripetal (radial) acceleration, ac or ar
Figure 6.3 shows a particle moving with constant speed in a circular path of radius, r with centre at O. The particle moves from A to B in a time, t.
v1
The arc length AB is given by
v2
Δs rΔ Δs Δ r
(1)
The velocities of the particle at A and B are v1 and v2 respectively where
v1 v2 v
Figure 6.3
8
PHYSICS
CHAPTER 6
Let PQ and PR represent the velocity vectors v1 and v2 respectively, as shown in Figure 6.4. P
Figure 6.4
v1
v2
Q
Δv v2 v1
R Then QR represent the change in velocity vector v of the particle in time interval t. Since the angle between PQ and PR is small hence
QR PQ Δv vΔ Δv Δ v
(2)
By equating (1) and (2) then
Δs Δv r v
9
PHYSICS
CHAPTER 6
Dividing by time, t, thus
1 Δs 1 Δv r Δt v Δt v a r v v2 2 OR ac r v ac r where
ac : centripeta l accelerati on v : linear(tangential) velocity r : radius of circular path
ω : angular ve locity (angular frequency) 10
PHYSICS
CHAPTER 6
The centripetal acceleration is defined as the acceleration of an object moving in circular path whose direction is towards the centre of the circular path and whose magnitude is equal to the square of the speed divided by the radius. The direction of centripetal (radial) acceleration is always directed toward the centre of the circle and perpendicular to the linear (tangential) velocity as shown in Figure 6.5.
ac ac ac
ac ac ac
Figure 6.5
11
PHYSICS
CHAPTER 6
For uniform circular motion, the magnitude of the centripetal acceleration always constant but its direction continuously changes as the object moves around the circular path. 2r Because of
v
T
therefore we can obtain the alternative expression of centripetal acceleration is 2 r 2 ac T
r
4 2 r ac 2 T
12
PHYSICS
CHAPTER 6
Example 6.1 : A motorbike moving at a constant speed 20.0 m s1 in a circular track of radius 25.0 m. Calculate a. the centripetal acceleration of the motorbike, b. the time taken for the motorbike to complete one revolution. 1 Solution : v 20.0 m s ; r 25.0 m a. From the definition of the centripetal acceleration, thus
v2 ac r
ac
2 20.0
25.0
b. From the alternate formula of the centripetal acceleration, hence
4 2 r ac 2 T
4 2 25.0 16.0 T2
OR
2r v T
13
PHYSICS
CHAPTER 6
Example 6.2 :
Figure 6.6
A car initially travelling eastward turns north by travelling in a circular path at uniform speed as shown in Figure 6.6. The length of the arc ABC is 235 m and the car completes the turn in 36.0 s. Determine a. the acceleration when the car is at B located at an angle of 35.0, b. the car’s speed, c. its average acceleration during the 36.0 s interval. 14
PHYSICS
CHAPTER 6
Solution : s ABC 235 m, t 36.0 s a. The period of the car is given by
T 4t 436.0 T 144 s
The radius of the circular path is
s ABC rθ π 235 r 2
Therefore the magnitude of the centripetal acceleration is
4 2 r ac 2 T
4π 2 150 ac 1442
15
PHYSICS
CHAPTER 6
Solution : s ABC 235 m, t 36.0 s b. From the definition of the speed, thus
s s ABC v t t 235 v 36.0
c. 1st method : By using the triangle method for vector addition, thus the change in the velocity isgiven by
vA θ v vC v A θ
vC
v
v
vC 2 v A 2 6.532 6.532
16
PHYSICS
CHAPTER 6
Solution : s ABC 235 m, t 36.0 s Therefore the magnitude of the average acceleration is
v aav t 9.24 aav 36.0
1 vC
and its direction : θ tan v A 1 6.53 θ tan 6.53
17
PHYSICS
CHAPTER 6
Solution : s ABC 235 m, t 36.0 s c. 2nd method : v x vCx x-component : aav
x
aav x
y-component :
t 0 6.53 36.0
aav y
aav y
v y
v Ax t
vCy v Ay
t t 6.53 0 36.0
18
PHYSICS
CHAPTER 6
Solution : s ABC 235 m, t 36.0 s Therefore the magnitude of the average acceleration is
and
aav
aav
0.181 0.181
2 aav x
2 aav y
2
2
aav y θ tan a av x 1 0.181 θ tan 0.181 1
19
PHYSICS
CHAPTER 6
Example 6.3 : A boy whirls a marble in a horizontal circle of radius 2.00 m and at height 1.65 m above the ground. The string breaks and the marble flies off horizontally and strikes the ground after traveling a horizontal distance of 13.0 m. Calculate a. the speed of the marble in the circular path, b. the centripetal acceleration of the marble while in the circular motion. (Given g = 9.81 m s-2) Solution :
u
r =2.00 m
u 1.65 m
1.65 m Before
13.0 m After 20
PHYSICS
CHAPTER 6
Solution : a. From the diagram :
u x u; u y 0 sx 13.0 m ; s y 1.65 m
The time taken for the marble to strike the ground is
1 2 s y u y t gt 2 1 1.65 0 9.81t 2 2 The initial speed of the marble after the string breaks is equal to the tangential speed of the marble in the horizontal circle. Therefore s u t
13.0 u0.580 x
x
21
PHYSICS
CHAPTER 6
Solution : b. From the definition of the centripetal acceleration, thus 2
2
v u ac r r 2 22.4 ac 2.00
22
PHYSICS
CHAPTER 6
6.3 Centripetal force 6.3.1 Equation of centripetal force
From Newton’s second law of motion, a force must be associated with the centripetal acceleration. This force is known as the centripetal force and is given by
F Fnett ma
where a ac and F Fc 2 v 2 and ac r v Fc mac r
mv2 Fc mr 2 mv r where
Fc : centripeta l force 25
PHYSICS
CHAPTER 6
The centripetal force is defined as a force acting on a body causing it to move in a circular path of magnitude
mv 2 Fc r
and its always directed towards the centre of the circular path. Its direction is in the same direction of the centripetal acceleration as shown in Figure 6.8.
v
ac
Fc Fc
Figure 6.8
ac
v
Fc ac
v
26
PHYSICS
CHAPTER 6
27
PHYSICS
CHAPTER 6
If the centripetal force suddenly stops to act on a body in the circular motion, the body flies off in a straight line with the constant tangential (linear) speed as show in Figure 6.9.
Fc v ac
ac a c Fc v
v
Fc
Simulation 6.1
F c 0 ac 0
F c 0 ac 0
Note : v v In uniform circular motion, the nett force on the system is centripetal force. The work done by the centripetal force is zero but the kinetic energy of the body is not zero and given by Figure 6.9
1 2 1 2 2 K mv mr 2 2 28
PHYSICS
As a car makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion.
CHAPTER 6
As a bucket of water is tied to a string and spun in a circle, the tension force acting upon the bucket provides the centripetal force required for circular motion.
As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion
29
PHYSICS
CHAPTER 6
Without a centripetal force, an object in motion continues along a straight-line path.
With a centripetal force, an object in motion will be accelerated and change its direction.
30
PHYSICS
CHAPTER 6
Note that the centripetal force is proportional to the square of the velocity, implying that a doubling of speed will require four times the centripetal force to keep the motion in a circle. If the centripetal force must be provided by friction alone on a curve, an increase in speed could lead to an unexpected skid if friction is insufficient. 31
PHYSICS
CHAPTER 6
32
PHYSICS
CHAPTER 6
6.3.2 Examples of uniform circular motion Conical Pendulum
Example 6.4 : Figure 6.10 shows a conical pendulum with a bob of mass 80.0 kg on a 10.0 m long string making an angle of 5.00 to the vertical. a. Sketch a free body diagram of the bob. b. Determine i. the tension in the string, ii. the speed and the period of the bob, iii. the radial acceleration of the bob. (Given g =9.81 m s2) Figure 6.10 33
PHYSICS
CHAPTER 6
Solution : m 80.0 kg; l 10.0 m; θ a. The free body diagram of the bob :
5.00
T θ T cos θ ac
T sin θ mg b. i. From the diagram,
F
y
0
T cos θ mg 34
PHYSICS Solution : b. ii.
CHAPTER 6 m 80.0 kg; l 10.0 m; θ 5.00 The centripetal force is contributed by the horizontal component of the tension. F F
r sin θ l
l
r l sin θ
r
x
c
mv 2 T sin θ r2 mv T sin θ l sin θ Tl sin 2 θ v m
v
788 10.0 sin 5.00
2
80.0 35
PHYSICS
CHAPTER 6
Solution : m 80.0 kg; l 10.0 m; θ 5.00 b. ii. and the period of the bob is given by
2r v T 2l sin θ v T
2 10.0 sin 5.00 0.865 T
iii. From the definition of the radial acceleration, hence
v2 ar r
v2 ar l sin θ 2 0.865 ar 10.0 sin 5.00
36
PHYSICS
CHAPTER 6
Motion rounds a curve on a flat (unbanked) track (for car, motorcycle, bicycle, etc…) Picture 6.1
Example 6.5 :
A car of mass 2000 kg rounds a circular turn of radius 20 m. The road is flat and the coefficient of friction between tires and the road is 0.70. a. Sketch a free body diagram of the car. b. Determine the maximum car’s speed without skidding. (Given g = 9.81 m s-2) Solution : m 2000 kg; r 20 m; μ 0.70 a. The free body diagram of the car :
ac
Centre of circle
f
N
mg 37
PHYSICS
CHAPTER 6
Solution : m 2000 kg; r 20 m; μ 0.70 b. From the diagram in (a), N mg y-component : Fy 0
x-component : The centripetal force is provided by the frictional force between the wheel (4 tyres) and the road. Therefore mv 2
F
x
r2 mv f r2 mv μmg r v μrg
38
PHYSICS
CHAPTER 6
Motion in a horizontal circle
Example 6.6 : A ball of mass 150 g is attached to one end of a string 1.10 m long. The ball makes 2.00 revolution per second in a horizontal circle. a. Sketch the free body diagram for the ball. b. Determine i. the centripetal acceleration of the ball, ii. the magnitude of the tension in the string. Solution : m 0.150 kg; l r 1.10 m; f 2.00 Hz a. The free body diagram for the ball :
ac
T
r
mg 39
PHYSICS
CHAPTER 6
Solution : m 0.150 kg; l r 1.10 m; b. i. The linear speed of the ball is given by
f 2.00 Hz
2r v 2rf T v 21.10 2.00
Therefore the centripetal acceleration is
13.8
2
2
v ac r
ac
1.10
ii. From the diagram in (a), the centripetal force enables the ball to move in a circle is provided by the tension in the string. Hence Fx Fc mac
T ma c
40
PHYSICS
CHAPTER 6
Motion in a vertical circle
B
Example 6.7 :
v 3.00 m
v Figure 6.12
A A small remote control car with mass 1.20 kg moves at a constant speed of v = 15.0 m s1 in a vertical circle track of radius 3.00 m as shown in Figure 6.12. Determine the magnitude of the reaction force exerted on the car by the track at a. point A, b. point B. (Given g = 9.81 m s2) 41
PHYSICS
CHAPTER 6
Solution : m 1.20 kg; r 3.00 m; v 15.0 m s 1 a. The free body diagram of the car at point A :
ac
NA
mg
mv2 N A mg r 2 1.20 15.0 N A 1.20 9.81 3.00
mv 2 F r
42
PHYSICS
CHAPTER 6
Solution : m 1.20 kg; r 3.00 m; v 15.0 m s 1 b. The free body diagram of the car at point B :
NB mg ac
2 mv mv N B mg F r r 2 1.20 15.0 N B 1.20 9.81 3.00 2
43
PHYSICS
CHAPTER 6
Example 6.8 :
v
v Figure 6.13
A rider on a Ferris wheel moves in a vertical circle of radius, r = 8 m at constant speed, v as shown in Figure 6.13. If the time taken to makes one rotation is 10 s and the mass of the rider is 60 kg, Calculate the normal force exerted on the rider a. at the top of the circle, b. at the bottom of the circle. (Given g = 9.81 m s-2)
44
PHYSICS
CHAPTER 6
Solution : m 60 kg; r 8 m; T 10 s a. The constant speed of the rider is
2r v T
2π8 v 10
The free body diagram of the rider at the top of the circle :
Nt
ac
mg
mv 2 F r mv 2 mg N t r 2 60 5.03 60 9.81 N t 8
45
PHYSICS
CHAPTER 6
Solution : m 60 kg; r 8 m; T 10 s b. The free body diagram of the rider at the bottom of the circle :
mv 2 F r mv 2 N b mg r 2 60 5.03 N b 60 9.81 8
ac
Nb
mg
46
PHYSICS
CHAPTER 6
Example 6.9 : 3.0 m s1 A 3.0 m s1 D
Figure 6.14
E
3.0 m s1
A sphere of mass 5.0 kg is tied to an inelastic string. It moves in a vertical circle of radius 55 cm at a constant speed of 3.0 m s1 as shown in Figure 6.14. By the aid of the free body diagram, determine the tension in the string at points A, D and E. (Given g = 9.81 m s-2) 47
PHYSICS
CHAPTER 6
Solution : m 5.0 kg; r 0.55 m; v 3.0 m s 1 The free body diagram of the sphere at : Point A, A mv 2 mv2
TA
ac Point D,
mg
ac
TD
F
r
TA mg
r 2 5.03.0 TA 5.09.81 0.55
mv 2 TD r 2 D 5.0 3.0 TD 0.55 mg 48
PHYSICS
CHAPTER 6
Solution : m 5.0 kg; r 0.55 m; v 3.0 m s 1 The free body diagram of the sphere at : Point E,
ac
TE E
mv2 TE mg r 2 5.03.0 TE 5.09.81 0.55
mg
Caution : For vertical uniform circular motion only, the normal force or tension is maximum at the bottom of the circle. the normal force or tension is minimum at the top of the circle. 49
PHYSICS
CHAPTER 6
Exercise 6.2 : Use gravitational acceleration, g = 9.81 m s2 1. A cyclist goes around a curve of 50 m radius at a speed of 15 m s1. The road is banked at an angle to the horizontal and the cyclist travels at the right angle with the surface of the road. The mass of the bicycle and the cyclist together equals 95 kg. Calculate a. the magnitude of the centripetal acceleration of the cyclist, b. the magnitude of the normal force which the road exerts on the bicycle and the cyclist, c. the angle . ANS. : 4.5 m s2; 1.02 kN; 24.6
50
PHYSICS
CHAPTER 6
Exercise 6.2 : 2. A ball of mass 0.35 kg is attached to the end of a horizontal cord and is rotated in a circle of radius 1.0 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 80 N, determine a. the maximum speed of the ball, b. the minimum period of the ball. ANS. : 15.1 m s1; 0.416 s 3. A small mass, m is set on the surface of a sphere as shown in Figure 6.14. If the coefficient of static friction is s = 0.60, calculate the angle would the mass start sliding. ANS. : 31
m θ O
Figure 6.14
51
PHYSICS
CHAPTER 6
Exercise 6.2 : 4. A ball of mass 1.34 kg is connected by means of two massless string to a vertical rotating rod as shown in Figure 6.15. The strings are tied to the rod and are taut. The tension in the upper string is 35 N. a. Sketch a free body diagram for the ball. b. Calculate i. the magnitude of the tension in the lower string, ii. the nett force on the ball, iii. the speed of the ball. ANS. : 8.74 N; 37.9 N (radially inward); 6.45 m s1
Figure 6.15
52
PHYSICS
CHAPTER 6
THE END… Next Chapter… CHAPTER 7 : Gravitation
53
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PHYSICS
CHAPTER 7
CHAPTER 7: Gravitation (2 Hours)
1
PHYSICS
CHAPTER 7
In this chapter, we learns about 7.1 Gravitational force and field strength 7.2 Gravitational potential 7.3 Satellite motion in a circular orbit
2
PHYSICS
CHAPTER 7
7.1 Gravitational Force and Field Strength 7.1.1 Newton’s law of gravitation 7.1.2 Gravitational Field 7.1.3 Gravitational force and field strength
3
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PHYSICS CHAPTER 7 Learning Outcome: 7.1 Newton’s law of gravitation (1 hour) At the end of this chapter, students should be able to: State and use the Newton’s law of gravitation,
m1m2 Fg G 2 r
4
PHYSICS
CHAPTER 7
7.1.1 Newton’s law of gravitation
States that a magnitude of an attractive force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. OR mathematically,
Fg m1m2 and m1m2 Fg 2 r
1 Fg 2 r
m1m2 Fg G 2 r
where
Fg : Gravitatio nal force m1 , m2 : masses of particle 1 and 2 r : distance between particle 1 and 2
G : Universal gravitatio nal Constant 6.67 x10 11 N m 2 kg 2 5
PHYSICS
CHAPTER 7
The statement can also be shown by using the Figure 7.1.
m1
m2
F12
F21 r
Figure 7.1
m1m2 F21 F12 Fg G 2 r where
F21 : Gravitatio nal force by particle 2 on particle 1 F12 : Gravitatio nal force by particle 1 on particle 2 Simulation 7.1 6
PHYSICS
CHAPTER 7
Figures 7.2a and 7.2b show the gravitational force, Fg varies with the distance, r.
Fg
Fg
gradient Gm1m2
0
r 0
1 r2
Figure 7.2b Figure 7.2a Notes: Every spherical object with constant density can be reduced to a point mass at the centre of the sphere. The gravitational forces always attractive in nature and the forces always act along the line joining the two point masses. 7
PHYSICS
CHAPTER 7
Example 7.1 : A spaceship of mass 9000 kg travels from the Earth to the Moon along a line that passes through the Earth’s centre and the Moon’s centre. The average distance separating Earth and the Moon is 384,000 km. Determine the distance of the spaceship from the Earth at which the gravitational force due to the Earth twice the magnitude of the gravitational force due to the Moon. (Given the mass of the Earth, mE=6.001024 kg, the mass of the
Moon, mM=7.351022 kg and the universal gravitational constant,
G=6.671011 N m2 kg2)
8
PHYSICS
CHAPTER 7
Solution :
mE 6.00 10 24 kg; mM 7.35 10 22 kg; ms 900 0 kg; rEM 3.84 108 m
mE
m FEs s FMs rEM x x
mM
rEM
FEs 2FMs GmE ms GmM ms 2 2 2 x rEM x mE x2 2 rEM x 2mM x2 6.00 10 24 2 22 8 2 7.35 10 3.84 10 x Given
x 3.32 108 m
9
PHYSICS
CHAPTER 7
Example 7.2 :
C
50 g
6 cm 3.2 kg
A
8 cm
2.5 kg
B
Figure 7.3 Two spheres of masses 3.2 kg and 2.5 kg respectively are fixed at points A and B as shown in Figure 7.3. If a 50 g sphere is placed at point C, determine a. the resultant force acting on it.
b. the magnitude of the sphere’s acceleration. (Given G = 6.671011 N m2 kg2) 10
PHYSICS
CHAPTER 7 mA 3.2 kg; mB 2.5 kg; mC 50 10 3 kg rBC 6 10 2 m; rAC 10 10 2 m C θ FA sin θ 0.6 cos θ 0.8 2 10 10 m FB 2
Solution : a.
6 10
m
θ A
8 10 - 2 m
B
The magnitude of the forces on mC,
FA
GmA mC rAC
2
6.67 10 3.250 10 10 10 11
3
2 2
FA 1.07 10 9 N
11
PHYSICS
CHAPTER 7 mA 3.2 kg; mB 2.5 kg; mC 50 10 3 kg rBC 6 10 2 m; rAC 10 10 2 m
Solution :
FB
GmB mC rBC
2
6.67 10 2.550 10 6 10 11
3
2 2
FB 2.32 10 9 N Force
FA FB
x-component (N)
FA cos θ
y-component (N)
FA sin θ
1.07 10 0.8 8.56 10 10
1.07 10 9 0.6 6.42 10 10
0
FB 2.32 10 9
9
12
PHYSICS
CHAPTER 7
Solution :
F
Fx 8.56 10 10 N y
6.42 10 10 2.32 10 9 2.96 10 9 N
The magnitude of the nett force is
F F F 2
2
x
y
8.56 10 2.96 10 10 2
9 2
F 3.08 109 N and its direction is 9 F 2 .96 10 y 1 1 θ tan tan 10 Fx 8 .56 10 θ 73.9 (254 from +x axis anticlockwise) 13
PHYSICS
CHAPTER 7
Solution : b. By using the Newton’s second law of motion, thus
F m a 3.08 10 50 10 a C
9
3
3.08 10 9 a 50 10 3
a 6.16 10 8 m s 2 and the direction of the acceleration in the same direction of the nett force on the mC i.e. 254 from +x axis anticlockwise.
14
PHYSICS
CHAPTER 7
7.1.2 Gravitational Field
is defined as a region of space surrounding a body that has the property of mass where the attractive force is experienced if a test mass placed in the region. Field lines are used to show gravitational field around an object with mass. For spherical objects (such as the Earth) the field is radial as shown in Figure 7.4.
M
Figure 7.4
15
PHYSICS
CHAPTER 7
The gravitational field in small region near the Earth’s surface are uniform and can be drawn parallel to each other as shown in Figure 7.5.
Figure 7.5 The field lines indicate two things: The arrows – the direction of the field Note: The spacing – the strength of the field The gravitational field is a conservative field in which the work done New in moving a body from one point to another is independent of the path taken. 16
PHYSICS
CHAPTER 7
Exercise 7.1 : Given G = 6.671011 N m2 kg2 1. Four identical masses of 800 kg each are placed at the corners of a square whose side length is 10.0 cm. Determine the nett gravitational force on one of the masses, due to the other three. ANS. : 8.2103 N; 45 2. Three 5.0 kg spheres are located in the xy plane as shown in Figure 7.6.Calculate the magnitude of the nett gravitational force on the sphere at the origin due to the other two spheres. ANS. : 2.1108 N
Figure 7.6 17
PHYSICS
CHAPTER 7
Exercise 7.1 : 3.
Figure 7.7 In Figure 8.7, four spheres form the corners of a square whose side is 2.0 cm long. Calculate the magnitude and direction of the nett gravitational force on a central sphere with mass of m5 = 250 kg. ANS. : 1.68102 N; 45 18
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PHYSICS CHAPTER 7 Learning Outcome: 7.1.3 Gravitational force and field strength At the end of this chapter, students should be able to: Define gravitational field strength as gravitational force per unit mass, F
ag
g
m
Derive and use the equation for gravitational field strength.
M ag G 2 r
Sketch a graph of ag against r and explain the change in ag with altitude and depth from the surface of the earth.
19
PHYSICS
CHAPTER 7
Gravitational field strength, ag
7.1.3 is defined as the gravitational force per unit mass of a body (test mass) placed at a point. OR
Fg ag m
where
Fg : Gravitatio nal force ag : Gravitatio nal field strength
m : mass of a body (test mass)
It is a vector quantity. The S.I. unit of the gravitational field strength is N kg1 or m s2.
20
PHYSICS
CHAPTER 7
It is also known as gravitational acceleration (the free-fall acceleration). Its direction is in the same direction of the gravitational force. Another formula for the gravitational field strength at a point is given by
Fg
GMm and F ag g r2 m 1 GMm ag 2 m r GM ag 2 r
where
M : mass of the point mass r : distance between test mass and point mass 21
PHYSICS
CHAPTER 7
Figure 7.8 shows the direction of the gravitational field strength on a point S at distance r from the centre of the planet.
GM ag 2 r
M
Figure 7.8
r
22
PHYSICS
CHAPTER 7
The gravitational field in the small region near the Earth’s surface( r R) are uniform where its strength is 9.81 m s2 and its direction can be shown by using the Figure 7.9.
GM ag g 2 R
Figure 7.9 where
R : radius of the Earth g : gravitatio nal acceleration 9.81 m s 2 23
PHYSICS
CHAPTER 7
Example 7.3 : Determine the Earth’s gravitational field strength a. on the surface. b. at an altitude of 350 km. (Given G = 6.671011 N m2 kg2, mass of the Earth,
M = 6.00 1024 kg and radius of the Earth, R = 6.40 106 m) Solution : a.
g
R
r M
r R 6.40 10 6 m; ag g The gravitational field strength is
GM 6.67 10 11 6.00 10 24 g 2 6 2 R 6.40 10
g 9.77 N kg
1
2
OR 9.77 m s (Towards the centre of the Earth)
24
PHYSICS
CHAPTER 7
Solution : b.
r Rh
ag h
r
6.40 10 6 350 10 3 r 6.75 10 6 m
The gravitational field strength is given by
R M
GM ag 2 r
6.67 10 6.00 10 6.75 10 11
24
6 2
ag 8.78 m s 2
(Towards the centre of the Earth)
25
PHYSICS
CHAPTER 7
Example 7.4 : The gravitational field strength on the Earth’s surface is 9.81 N kg1. Calculate a. the gravitational field strength at a point C at distance 1.5R from the Earth’s surface where R is the radius of the Earth. b. the weight of a rock of mass 2.5 kg at point C. Solution : g 9.81 N kg 1 a. The gravitational field strength on the Earth’s surface is
GM g 2 9.81 N kg 1 R The distance of point C from the Earth’s centre is
r R 1.5R 2.5R
26
PHYSICS
CHAPTER 7
Solution : a. Thus the gravitational field strength at point C is given by
ag
GM rC
2
ag
GM
2.5R 2
1 GM 2 6.25 R 1 9.81 1.57 N kg 1 ag 6.25
b. Given m 2.5 kg The weight of the rock is
(Towards the centre of the Earth)
W ma g 2.51.57 W 3.93 N (Towards the centre of the Earth) 27
PHYSICS
CHAPTER 7
Example 7.5 :
5 km
B A
Figure 7.10
Figure 8.10 shows an object A at a distance of 5 km from the object B. The mass A is four times of the mass B. Determine the location of a point on the line joining both objects from B at which the nett gravitational field strength is zero.
28
PHYSICS
CHAPTER 7
Solution :r
5 10 3 m; M A 4M B A
a g1 C a g 2
rx
a
r
B
x
0 a g1 a g 2 GM A GM B 2 2 x r x 4M B MB 2 2 x 5 10 3 x
At point C,
g nett
x 1.67 10 3 m
29
PHYSICS
CHAPTER 7
7.1.4 Variation of gravitational field strength on the distance from the centre of the Earth Outside the Earth ( r > R) Figure 8.11 shows a test mass which is outside the Earth and at a distance r from the centre.
M
r
R
Figure 8.11 The gravitational field strength outside the Earth is
GM ag 2 r
1 ag 2 r 30
PHYSICS
CHAPTER 7
On the Earth ( r = R) Figure 7.12 shows a test mass on the Earth’s surface.
M
r
R Figure 7.12
The gravitational field strength on the Earth’s surface is
GM ag 2 g 9.81 m s 2 R 31
PHYSICS
CHAPTER 7
Inside the Earth ( r < R) Figure 7.13 shows a test mass which is inside the Earth and at distance r from the centre.
M M'
where
r
R
M ' : the mass of sphericalportion of the Earth of radius, r
Figure 7.13
The gravitational field strength inside the Earth is given by
GM ' ag 2 r 32
PHYSICS
CHAPTER 7
By assuming the Earth is a solid sphere and constant density, hence
M ' V ' M V
M ' 43 r 3 r3 4 3 3 M 3 R R 3 r M ' 3 M R
Therefore the gravitational field strength inside the Earth is
r3 G 3 M R ag r2 GM ag 3 r R
ag r 33
PHYSICS
CHAPTER 7
The variation of gravitational field strength, ag as a function of distance from the centre of the Earth, r is shown in Figure 7.14.
R
ag GM ag 2 g R
ag r 0
R
1 ag 2 r r Figure 7.14
34
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PHYSICS CHAPTER 7 Learning Outcome: 7.2 Gravitational potential (½ hour) At the end of this chapter, students should be able to: Define gravitational potential in a gravitational field. Derive and use the formulae,
GM V r
Sketch the variation of gravitational potential, V with distance, r from the centre of the earth.
35
PHYSICS 7.2
CHAPTER 7 Gravitational potential
7.2.1 Work done by the external force Consider an external force, F F m dr is required to bring a test mass, m from r1 to r2 , F as shown in Figure 7.18.
g
At the distance r2 from the centre of the Earth,
r1 r 2
F Fg
The work done by the external force through the small displacement
dr is
dW Fdr cos 0 dW Fg dr
M
Figure 7.18
36
PHYSICS
CHAPTER 7
Therefore the work done by the external force to bring test mass, m from r1 to r2 is
GMm and F dW F dr g g 2 r1 r r2 GMm W dr 2 r1 r r2 1 W GMm 2 dr r1 r r 1 W GMm r r r2
2
1
where
Wr1 r2
1 1 r : initial distance GMm 1 r : final distance r r 2 1 2 37
PHYSICS 7.2.2
CHAPTER 7 Gravitational potential, V
at a point is defined as the work done by an external force in bringing a test mass from infinity to a point per unit the test mass. OR mathematically, V is written as:
W V m
where
m : mass of the test mass
V : gravitatio nal potential at a point
W : work done in bringing a test mass from infinity t o a point
It is a scalar quantity. Its dimension is given by
W V m
ML2T 2 V M
V L T 2
2
38
PHYSICS
CHAPTER 7
The S.I unit for gravitational potential is m2 s2 or J kg1. Another formula for the gravitational potential at a point is given by
1 1 W V and W GMm m r1 r2 GMm 1 1 where r1 and r2 r V m r1 r2 GMm 1 1 V m r
GM where V r : distance between the point r and the point mass, M 39
PHYSICS
CHAPTER 7
The gravitational potential difference between point A and B (VAB) in the Earth’s gravitational field is defined as the work done in bringing a test mass from point B to point A per unit the test mass. OR mathematically, VAB is written as:
WBA VAB VA -VB m where
WBA : work done in bringing the test mass from point B to point A.
VA : gravitatio nal potential at point A VB : gravitatio nal potential at point B 40
PHYSICS
CHAPTER 7
Figure 7.19 shows two points A and B at a distance rA and rB from the centre of the Earth respectively in the Earth’s gravitational field.
A
rA
B
rB M
Figure 7.19
The gravitational potential difference between the points A and B is given by
VAB VA VB GM GM VAB rA rB 1 1 VAB GM rB rA 41
PHYSICS
CHAPTER 7
The gravitational potential difference between point B and A in the Earth’s gravitational field is given by
VBA
WAB VB VA m
The variation of gravitational potential, V when the test mass, m move away from the Earth’s surface is illustrated by the graph in Figure 7.20.
V
0
GM R
R
r
1 V r
Figure 7.20 Note: The Gravitational potential at infinity is zero. V 0
42
PHYSICS
CHAPTER 7
Example 7.7 : When in orbit, a satellite attracts the Earth with a force of 19 kN and the satellite’s gravitational potential due to the Earth is 5.45107 J kg1. a. Calculate the satellite’s distance from the Earth’s surface. b. Determine the satellite’s mass. (Given G = 6.671011 N m2 kg2, mass of the Earth,
M = 5.981024 kg and radius of the Earth , R = 6.38106 m) Solution : Fg
19 103 N; V 5.45 10 7 J kg 1
Fg
h
r
R 43
PHYSICS
CHAPTER 7
Solution : F 19 10 3 N; V 5.45 10 7 J kg 1 g a. By using the formulae of gravitational potential, thus
GM V r
5.45 10 7
6.67 10 5.98 10 11
r 7.32 10 6 m
24
r
Therefore the satellite’s distance from the Earth’s surface is
r hR 7.32 10 6 h 6.38 10 6 h 9.4 10 5 m
44
PHYSICS
CHAPTER 7
Solution : F 19 10 3 N; V 5.45 10 7 g b. From the Newton’s law of gravitation, hence
J kg 1
GMm Fg 2 r 19 10 3
6.67 10 5.98 10 m 7.32 10 11
24
6 2
m 2552 kg
45
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PHYSICS CHAPTER 7 Learning Outcome: 7.3 Satellite motion in a circular orbit (½ hour) At the end of this chapter, students should be able to: Explain satellite motion with: velocity,
GM v r
period,
r3 T 2 GM
46
PHYSICS
CHAPTER 7
7.3
Satellite motion in a circular orbit
7.3.1 Tangential (linear/orbital) velocity, v Consider a satellite of mass, m travelling around the Earth of mass, M, radius, R, in a circular orbit of radius, r with constant tangential (orbital) speed, v as shown in Figure 7.22.
Figure 7.22 47
PHYSICS
CHAPTER 7
The centripetal force, Fc is contributed by the gravitational force of attraction, Fg exerted on the satellite by the Earth.
Fg Fc mac GMm mv2 2 r r
Hence the tangential velocity, v is given by
GM v r where r : distance of the satellite from the centre of the Earth M : mass of the Earth G : universal gravitatio nal constant 48
PHYSICS
CHAPTER 7
For a satellite close to the Earth’s surface, 2 and
rR
Therefore
GM gR
v gR
The relationship between tangential velocity and angular velocity is 2 r
v r
T
Hence , the period, T of the satellite orbits around the Earth is given by
2r GM T r
3
r T 2 GM 49
PHYSICS 7.3.2
CHAPTER 7 Synchronous (Geostationary) Satellite
Figure 8.23 shows a synchronous (geostationary) satellite which stays above the same point on the equator of the Earth.
Figure 8.23 The satellite have the following characteristics: It revolves in the same direction as the Earth. It rotates with the same period of rotation as that of the Earth (24 hours). It moves directly above the equator. The centre of a synchronous satellite orbit is at the centre of the Earth. It is used as a communication satellite. 50
PHYSICS
CHAPTER 7
Example 7.12 : The weight of a satellite in a circular orbit round the Earth is half of its weight on the surface of the Earth. If the mass of the satellite is 800 kg, determine a. the altitude of the satellite, b. the speed of the satellite in the orbit, (Given G = 6.671011 N m2 kg2, mass of the Earth,
M = 6.001024 kg, and radius of the Earth , R = 6.40106 m)
51
PHYSICS
CHAPTER 7
Solution : a. The satellite orbits the Earth in the circular path, thus
b. The speed of the satellite is given by
52
PHYSICS
CHAPTER 7
Example 7.13 : The radius of the Moon’s orbit around the Earth is 3.8 108 m and the period of the orbit is 27.3 days. The masses of the Earth and Moon are 6.0 1024 kg and 7.4 1022 kg respectively. Calculate the total energy of the Moon in the orbit. 6 2 Solution : r 8.50 10 m; m 120 kg; g 9.50 m s The period of the satellite is
T 3.53600 T 12600 s The tangential speed of the satellite is 2r v T 2 8.50 106 v 12600 3 1 v 4.24 10 m s
53
PHYSICS
CHAPTER 7
Solution : r 8.50 10 m; m 120 kg; g 9.50 m s A satellite orbits the planet in the circular path, thus 6
2
Fg Fc 2 GMm mv 2 r r GM 2 2 and GM gR v r 2 gR 2 v r 2 9.50 R 3 2 4.24 10 8.50 106 R 4.01106 m
54
PHYSICS
CHAPTER 7
Exercise 7.2 : Given G = 6.671011 N m2 kg2 1. A rocket is launched vertically from the surface of the Earth at speed 25 km s-1. Determine its speed when it escapes from the gravitational field of the Earth. (Given g on the Earth = 9.81 m s2, radius of the Earth , R = 6.38 106 m) ANS. : 2.24104 m s1 2. A satellite revolves round the Earth in a circular orbit whose radius is five times that of the radius of the Earth. The gravitational field strength at the surface of the Earth is 9.81 N kg1. Determine a. the tangential speed of the satellite in the orbit, b. the angular frequency of the satellite. (Given radius of the Earth , R = 6.38 106 m) ANS. : 3538 m s1 ; 1.11104 rad s1
55
PHYSICS
CHAPTER 7
Exercise 7.2 : 3. A geostationary satellite of mass 2400 kg is placed 35.92 Mm from the Earth’s surface orbits the Earth along a circular path. Determine a. the angular velocity of the satellite, b. the tangential speed of the satellite, c. the acceleration of the satellite, d. the force of attraction between the Earth and the satellite, e. the mass of the Earth. (Given radius of the Earth , R = 6.38 106 m) ANS. : 7.27105 rad s1; 3.08103 m s1; 0.224 m s2; 537 N ; 6.001024 kg
56
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PHYSICS
CHAPTER 7
THE END… Next Chapter… CHAPTER 8 : Simple Harmonic Motion
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PHYSICS
CHAPTER 8
CHAPTER 8: Rotational of rigid body (8 Hours)
1
PHYSICS CHAPTER 8 Learning Outcome: http//:kms.matrik.edu.my/
8.1 Rotational Kinematics (2 hour) At the end of this chapter, students should be able to: a) Define and describe: angular displacement (θ) average angular velocity (ωav)
instantaneous angular velocity (ω) average angular acceleration (αav)
instantaneous angular acceleration (α). b) Relate parameters in rotational motion with their corresponding quantities in linear motion. Write and use :
2
v s=rθ ; v=rω ; a t = rα ; a c =rω 2 = r
c) Use equations for rotational motion with constant angular acceleration.
ω=ω0 αt
1 2 θ=ω 0 t αt 2
2
ω =ω22 2 αθ 0
PHYSICS
CHAPTER 8
8.1 Parameters in rotational motion 8.1.1 Angular displacement,θ
is defined as an angle through which a point or line has been rotated in a specified direction about a specified axis. The S.I. unit of the angular displacement is radian (rad). (rad) Figure 8.1 shows a point P on a rotating compact disc (CD) moves through an arc length s on a circular path of radius r about a fixed axis through point O.
Figure 8.1 3
PHYSICS
CHAPTER 8
From Figure 8.1, thus
s θ= r where
OR
s=rθ
θ : angle angular displacement in radian s: arc length r : radius of the circle
Others unit for angular displacement is degree (°) and revolution (rev). (rev) Conversion factor :
1 rev=2π rad=360
°
Sign convention of angular displacement : Positive – if the rotational motion is anticlockwise. anticlockwise Negative – if the rotational motion is clockwise. clockwise 4
PHYSICS
CHAPTER 8
8.1.2 Angular velocity Average angular velocity, ωav
is defined as the rate of change of angular displacement. displacement Equation :
θ 2−θ1 Δθ ω av = = t 2−t 1 Δt
where
θ 2 : final angular displacement in radian θ1 : initial angular displacement in radian
Δt : time interval Instantaneous angular velocity, ω
is defined as the instantaneous rate of change of angular displacement. displacement Equation : Δθ dθ
ω=limit Δt 0
Δt
=
dt 5
PHYSICS
CHAPTER 8
It is a vector quantity. quantity The unit of angular velocity is radian per second (rad s-1) Others unit is revolution per minute (rev min−1 or rpm) Conversion factor:
2π −1 π −1 1 rpm = rad s = rad s 60 30 Note : Every part of a rotating rigid body has the same angular velocity. velocity Direction of the angular velocity Its direction can be determine by using right hand grip rule where
Thumb Curl fingers
: direction of angular velocity : direction of rotation 6
PHYSICS
CHAPTER 8
Figures 8.2 and 8.3 show the right hand grip rule for determining the direction of the angular velocity.
ω
Figure 8.2
ω Figure 8.3 7
PHYSICS
CHAPTER 8
Example 8.1 : The angular displacement,θ of the wheel is given by 2
θ=5t −t
where θ in radians and t in seconds. The diameter of the wheel is 0.56 m. Determine a. the angle, θ in degree, at time 2.2 s and 4.8 s, b. the distance that a particle on the rim moves during that time interval, c. the average angular velocity, in rad s−1 and in rev min−1 (rpm), between 2.2 s and 4.8 s, d. the instantaneous angular velocity at time 3.0 s.
8
PHYSICS Solution :
CHAPTER 8 d 0 .56 r= = =0 . 28 m 2 2
a. At time, t1 =2.2 s :
2
θ1 =5 2. 2 − 2. 2
θ1 =22 rad
At time, t2 =4.8 s :
2
θ 2 =5 4 . 8 − 4 . 8
θ 2 =110 rad
9
PHYSICS Solution :
CHAPTER 8 d 0 .56 r= = =0 . 28 m 2 2
b. By applying the equation of arc length,
s=rθ
Therefore
Δs=rΔθ=r θ 2 −θ1
Δs=0.28 110−22 c. The average angular velocity in rad s−1 is given by
Δθ θ 2 −θ 1 ω av = = Δt t 2 −t 1
ω av =
110−22
4 . 8−2. 2 10
PHYSICS
CHAPTER 8
Solution : c. and the average angular velocity in rev min−1 is
33 . 9 rad 1 rev 60 s ω av = 1s 2π rad 1 min
d. The instantaneous angular velocity as a function of time is
dθ ω= dt d 2 ω= 5t −t dt ω=10 t −1
At time, t =3.0 s :
ω=10 3.0 −1 11
PHYSICS
CHAPTER 8
Example 8.2 : A diver makes 2.5 revolutions on the way down from a 10 m high platform to the water. Assuming zero initial vertical velocity, calculate the diver’s average angular (rotational) velocity during a dive. (Given g = 9.81 m s−2) Solution :
u y =0 θ 0 =0
10 m
θ1 =2. 5 rev
water
12
PHYSICS
CHAPTER 8
Solution : θ1 =2. 5×2π=5π rad From the diagram, s =−10 m y Thus 1 2
s y=u y t− gt 2 1 2 −10=0− 9 . 81 t 2
Therefore the diver’s average angular velocity is
ω av =
θ1 −θ 0 t
5π−0 ω av = 1 . 43 13
PHYSICS
CHAPTER 8
8.1.3 Angular acceleration Average angular acceleration, αav
is defined as the rate of change of angular velocity. velocity Equation : ω −ω
Δω α av = = t 2 −t 1 Δt where ω 2 : final angular velocity ω 1 : initial angular velocity 2
1
Δt : time interval Instantaneous angular acceleration, α
is defined as the instantaneous rate of change of angular velocity. velocity Equation : Δω dω
α=limit Δt 0
Δt
=
dt 14
PHYSICS
CHAPTER 8
It is a vector quantity. quantity The unit of angular acceleration is rad s−2. Note:
If the angular acceleration, α is positive, positive then the angular velocity, ω is increasing. increasing
If the angular acceleration, α is negative, negative then the angular velocity, ω is decreasing. decreasing Direction of the angular acceleration
If the rotation is speeding up, up α and ω in the same direction as shown in Figure 8.4.
ω Figure 8.4
α
15
PHYSICS
CHAPTER 8
If the rotation is slowing down, down α and ω have the opposite direction as shown in Figure 8.5.
α
ω Figure 8.5
Example 8.3 : The instantaneous angular velocity, ω of the flywheel is given by 3 2
ω=8t −t
where ω in radian per second and t in seconds. Determine a. the average angular acceleration between 2.2 s and 4.8 s, b. the instantaneous angular acceleration at time, 3.0 s. 16
PHYSICS
CHAPTER 8
Solution : a. At time, t1 =2.2 s :
3
2
ω 1 =8 2 . 2 − 2 . 2 −1
ω 1 =80 . 3 rad s
At time, t2 =4.8 s :
3
2
ω 2 =8 4 . 8 − 4 . 8
Therefore the average angular acceleration is
α av =
ω 2 −ω 1 t 2 −t 1
862−80 .3 α av = 4 . 8−2. 2 17
PHYSICS
CHAPTER 8
Solution : b. The instantaneous angular acceleration as a function of time is
dω α= dt d 3 2 α= 8t −t dt
At time, t =3.0 s :
2
α=24 3 . 0 −2 3 . 0
18
PHYSICS
CHAPTER 8
Exercise 8.1 : 1. If a disc 30 cm in diameter rolls 65 m along a straight line without slipping, calculate a. the number of revolutions would it makes in the process, b. the angular displacement would be through by a speck of gum on its rim. ANS. : 69 rev; 138π rad 2. During a certain period of time, the angular displacement of a swinging door is described by
θ=5.0010 .0t2.00 t
2
where θ is in radians and t is in seconds. Determine the angular displacement, angular speed and angular acceleration a. at time, t =0, b. at time, t =3.00 s. ANS. : 5.00 rad, 10.0 rad s−1, 4.00 rad s−2; 53.0 rad, 22.0 rad s−1, 4.00 rad s−2 19
PHYSICS
CHAPTER 8
8.1.2 Relationship between linear and rotational motion 8.1.2 Relationship between linear velocity, v and angular velocity, ω
When a rigid body is rotates about rotation axis O , every particle in the body moves in a circle as shown in the Figure 8.6.
y
v P r O Figure 8.6
θ
s x
20
PHYSICS
CHAPTER 8
Point P moves in a circle of radius r with the tangential velocity v where its magnitude is given by
ds v= dt
and
s=rθ
dθ v=r dt v=rω
The direction of the linear (tangential) velocity always tangent to the circular path. path Every particle on the rigid body has the same angular speed (magnitude of angular velocity) but the tangential speed is not the same because the radius of the circle, r is changing depend on the position of the particle. particle Simulation 7.1 21
PHYSICS
CHAPTER 8
8.1.2 Relationship between tangential acceleration, at and angular acceleration, α
If the rigid body is gaining the angular speed then the tangential velocity of a particle also increasing thus two component of acceleration are occurred as shown in Figure 8.7.
y
at a O
P
ac x
Figure 8.7 22
PHYSICS
CHAPTER 8
The components are tangential acceleration, at and centripetal acceleration, ac given by
but
dv and v=rω at= dt dω a t =rα a t =r dt 2 v 2 a c = =rω =vω r
The vector sum of centripetal and tangential acceleration of a particle in a rotating body is resultant (linear) acceleration, a given by Vector form t c
a =a a
and its magnitude,
∣a∣= a 2t a 2c
23
PHYSICS
CHAPTER 8
8.1.3 Rotational motion with uniform angular acceleration
Table 8.1 shows the symbols used in linear and rotational kinematics. Linear motion
Quantity
Rotational motion
s
Displacement
u
Initial velocity
θ ω0
v
Final velocity
ω
a
Acceleration
α
t
Time
t
Table 8.1 24
PHYSICS
CHAPTER 8
Table 8.2 shows the comparison of linear and rotational motion with constant acceleration. Linear motion
Rotational motion
a=constant
α =constant
v=uat
ω=ω0 αt
1 2 s=ut at 2
1 2 θ=ω 0 t αt 2
2
2
2
v =u 2 as 1 s= vu t 2 where θ in radian.
ω
2 =ω 0 2 αθ
1 θ= ωω 0 t 2 Table 8.2 25
PHYSICS
CHAPTER 8
Example 8.4 : A car is travelling with a velocity of 17.0 m s−1 on a straight horizontal highway. The wheels of the car has a radius of 48.0 cm. If the car then speeds up with an acceleration of 2.00 m s−2 for 5.00 s, calculate a. the number of revolutions of the wheels during this period, b. the angular speed of the wheels after 5.00 s. −1 −2 Solution : u=17.0 m s ,r=0.48 m ,a=2.00 m s ,t=5.00 s a. The initial angular velocity is
u=rω0 17.0=0.48 ω0
and the angular acceleration of the wheels is given by
a=rα
2 . 00=0 . 48 α 26
PHYSICS
CHAPTER 8 −1 −2 Solution : u=17.0 m s ,r=0.48 m ,a=2.00 m s ,t=5.00 s a. By applying the equation of rotational motion with constant angular acceleration, thus 1 2
θ=ω 0 t αt 2 1 2 θ= 35 . 4 5 . 00 4 . 17 5 . 00 2 θ=229 rad
therefore
b. The angular speed of the wheels after 5.00 s is
ω=ω0 αt
ω=35. 4 4.17 5.00 27
PHYSICS
CHAPTER 8
Example 8.5 : The wheels of a bicycle make 30 revolutions as the bicycle reduces its speed uniformly from 50.0 km h-1 to 35.0 km h-1. The wheels have a diameter of 70 cm. a. Calculate the angular acceleration. b. If the bicycle continues to decelerate at this rate, determine the time taken for the bicycle to stop. 0 .70 Solution : θ=30×2π=60π rad, r= =0 . 35 m ,
2
3
50. 0 km 10 m 1 h u= =13. 9 m s−1 , 1h 1 km 3600 s 3
35.0 km 10 m 1 h −1 v= =9. 72 m s 1h 1 km 3600 s 28
PHYSICS
CHAPTER 8
Solution : a. The initial angular speed of the wheels is
u=rω0 13.9=0.35 ω 0
and the final angular speed of the wheels is
v=rω
9 . 72=0 . 35 ω therefore 2 2 ω =ω 0 2 αθ 2 2 27 . 8 =39 . 7 2α 60π b. The car stops thus ω=0 and ω=ω0 αt Hence
ω 0 =27 . 8 rad s
0=27 .8 −2.13 t 29
−1
PHYSICS
CHAPTER 8
Example 8.6 : A blade of a ceiling fan has a radius of 0.400 m is rotating about a fixed axis with an initial angular velocity of 0.150 rev s -1. The angular acceleration of the blade is 0.750 rev s-2. Determine a. the angular velocity after 4.00 s, b. the number of revolutions for the blade turns in this time interval, c. the tangential speed of a point on the tip of the blade at time,
t =4.00 s, d. the magnitude of the resultant acceleration of a point on the tip −1 of the blade at t =4.00 s. r=0 . 400 m , ω 0 =0. 150×2π=0 .300π rad s , Solution : α=0.750×2π =1. 50π rad s−2
a. Given tω=ω =4.00 0s,αt thus
ω= 0.300π −11.50π 4 .00 ω=19.8 rad s 30
PHYSICS
CHAPTER 8
Solution : b. The number of revolutions of the blade is
1 2 θ=ω 0 t αt 2 1 2 θ= 0 . 300 π 4 . 00 1 .50 π 4 . 00 2 θ=41. 5 rad
c. The tangential speed of a point is given by
v=rω
v= 0.400 19.8 31
PHYSICS
CHAPTER 8
Solution : d. The magnitude of the resultant acceleration is
c
a= a 2 a 2 t
a=
2 2
v r
2
rα 2 2
a=
7 . 92
0 . 400
2
0 . 400×1. 50π
32
PHYSICS
CHAPTER 8
Example 8.7 : Calculate the angular velocity of a. the second-hand, b. the minute-hand and c. the hour-hand, of a clock. State in rad s-1. d. What is the angular acceleration in each case? Solution : a. The period of second-hand of the clock is
2π ω= T
T = 60 s, hence
2π ω= 60
33
PHYSICS
CHAPTER 8
Solution : b. The period of minute-hand of the clock is T = 60 min = 3600 s, hence
2π ω= 3600
c. The period of hour-hand of the clock is hence
2π ω= 4 . 32×10 4
T = 12 h = 4.32 ×104 s,
d. The angular acceleration in each cases is 34
PHYSICS
CHAPTER 8
Example 8.8 : A coin with a diameter of 2.40 cm is dropped on edge on a horizontal surface. The coin starts out with an initial angular speed of 18 rad s−1 and rolls in a straight line without slipping. If the rotation slows down with an angular acceleration of magnitude 1.90 rad s−2, calculate the distance travelled by the coin before coming to rest. −1 Solution : ω =18 rad s−1
ω=0 rad s
0
−2
α=−1.90 rad s
−2
d=2.40×10 m The radius of the coin is
s
d −2 r= =1 .20×10 m 2 35
PHYSICS
CHAPTER 8
Solution : The initial speed of the point at the edge the coin is
u=rω0 −2 u= 1.20×10 18
v=0 m s
−1
and the final speed is The linear acceleration of the point at the edge the coin is given by
a=rα
−2 a= 1.20×10 −1.90
Therefore the distance by the coin is 2 travelled 2
v =u 2 as 2 0= 0 .216 2 −2 . 28×10−2 s 36
PHYSICS
CHAPTER 8
Exercise 8.2 : 1. A disk 8.00 cm in radius rotates at a constant rate of 1200 rev min-1 about its central axis. Determine a. its angular speed, b. the tangential speed at a point 3.00 cm from its centre, c. the radial acceleration of a point on the rim, d. the total distance a point on the rim moves in 2.00 s. ANS. : 126 rad s−1; 3.77 m s−1; 1.26 × 103 m s−2; 20.1 m 2. A 0.35 m diameter grinding wheel rotates at 2500 rpm. Calculate a. its angular velocity in rad s−1, b. the linear speed and the radial acceleration of a point on the edge of the grinding wheel. ANS. : 262 rad s−1; 46 m s−1, 1.2 × 104 m s−2 37
PHYSICS
CHAPTER 8
Exercise 8.2 : 3. A rotating wheel required 3.00 s to rotate through 37.0 revolution. Its angular speed at the end of the 3.00 s interval is 98.0 rad s-1. Calculate the constant angular acceleration of the wheel. ANS. : 13.6 rad s−2 4. A wheel rotates with a constant angular acceleration of 3.50 rad s−2. a. If the angular speed of the wheel is 2.00 rad s−1 at t =0, through what angular displacement does the wheel rotate in 2.00 s. b. Through how many revolutions has the wheel turned during this time interval? c. What is the angular speed of the wheel at t = 2.00 s? ANS. : 11.0 rad; 1.75 rev; 9.00 rad s−1 38
PHYSICS
CHAPTER 8
Exercise 8.2 : 5. A bicycle wheel is being tested at a repair shop. The angular velocity of the wheel is 4.00 rad s-1 at time t = 0 , and its angular acceleration is constant and equal −1.20 rad s-2. A spoke OP on the wheel coincides with the +x-axis at time t = 0 as shown in y Figure 8.8.
O
P
x
Figure 8.8
a. What is the wheel’s angular velocity at t = 3.00 s? b. What angle in degree does the spoke OP make with the positive x-axis at this time? ANS. : 0.40 rad s−1; 18° 39
http//:kms.matrik.edu.my/
PHYSICS CHAPTER 8 Learning Outcome: 8.2 Equilibrium of a uniform rigid body (2 hour) At the end of this chapter, students should be able to: Define and use torque.
State and use conditions for equilibrium of rigid body:
∑
Fx = 0 ,
∑
Fy = 0 ,
∑
τ= 0
40
PHYSICS 8.2.1
Torque (moment of a force),τ
CHAPTER 8
The magnitude of the torque is defined as the product of a force and its perpendicular distance from the line of action of the force to the point (rotation axis). axis) OR
τ = Fd
where
Because of
τ : magnitude of the torque F : magnitude of the force d : perpendicular distance (moment arm)
d = r sin θ where r : distance between the pivot point (rotation axis) and the point of application of force.
Thus
τ = Fr sin θ OR τ = r× F where θ : angle between F and r 41
PHYSICS
CHAPTER 8
It is a vector quantity. quantity The dimension of torque is
[τ ] = [ F ][ d ] =
ML2T − 2
The unit of torque is N m (newton metre), a vector product unlike the joule (unit of work), work) also equal to a newton metre, which is scalar product. product Torque is occurred because of turning (twisting) effects of the forces on a body. Sign convention of torque: Positive - turning tendency of the force is anticlockwise. anticlockwise Negative - turning tendency of the force is clockwise. clockwise The value of torque depends on the rotation axis and the magnitude of applied force. force
42
PHYSICS
CHAPTER 8
Case 1 : Consider a force is applied to a metre rule which is pivoted at one end as shown in Figures 8.9a and 8.9b.
F τ = Fd
(anticlockwise)
d
Figure 8.9a
Line of action of a force Pivot point (rotation axis)
d = r sin θ r
Figure 8.9b
θ
F
Point of action of a force
τ = Fd = Fr sin θ
(anticlockwise) 43
PHYSICS
CHAPTER 8
Case 2 : Consider three forces are applied to the metre rule which is pivoted at one end (point O) as shown in Figures 8.10.
F3 d 1 = r1 sin θ1 r2 O
d 2 = r2 sin θ2
θ2
θ1
F2
r1 Figure 8.10
F1
τ 1 = F1 d 1 = F1 r1 sin θ1 τ 2 = − F2 d 2 = − F2 r2 sin θ 2 τ 3 = F3 d 3 = F3 r3 sin θ3 = 0 Therefore the resultant (nett) torque is
∑ ∑
τ O = τ1 + τ 2 + τ 3 τ O = F1 d 1 − F2 d 2
Caution : If the line of action of a force is through the rotation axis τ = Fr sin θ and θ = 0 then τ= 0 Simulation 5.1 44
PHYSICS
CHAPTER 8
Example 8.9 : Determine a resultant torque of all the forces about rotation axis, O in the following problems. a. F = 10 N 2
F1 = 30 N
5m
5m 3m
6m
O 3m
10 m
F3 = 20 N 45
PHYSICS
CHAPTER 8
Example 8.9 : b.
F1 = 30 N
10 m 3m
β F3 = 20 N
6m
O
3m 5m
5m
F2 = 10 N
α F4 = 25 N
46
PHYSICS
F2 = 10 N
Solution : a. 5m
CHAPTER 8 F1 = 30 N
5m
d1 = 3 m O
6m
d2 = 5 m 10 m Force
F3 = 20 N The resultant torque:
F1 F2 F3
Torque (N m), τo=Fd=Frsinθ
− ( 30 )( 3) = − 90
+ (10)( 5) = + 50
0 47
PHYSICS
CHAPTER 8
Solution : b.
3m
β F3 = 20 N
F1 = 30 N
10 m
d3 d = 3 m 1
β r = 5m 5m
O
6m
sin β = 5m
F2 = 10 N Force
F1 F2 F3 F4
α
3 32 + 5 2
F4 = 25 N
Torque (N m), τo=Fd=Frsinθ
− ( 30 )( 3) = − 90 0
The resultant torque:
F3 r sin β = ( 20 )( 5)( 0.515) = 51.5
0
48
= 0.515
PHYSICS
CHAPTER 8
8.2 Equilibrium of a rigid body 8.2.1.1 Non-concurrent forces
is defined as the forces whose lines of action do not pass through a single common point. The forces cause the rotational motion on the body. The combination of concurrent and non-concurrent forces cause rolling motion on the body. (translational and rotational motion) Figure 8.11 shows an example of non-concurrent forces.
F1
F2
F4 Figure 8.11
49
F3
PHYSICS
CHAPTER 8
8.2.1.2 Equilibrium of a rigid body
Rigid body is defined as a body with definite shape that doesn’t change, so that the particles that compose it stay in fixed position relative to one another even though a force is exerted on it. it If the rigid body is in equilibrium, equilibrium means the body is translational and rotational equilibrium. equilibrium There are two conditions for the equilibrium of forces acting on a rigid body. The vector sum of all forces acting on a rigid body must be zero.
∑
F = Fnett = 0 OR
∑
Fx = 0 ,
∑
Fy = 0 ,
∑
Fz = 0 50
PHYSICS
CHAPTER 8
The vector sum of all external torques acting on a rigid body must be zero about any rotation axis. axis
∑
τ = τ nett = 0
This ensures rotational equilibrium. equilibrium This is equivalent to the three independent scalar equations along the direction of the coordinate axes,
∑
τx = 0 ,
∑
τy = 0,
∑
τz = 0
Centre of gravity, CG is defined as the point at which the whole weight of a body may be considered to act. act A force that exerts on the centre of gravity of an object will cause a translational motion. motion 51
PHYSICS
CHAPTER 8
Figures 8.14 and 8.15 show the centre of gravity for uniform (symmetric) object i.e. rod and sphere rod – refer to the midway point between its end. end
l l 2
CG
Figure 8.12
l 2
sphere – refer to geometric centre. centre
CG Figure 8.13 52
PHYSICS
CHAPTER 8
8.2.4 Problem solving strategies for equilibrium of a rigid body
The following procedure is recommended when dealing with problems involving the equilibrium of a rigid body: Sketch a simple diagram of the system to help conceptualize the problem. Sketch a separate free body diagram for each body. Choose a convenient coordinate axes for each body and construct a table to resolve the forces into their components and to determine the torque by each force. Apply the condition for equilibrium of a rigid body :
∑
Fx = 0 ;
∑
Fy = 0
and
∑
τ= 0
Solve the equations for the unknowns. 53
PHYSICS
CHAPTER 8
Example 8.10 : A 35 cm O
B
75 cm
W1
W2 Figure 8.14
A hanging flower basket having weight, W2 =23 N is hung out over the edge of a balcony railing on a uniform horizontal beam AB of length 110 cm that rests on the balcony railing. The basket is counterbalanced by a body of weight, W1 as shown in Figure 8.14. If the mass of the beam is 3.0 kg, calculate a. the weight, W1 needed, b. the force exerted on the beam at point O. (Given g =9.81 m s−2)
54
PHYSICS
CHAPTER 8 N
Solution : m = 3 kg; W2 = 23 N The free body diagram of the beam :
A W2
0.20 m
0.35 m
O
Let point O as the rotation axis. Force
W1 W2 mg N
y-comp. (N)
− W1 − 23
− ( 3)( 9.81) = − 29.4 N
CG
mg
0.55 m
0.55 m
Torque (N m), τo=Fd=Frsinθ
− W1 ( 0.75) = − 0.75W1
+ ( 23)( 0.35) = 8.05
− ( 29.4 )( 0.20 ) = − 5.88 0
B
0.75 m
55
W1
PHYSICS
CHAPTER 8
Solution : Since the beam remains at rest thus the system in equilibrium. a. Hence τ = 0
∑
O
− 0.75W1 + 8.05 − 5.88 = 0 b. and
∑
Fy = 0
− W1 − 23 − 29.4 + N = 0
− ( 2.89 ) − 23 − 29.4 + N = 0
56
PHYSICS
CHAPTER 8
Example 8.11 : A uniform ladder AB of length 10 m and mass 5.0 kg leans against a smooth wall as shown in Figure 8.15. The height of the A end A of the ladder is 8.0 m from the rough floor. a. Determine the horizontal and vertical forces the floor exerts on the end B of the ladder when a firefighter of mass 60 kg is 3.0 m from B. b. If the ladder is just on the verge of smooth slipping when the firefighter is 7.0 m wall B up the ladder , Calculate the coefficient rough floor of static friction between ladder and Figure 8.15 floor. (Given g =9.81 m s−2)
57
PHYSICS
Force
ml g mf g N1 N2 fs
CHAPTER 8
Solution : ml = 5.0 kg; m f = 60 kg a. The free body diagram of the ladder : Let point B as the rotation axis. A Torque (N m), x-comp. y-comp. (N) (N) τB=Fd=Frsinθ
0
− 49.1
0
− 589
N1
0
0
N2
− fs
0
( 49.1)( 5.0) sin β
α
N1
β
8 sin α = = 0.8 10 6 sin β = = 0.6 10
= 147 ( 589)( 3.0) sin β 8.0 m CG 10 m = 1060 ml g β − N 1 (10) sin α = − 8N1 mf g β α 5.0 m 0
0
6.0 m 58
fs
3.0 m
N2 B
PHYSICS
CHAPTER 8
Solution : Since the ladder in equilibrium thus
∑
τB = 0
147 + 1060 − 8 N 1 = 0
∑
N 1 = 151 N
Fx = 0
N1 − f s = 0 Horizontal force:
∑
Fy = 0
− 49.1 − 589 + N 2 = 0 Vertical force:
59
PHYSICS
Force
ml g mf g N1 N2 fs
CHAPTER 8
Solution : sin α = 0.8; sin β = 0.6 b. The free body diagram of the ladder : Let point B as the rotation axis. A Torque (N m), x-comp. y-comp. (N) (N) τB=Fd=Frsinθ β
0 0 N1
− 49.1 − 589
0
( 49.1)( 5.0) sin β
= 147 ( 589)( 7.0) sin β = 2474 − N 1 (10) sin α = − 8N1
0
N2
0
− μs N 2
0
0
α
mf g 8.0 m
N1
β
10 m
ml g β
5.0 m 6.0 m 60
7.0 m
N2 α fs
B
PHYSICS
CHAPTER 8
Solution : Consider the ladder stills in equilibrium thus
∑
τB = 0
∑
N 1 = 328 N Fy = 0
∑
N 2 = 638 N Fx = 0
147 + 2474 − 8 N 1 = 0
− 49.1 − 589 + N 2 = 0
N 1 − μs N 2 = 0 ( 328) − μs ( 638) = 0
61
PHYSICS
CHAPTER 8
Example 8.12 : A floodlight of mass 20.0 kg in a park is supported at the end of a 10.0 kg uniform horizontal beam that is hinged to a pole as shown in Figure 8.16. A cable at an angle 30° with the beam helps to support the light. a. Sketch a free body diagram of the beam. b. Determine i. the tension in the cable, ii. the force exerted on the beam by the pole.
Figure 8.16
(Given g =9.81 m s ) −2
62
PHYSICS
CHAPTER 8
Solution : m f = 20.0 kg; mb = 10.0 kg a. The free body diagram of the beam :
T
S
O
30 mb g
CG
0.5l
l
mf g
b. Let point O as the rotation axis. Force x-comp. (N) y-comp. (N)
mf g 0 mb g 0 T − T cos 30 Sx S
− (196 ) l
− 196
− ( 98.1)( 0.5l ) = − 49.1l
− 98.1 T sin 30
Sy
Torque (N m), τo=Fd=Frsinθ
Tl sin 30 = 0.5Tl 0 63
PHYSICS
CHAPTER 8
Solution : b. The floodlight and beam remain at rest thus i. τ = 0
∑
O
− 196l − 49.1l + 0.5Tl = 0 ii.
∑
Fx = 0
− T cos 30 + S x = 0
∑
S x = 424 N
Fy = 0
− 196 − 98.1 + T sin 30 + S y = 0
S y = 49.1 N 64
PHYSICS
CHAPTER 8
Solution : b. ii. Therefore the magnitude of the force is 2
2
S=
Sx + S y
S=
( 424) 2 + ( 49.1) 2
and its direction is given by − 1
Sy θ = tan Sx − 1 49.1 θ = tan 424 65
PHYSICS
CHAPTER 8
Exercise 8.3 : Use gravitational acceleration, g = 9.81 m s−2 F1 1. a B A
F2
b D
C Figure 8.17
γ
F3
Figure 8.17 shows the forces, F1 =10 N, F2= 50 N and F3= 60 N are applied to a rectangle with side lengths, a = 4.0 cm and b = 5.0 cm. The angle γ is 30°. Calculate the resultant torque about point D. ANS. : -3.7 N m 66
PHYSICS
CHAPTER 8
Exercise 8.3 : 2.
Figure 8.18
A see-saw consists of a uniform board of mass 10 kg and length 3.50 m supports a father and daughter with masses 60 kg and 45 kg, respectively as shown in Figure 8.18. The fulcrum is under the centre of gravity of the board. Determine a. the magnitude of the force exerted by the fulcrum on the board, b. where the father should sit from the fulcrum to balance the system. 67 ANS. : 1128 N; 1.31 m
PHYSICS
CHAPTER 8
Exercise 8.3 : 3.
Figure 8.19
A traffic light hangs from a structure as show in Figure 8.19. The uniform aluminum pole AB is 7.5 m long has a mass of 8.0 kg. The mass of the traffic light is 12.0 kg. Determine a. the tension in the horizontal massless cable CD, b. the vertical and horizontal components of the force exerted by the pivot A on the aluminum pole. 68 ANS. : 248 N; 197 N, 248 N
PHYSICS
CHAPTER 8
Exercise 8.3 : 4.
30.0 cm
50.0
15.0 cm
F Figure 8.20
A uniform 10.0 N picture frame is supported by two light string as shown in Figure 8.20. The horizontal force, F is applied for holding the frame in the position shown. a. Sketch the free body diagram of the picture frame. b. Calculate i. the tension in the ropes, ii. the magnitude of the horizontal force, F . ANS. : 1.42 N, 11.2 N; 7.20 N 69
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PHYSICS CHAPTER 8 Learning Outcome: 8.3 Rotational dynamics (1 hour) At the end of this chapter, students should be able to: Define the moment of inertia of a rigid body about an axis,
I=
n
∑
mi ri
2
i= 1
State and use torque,
τ = Iα 70
PHYSICS
CHAPTER 8
8.3.1 Centre of mass, moment of inertia and torque 8.3.1.1 Centre of mass (CM)
is defined as the point at which the whole mass of a body may be considered to be concentrated. concentrated
Its coordinate (xCM, yCM) is given the expression below:
n
n
∑ mi y i
∑ mi x i
x CM =
i =1 n
; y CM =
∑ mi
∑ mi
i=1
i=1
where
i=1 n
th
mi : mass of the i particle th x i : x coordinate of the i th particle y i : y coordinate of the i particle 71
PHYSICS
CHAPTER 8
Example 8.13 :
5m
Two masses, 3 kg and 5 kg are located on the y-axis at y =1 m and y =5 m respectively. Determine the centre of mass of this system. Solution : m 1 =3 kg; m 2 =5 kg
m2
3.5 m CM
2
∑ mi y i
y CM = i=12
=
m 1 y 1 m 2 y 2
∑ mi
m1 m 2
i=1
3 1 5 5 y CM = 35 1m
y= 0
m1 72
PHYSICS
CHAPTER 8
Example 8.14 : A system consists of three spheres have the following masses and coordinates : (1) 1 kg, (3,2) ; (2) 2 kg, (4,5) and (3) 3 kg, (3,0). Determine the coordinate of the centre of mass of the system. Solution : m1 =1 kg; m 2 =2 kg; m 3 =3 kg The x coordinate of the CM is 3
∑ mi x i
i=1 x CM= 3
∑ mi
=
m1 x 1 m 2 x 2m3 x 3 m1 m 2m3
i=1
1 3 2 4 3 3 x CM= 123 73
PHYSICS
CHAPTER 8
Solution : The y coordinate of the CM is 3
∑ mi y i
y CM = i=13
∑ yi
=
m 1 y 1 m 2 y 2 m3 y 3 m 1 m2 m3
i=1
1 2 2 5 3 0 y CM = 123 Therefore the coordinate of the CM is
74
PHYSICS
CHAPTER 8
8.3.1.2 Moment of inertia, I
Figure 8.21 shows a rigid body about a fixed axis O with angular velocity ω.
m1 mn
rn
r1 Or
3
ω
r 2 m2 m3
Figure 8.21
is defined as the sum of the products of the mass of each particle and the square of its respective distance from the rotation axis. axis 75
PHYSICS
CHAPTER 8 n
OR
I=m1 r 21 m 2 r 22 m3 r 23 . .. mn r 2n =∑ mi r 2i i=1
where
I : moment of inertia of a rigid body about rotation axis m : mass of particle r : distance from the particle to the rotation axis
It is a scalar quantity. quantity
Moment of inertia, I in the rotational kinematics is analogous to the mass, m in linear kinematics. The dimension of the moment of inertia is M L2. The S.I. unit of moment of inertia is kg m2.
The factors which affect the moment of inertia, I of a rigid body: a. the mass of the body, b. the shape of the body, c. the position of the rotation axis. axis 76
PHYSICS
CHAPTER 8
Moments of inertia of various bodies Table 8.3 shows the moments of inertia for a number of objects about axes through the centre of mass. Shape
Diagram
Hoop or ring or thin cylindrical shell
CM
I CM=MR
CM
1 2 I CM= MR 2
Solid cylinder or disk
Equation
77
2
PHYSICS
CHAPTER 8
Moments of inertia of various bodies Table 8.3 shows the moments of inertia for a number of objects about axes through the centre of mass. Shape Uniform rod or long thin rod with rotation axis through the centre of mass.
Solid Sphere
Diagram
Equation
CM
1 2 I CM = ML 12
CM
2 2 I CM= MR 5 78
PHYSICS
CHAPTER 8
Moments of inertia of various bodies Table 8.3 shows the moments of inertia for a number of objects about axes through the centre of mass. Shape
Hollow Sphere or thin spherical shell
Diagram
Equation
CM
2 2 I CM= MR 3
Table 8.3
79
PHYSICS
CHAPTER 8
Example 8.15 : Four spheres are arranged in a rectangular shape of sides 250 cm and 120 cm as shown in Figure 8.22.
2 kg
3 kg 60 cm
A
O
5 kg
250 cm
B
60 cm
4 kg
Figure 8.22
The spheres are connected by light rods . Determine the moment of inertia of the system about an axis a. through point O, b. along the line AB. 80
PHYSICS
CHAPTER 8
Solution : m1 =2 kg; m 2 =3 kg; a. rotation axis about point O,
m3 =4 kg; m 4 =5 kg m2
m1
r2
r1
r4
O
1.25 m
r3 m3
m4 Since r1= r2=
r3= r4= r thus
0.6 m
2
2
r= 0.6 1.25 =1.39 m
and the connecting rods 2 2 are light 2 therefore 2 I O=m1 r 1 m2 r 2 m3 r 3 m4 r 4
2
2
I O=r m 1 m2 m3 m 4 = 1 .39 2345 81
PHYSICS
CHAPTER 8
Solution : m1 =2 kg; m 2 =3 kg; b. rotation axis along the line AB,
m3 =4 kg; m 4 =5 kg m2
m1
r2
r1 A
r4
r3 m3
m4 r1= r2= r3= r4= r=0.6 m therefore
2 2 2 2 I AB =m1 r 1 m2 r 2m3 r 3 m4 r 4 2 I AB =r m1 m 2 m 3 m 4
2
I AB = 0 . 6 234 5 82
B
PHYSICS
CHAPTER 8
8.3.2 Torque,τ Relationship between torque,τ and angular acceleration, α
Consider a force, F acts on a rigid body freely pivoted on an axis through point O as shown in Figure 8.23.
mn
r1
an rn
Or
2
a 1 m1
m2
F
a2
Figure 8.23
The body rotates in the anticlockwise direction and a nett torque is produced. 83
PHYSICS
CHAPTER 8
A particle of mass, m1 of distance r1 from the rotation axis O will experience a nett force F1 . The nett force on this particle is F 1=m1 a 1 and a 1 =r 1 α
F 1=m1 r 1 α
° The torque on the mass m1 is τ 1=r 1 F 1 sin 90
2 τ 1=m1 r 1 α
The total (nett) torque on 2 the rigid2body is given by2
∑ τ=m1nr1 αm2 r2 α .. .m nrnα n
∑ τ=
2 m r ∑ ii α i =1
∑ τ=Iα
and
2 m r ∑ i i =I i=1
84
PHYSICS
CHAPTER 8
From the equation, the nett torque acting on the rigid body is proportional to the body’s angular acceleration. acceleration
Note :
Nett torque , ∑ τ =Iα is analogous to the
Nett force, ∑ F =ma
85
PHYSICS
CHAPTER 8
Example 8.16 : Forces, F1 = 5.60 N and F2 = 10.3 N are applied tangentially to a disc with radius 30.0 cm and the mass 5.00 kg as shown in Figure 8.24.
F 2
O
F 1
30.0 cm Figure 8.24
Calculate, a. the nett torque on the disc. b. the magnitude of angular acceleration influence by the disc. 1 2 ( Use the moment of inertia, I CM = MR )
2
86
PHYSICS
CHAPTER 8
Solution : R=0. 30 m ; M=5. 00 a. The nett torque on the disc is
kg
∑ τ=τ 1τ 2
∑ τ=−RF 1RF 2=R −F 1F 2
∑ τ= 0.30 −5.6010.3
b. By applying the relationship between torque and angular acceleration, 1 2
∑ τ=Iα
∑ τ=
MR α
2 1 2 1. 41= 5 . 00 0 . 30 α 2 87
PHYSICS
CHAPTER 8
Example 8.17 : A wheel of radius 0.20 m is mounted on a frictionless horizontal axis. The moment of inertia of the wheel about the axis is 0.050 kg m2. A light string wrapped around the wheel is attached to a 2.0 kg block that slides on a horizontal frictionless surface. A horizontal force of magnitude P = 3.0 N is applied to the block as shown in Figure 8.25. Assume the string does not slip on the wheel.
Figure 8.25
a. Sketch a free body diagram of the wheel and the block. b. Calculate the magnitude of the angular acceleration of the wheel. 88
PHYSICS
CHAPTER 8
Solution : R=0.20 m ; I=0.050 kg m 2 ; a. Free body diagram : for wheel,
T
S
W for block,
P=3.0 N; m=2.0 kg
N
T
a
P
W b 89
PHYSICS
CHAPTER 8
Solution : R=0.20 m ; I=0.050 kg m 2 ; b. For wheel,
∑ τ=Iα
For block,
RT =Iα
∑ F=ma
P=3.0 N; m=2.0 kg
Iα T= R P−T =ma
(1) (2)
By substituting eq. (1) into eq. (2), thus
Iα P− =ma and a=Rα R Iα P− =mRα R 90
PHYSICS
CHAPTER 8
Example 8.18 : An object of mass 1.50 kg is suspended from a rough pulley of radius 20.0 cm by light string as shown in Figure 8.26. The pulley has a moment of inertia 0.020 kg m2 about the axis of the pulley. The object is released from rest and the pulley rotates without encountering frictional force. Assume that the string does not slip on the pulley. After 0.3 s, determine a. the linear acceleration of the object, b. the angular acceleration of the pulley, c. the tension in the string, d. the liner velocity of the object, e. the distance travelled by the object. (Given g = 9.81 m s-2)
R
1.50 kg Figure 8.26
91
PHYSICS
CHAPTER 8
Solution : a. Free body diagram : for pulley,
S
W for block,
T
a
∑ τ=Iα
a RT =Iα and α= R a RT =I R T Ia T= 2 (1) R
∑ F=ma
mg−T =ma
(2)
m g 92
PHYSICS
CHAPTER 8
Solution :
2
R=0.20 m ; I=0.020 kg m ; m=1.50 kg;
u=0; t =0 . 3 s
a. By substituting eq. (1) into eq. (2), thus
Ia mg− 2 =ma R 0 .020 a 1 .50 9 .81 − =1 .50 a 2 0 . 20
b. By using the relationship between a and α, hence
a=Rα 7 . 36=0 . 20 α
93
PHYSICS Solution :
CHAPTER 8 2
R=0.20 m ; I=0.020 kg m ; m=1.50 kg;
u=0; t =0 . 3 s c. From eq. (1), thus
Ia T= 2 R
0 . 020 7 . 36 T= 2 0 .20
d. By applying the equation of liner motion, thus
v=uat
v=0 7.36 0.3 e. The distance travelled by the object in 0.3 s is
1 2 s=ut at 2 1 2 s=0 7 . 36 0 .3 2
94
PHYSICS
CHAPTER 8
Exercise 8.4 : Use gravitational acceleration, g = 9.81 m s−2 1. Three odd-shaped blocks of chocolate have following masses and centre of mass coordinates: 0.300 kg, (0.200 m,0.300 m); 0.400 kg, (0.100 m. -0.400 m); 0.200 kg, (-0.300 m, 0.600 m). Determine the coordinates of the centre of mass of the system of three chocolate blocks. ANS. : (0.044 m, 0.056 m) 2. Figure 8.27 shows four masses that are held at 70 g 40 cm the corners of a square by a very light 80 cm frame. Calculate the moment of inertia B of the system about an axis perpendicular A to the plane 150 g 150 g a. through point A, and 80 cm b. through point B. Figure 8.27 70 g ANS. : 0.141 kg m2; 0.211 kg m2 95
PHYSICS
CHAPTER 8
Exercise 8.4 : 3. A 5.00 kg object placed on a frictionless horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object as in Figure 8.28. The pulley has a radius of 0.250 m and moment of inertia I. The block on the table is moving with a constant acceleration of 2.00 m s−2. a. Sketch free body diagrams of both objects and pulley. b. Calculate T1 and T2 the tensions
2.00 m s−2 T1 T2
Figure 8.28
in the string. c. Determine I. ANS. : 10.0 N, 70.3 N; 1.88 kg m2
96
PHYSICS CHAPTER 8 Learning Outcome: http//:kms.matrik.edu.my/
8.4 Work and Energy of Rotational Motion (2 hours) At the end of this chapter, students should be able to: Solve problem related to : kinetic energy,
1 2 K r = Iω 2
work,
power,
W =τθ P=τω
97
PHYSICS
CHAPTER 8
8.4 Rotational kinetic energy and power 8.4.1 Rotational kinetic energy, Kr
Consider a rigid body rotating about the axis OZ as shown in Figure 8.29. Z
vn
mn
rn Or
r1
v 1 m1 r2
v2
m2 v3 3 m3
Figure 8.29
Every particle in the body is in the circular motion about point O. 98
PHYSICS
CHAPTER 8
The rigid body has a rotational kinetic energy which is the total of kinetic energy of all the particles in the body is given by 1 1 2 1 2 1 2 2 K r = m1 v 1 m2 v 2 m3 v 3 .. . mn v n
2 2 2 2 1 1 2 2 1 2 2 1 2 2 2 2 K r = m1 r 1 ω m2 r 2 ω m3 r 3 ω . .. mn r n ω 2 2 2 2 1 2 2 2 2 2 K r = ω m 1 r 1 m2 r 2m3 r 3 . . .mn r n 2 n n 1 K r = ω 2 ∑ mi r 2i and ∑ mi r 2i =I 2 i=1 i=1
1 2 K r = Iω 2 99
PHYSICS
CHAPTER 8
From the formula for translational kinetic energy,
Ktr
1 2 K tr = mv 2
After comparing both equations thus
ω is analogous to v I is analogous to m
For rolling body without slipping, slipping the total kinetic energy of the body, K is given by
K =K tr K r where
K tr : translational kinetic energy K r : rotational kinetic energy 100
PHYSICS
CHAPTER 8
Example 8.19 : A solid sphere of radius 15.0 cm and mass 10.0 kg rolls down an inclined plane make an angle 25° to the horizontal. If the sphere rolls without slipping from rest to the distance of 75.0 cm and the inclined surface is smooth, calculate a. the total kinetic energy of the sphere, b. the linear speed of the sphere, c. the angular speed about the centre of mass. 2 2 (Given the moment of inertia of solid sphere is I CM = mR and 5 the gravitational acceleration, g = 9.81 m s−2)
101
PHYSICS
CHAPTER 8
Solution :
R=0.15 m ; m=10.0 kg s=0 . 75 m
R
h=ssin 25 v
CM
25
°
a. From the principle of conservation of energy,
∑ E i =∑ E f
mgh= K ° K =mgs sin 25 ° K =10.0 9.81 0.75 sin 25 102
°
PHYSICS
CHAPTER 8
Solution : R=0.15 m ; m=10.0 kg b. The linear speed of the sphere is given by 1 1 2 2 K =K tr K r K = mv Iω and
v ω= R
2 2 1 2 1 2 2 v K = mv mR 2 2 5 R 7 2 K = mv 10 7 2 31.1= 10. 0 v 10
2
c. By using the relationship between v and ω, thus
v=Rω
2 .11=0 . 15 ω
103
PHYSICS
CHAPTER 8
Example 8.20 : The pulley in the Figure 8.30 has a radius of 0.120 m and a moment of inertia 0.055 g cm2. The rope does not slip on the pulley rim. Calculate the speed of the 5.00 kg block just before it strikes the floor. (Given g = 9.81 m s−2)
5.00 kg 7 . 00 m
2.00 kg Figure 8.30
104
PHYSICS
CHAPTER 8
Solution : m1 =5 . 00 kg ;m 2=2. 00 kg; The moment of inertia of the pulley,
R=0 .120 m; h=7 . 00 m
−3 −4 2 10 kg 10 m 2 −9 2 I= 0 . 055 g 1 cm = 5 . 5×10 kg m 1g 1 cm 2
m1 7. 00 m
m2 m2
Initial
∑ E i =U 1
v
m1 Final
v
7 . 00 m
∑ E f = K tr 1105K tr 2K rU 2
PHYSICS
CHAPTER 8
Solution : m 1 =5 . 00 kg ;m 2=2. 00 kg;
−9
R=0 .120 m;
h=7.00 m; I =5.5×10 kg m
2
By using the principle of conservation of energy, thus
∑ E i =∑ E f
U 1 =K tr1 K tr 2 K r U 2 1 2 1 2 1 2 m1 gh= m1 v m2 v Iω m2 gh 2 2 2 2 1 2 1 v m1−m2 gh= 2 v m1m2 2 I R 1 2 1 v −9 5 . 00−2 . 00 9 . 81 7 . 00 = v 5 . 002. 00 5 . 5×10 2 2 0 . 120
106
2
PHYSICS
CHAPTER 8
8.4.2 Work, W
Consider a tangential force, F acts on the solid disc of radius R freely pivoted on an axis through O as shown in Figure 8.31.
R
dθ
ds
R
O
F
Figure 8.31
The work done by the tangential force is given by
dW =Fds and ds=Rdθ dW =FRdθ θ
∫ dW=∫θ τdθ 2 1
θ
W =∫θ τdθ 2 1
107
PHYSICS
CHAPTER 8
If the torque is constant thus θ
W =τ ∫θ dθ 2
1
W =τ θ2 −θ 1
W =τΔθ
is analogous to the
W =Fs
τ : torque Δθ : change in angular displacement
where
W : work done
Work-rotational kinetic energy theorem states
W =ΔK r = K r f − K r i 1 2 1 2 W = Iω − Iω 0 2 2
108
PHYSICS
CHAPTER 8
8.4.3 Power, P
From the definition of instantaneous power,
dW P= dt τdθ P= dt P=τω
and
dW =τdθ
and
dθ =ω dt
is analogous to the
P=Fv
Caution : The unit of kinetic energy, work and power in the rotational kinematics is same as their unit in translational kinematics.
109
PHYSICS
CHAPTER 8
Example 8.21 : A horizontal merry-go-round has a radius of 2.40 m and a moment of inertia 2100 kg m2 about a vertical axle through its centre. A tangential force of magnitude 18.0 N is applied to the edge of the merry-go- round for 15.0 s. If the merry-go-round is initially at rest and ignore the frictional torque, determine a. the rotational kinetic energy of the merry-go-round, b. the work done by the force on the merry-go-round, c. the average power supplied by the force. (Given g = 9.81 m s−2) Solution :
R=2 . 40 m
F 110
PHYSICS Solution :
CHAPTER 8 2
R=2.40 m ; I=2100 kg m ; F=18.0 N; t=15 . 0 s; ω 0 =0
a. By applying the relationship between nett torque and angular acceleration, thus
∑ τ=Iα
RF=Iα
2.40 18.0 =2100α
Use the equation of rotational motion with uniform angular ω=ω0 αt acceleration,
15.0 ω=0 2.06×10 −1 ω=0.309 rad s −2
Therefore the rotational kinetic energy for 15.0 s is
1 2 K r = Iω 2
1 2 K r = 2100 0 .309 2 111
PHYSICS
CHAPTER 8
Solution :
2
R=2.40 m ; I=2100 kg m ; F=18.0 N; t=15 . 0 s; ω 0 =0
b. The angular displacement, θ for 15.0 s is given by
1 2 θ=ω 0 t αt 2
1 2 −2 θ=0 2. 06×10 15 . 0 2
By applying the formulae of work done in rotational motion, thus
W =τθ
W =RFθ
W =2.40 18.0 2.32 c. The average power supplied by the force is
W P av = t
100 P av = 15.0
112
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PHYSICS CHAPTER 8 Learning Outcome: 8.5 Conservation of angular momentum (1 hour) At the end of this chapter, students should be able to: Define and use angular momentum,
L=Iω
State and use the principle of conservation of angular momentum
113
PHYSICS
CHAPTER 8
8.5 Conservation of angular momentum
L 8.5.1 Angular momentum,
is defined as the product of the angular velocity of a body and its moment of inertia about the rotation axis. axis OR is analogous to the p=mv
L=Iω
where
L : angular momentum
I : moment of inertia of a body ω : angular velocity of a body
It is a vector quantity. Its dimension is M L2 T−1 The S.I. unit of the angular momentum is kg m2 s−1.
114
PHYSICS
CHAPTER 8
The relationship between angular momentum, L with linear momentum, p is given by vector notation : magnitude form :
L =r ×p =r ×m v
L=rpsin θ=mvr sin θ where
r : distance from the particle to the rotation axis θ:the angle between {r with {v ¿ ¿
Newton’s second law of motion in term of linear momentum is
d p ∑ F = F nett = dt
hence we can write the Newton’s second law in angular form as
d L ∑ τ =τ nett = dt
and states that a vector sum of all the torques acting on a rigid body is proportional to the rate of change of angular momentum. momentum 115
PHYSICS
CHAPTER 8
8.5.2 Principle of conservation of angular momentum
states that a total angular momentum of a system about an rotation axis is constant if no external torque acts on the system. system OR
Iω=constant
Therefore
If the
∑ τ =0
d L ∑ τ = dt =0 d L=0 and dL=∑ L f −∑ Li
∑ L i=∑ L f 116
PHYSICS
CHAPTER 8
Example 8.22 : A 200 kg wooden disc of radius 3.00 m is rotating with angular speed 4.0 rad s-1 about the rotation axis as shown in Figure 8.32 . A 50 kg bag of sand falls onto the disc at the edge of the wooden disc.
R
ω0
Before
ω
R
Figure 8.32
After
Calculate, a. the angular speed of the system after the bag of sand falling onto the disc. (treat the bag of sand as a particle) b. the initial and final rotational kinetic energy of the system. Why the rotational kinetic energy is not the same? 1 2 MR ) (Use the moment of inertia of disc is
2
117
PHYSICS
CHAPTER 8 −1
Solution : R=3 . 00 m ;ω 0 =4 . 0 rad s a. The moment of inertia of the disc,
; mw =200 kg; mb =50 kg
1 2 2 1 I w = mw R = 200 3. 00 2 2 2 I w =900 kg m
The moment of inertia of the bag of sand, 2
2
I b =m b R = 50 3 . 00
I b =450 kg m
2
By applying the principle of conservation of angular momentum,
∑ L i=∑ L f
I w ω 0 = I w I b ω
900 4.0 = 900450 ω 118
PHYSICS
CHAPTER 8 −1
Solution : R=3 . 00 m ;ω 0 =4 . 0 rad s b. The initial rotational kinetic energy,
; mw =200 kg; mb =50 kg
1 2 2 1 K r i = 2 I w ω 0= 2 900 4 . 0 K r i =7200 J
The final rotational kinetic energy,
1 2 2 1 K r f = 2 I w I b ω = 2 900450 2 . 67
thus K r i ≠ K r f It is because the energy is lost in the form of heat from the friction between the surface of the disc with the bag of sand.
119
PHYSICS
CHAPTER 8
Example 8.23 : A raw egg and a hard-boiled egg are rotating about the same axis of rotation with the same initial angular velocity. Explain which egg will rotate longer. Solution : The answer is hard-boiled egg. egg
120
PHYSICS
CHAPTER 8
Solution : Reason Raw egg : When the egg spins, its yolk being denser moves away from the axis of rotation and then the moment of inertia of the egg increases 2 because of I=mr From the principle of conservation of angular momentum,
Iω=constant
If the I is increases hence its angular velocity, ω will decreases. Hard-boiled egg : The position of the yolk of a hard-boiled egg is fixed. When the egg is rotated, its moment of inertia does not increase and then its angular velocity is constant. Therefore the egg continues to spin.
121
PHYSICS
CHAPTER 8
Example 8.24 : A student on a stool rotates freely with an angular speed of 2.95 rev s−1. The student holds a 1.25 kg mass in each outstretched arm that is 0.759 m from the rotation axis. The moment of inertia for the system of student-stool without the masses is 5.43 kg m2. When the student pulls his arms inward, the angular speed increases to 3.54 rev s−1. a. Determine the new distance of each mass from the rotation axis. b. Calculate the initial and the final rotational kinetic energy of the system. 2 .95 rev 2π rad Solution : ω 0 = =18 . 5 rad s−1
1s
1 rev
3 . 54 rev 2π rad ω= =22. 2 rad s−1 1s 1 rev 122
PHYSICS
CHAPTER 8
Solution :
2
−1
m=1. 25 kg ; ω0 =18. 5 rad s ; I ss =5. 43 kg m ; −1 r b =0 .759 m ; ω=22. 2 rad s ;
ω0
rb
ω
rb m
m
Before
ra ra
After 123
PHYSICS
CHAPTER 8
Solution :
2
−1
m=1. 25 kg ; ω0 =18. 5 rad s ; I ss =5. 43 kg m ; −1 r b =0 .759 m ; ω=22. 2 rad s ;
a. The moment of inertia of the system initially is
I i =I ss I m
2 mr
I i =I ss mr 2 mr
=I ss
b
b
2
b2
2
I i = 5 . 43 2 1 .25 0 .759 =6 . 87 kg m2
The moment of inertia of the system finally is
I f =I ss 2 mr 2 a = 5 . 43 2 1. 25 r 2 I f =5 . 432 .5r 2 a a
By using the principle of conservation of angular momentum, thus i f
∑ L =∑ L I i ω 0 =I f ω
6 . 87 18 .5 =5 . 432 .5r
a
2
22 .2 124
PHYSICS
CHAPTER 8 2
−1
Solution : m=1. 25 kg ; ω0 =18. 5 rad s
; I ss =5. 43 kg m ; r b =0 .759 m ; ω=22. 2 rad s ; −1
b. The initial rotational kinetic energy is given by
1 K r i = 2 I i ω02 1 2 = 6 .87 18 .5 2
K r i =1 .18×10
3
J
and the final rotational kinetic energy is
1 2 K r f = 2 I f ω
1 2 2 = 5 . 432 .5 0 .344 22. 2 2 125
PHYSICS
CHAPTER 8
Exercise 8.5 : Use gravitational acceleration, g = 9.81 m s−2 1. A woman of mass 60 kg stands at the rim of a horizontal turntable having a moment of inertia of 500 kg m2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about the frictionless vertical axle through its centre. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m s−1 relative to the Earth. a. In the what direction and with what value of angular speed does the turntable rotate? b. How much work does the woman do to set herself and the turntable into motion? ANS. : 0.360 rad s−1 ,U think; 99.9 J
126
PHYSICS
CHAPTER 8
Exercise 8.5 : 2. Determine the angular momentum of the Earth a. about its rotation axis (assume the Earth is a uniform solid sphere), and b. about its orbit around the Sun (treat the Earth as a particle orbiting the Sun). Given the Earth’s mass = 6.0 x 1024 kg, radius = 6.4 x 106 m and is 1.5 x 108 km from the Sun. ANS. : 7.1 x 1033 kg m2 s−1; 2.7 x 1040 kg m2 s−1 3. Calculate the magnitude of the angular momentum of the second hand on a clock about an axis through the centre of the clock face. The clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a thin rod rotating with angular velocity about one end. (Given the moment of inertia of 1 2 thin rod about the axis through the CM is ML ) 12 ANS. : 4.71 x 10−6 kg m2 s−1 127
PHYSICS
CHAPTER 8
Summary: Linear Motion
ds v= dt dv a= dt m
∑ F=ma p=mv W =Fs P=Fv
Relationship
v=rω a=rα n
Rotational Motion
dθ ω= dt dω α= dt
I=∑ mi r i2 i=1
τ =rF sin θ L=rpsin θ
I
∑ τ=Iα L=Iω
W =τθ P=τω
128
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PHYSICS
CHAPTER 8
THE END… Next Chapter…
CHAPTER 9 : Simple Harmonic Motion (SHM)
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