EP015 Note #KMKK

4.2 NEWTON’S LAW

1. If a nonzero net force is acting on an object, which of the following must we assume
regarding the object’s condition?
A. The object is at rest
B. The object is moving with constant velocity
C. The object is being accelerated
D. The object is losing mass

2 An object of mass m is moving at constant speed up an inclined plane by a force F as
shown in FIGURE . The frictional force between the object and the inclined plane is…

F

θ

A. mg sinθ FIGURE
B. mg cosθ
C. F - mg sinθ
D. F - mg cosθ

3. State :
i) Newton’s First Law
ii) Newton’s Second Law
iii) Newton’s Third Law

50

300

FIGURE
4. a) A 3.0 kg cube is placed on a rough plane. The plane is then slowly tilted until

the cube starts to move from rest. This occurred when the angle of inclination is
25°. Calculate static friction coefficient between the cube and the rough plane.

(0.466)
b) A wooden block of mass 2.0 kg slides down with constant velocity on a inclined

rough plane of 30° from the horizontal axis.
i) Sketch a free body diagram to show forces acting on the wooden block
ii) Calculate the kinetic frictional force between the wooden block and the

rough inclined plane.
(9.81 N)

5. 4 kg
A

rough
plane

300

B
1 kg

FIGURE

51

a) A 4.0 kg block A on a rough 30°inclined plane is connected to a freely hanging
1.0 kg block B by a mass-less cable passing over the frictionless pulley as shown
in FIGURE . When the objects are released from rest, object A slides down the
inclined plane with a friction force of 6.0 N. Calculate
i) the acceleration of the objects
(4.69 m s-2)
ii) the tension in the cable.
(5.12 N)

b) The weight of an object W, tension T1 and tension T2 are in static equilibrium as
shown in FIGURE 3.4. If W = 40 N, calculate T1 and T2.
(T1 = 101.3 N, T2 = 75.2 N)

500
T1

60°

T2

P

W

6. a) State the difference between static equilibrium and dynamic equilibrium.
b) Explain why it is easy to push a moving object than to push a stationary object of
the same mass.

7. A student of mass 50 kg stands inside a lift. Determine the apparent weight of the student

if the lift moves
(assume g = 9.81 m s-1 )

(a) downwards with constant speed,

(490.5 N)

(b) downwards with a constant acceleration of 0.50 m s-2,

(465.5 N)

(c) upwards with a constant acceleration 0.50 m s-2

(515.5 N)

52

53

54

55

56

57

58

59

60

CHAPTER 5
WORK, ENERGY & POWER

LEARNING OUTCOMES

5.1 Work

a) Define and apply work done by a constant force, = ⃗ ∙ ⃗ ∙

b) Apply work done by a constant force-displacement graph..

5.2 Energy and conservation of energy

a) State the principle of conservation of energy

b) Apply the pinciple of conservation of energy (mechanical energy and heat energy due to
friction)

c) State and apply the work energy theorem , W = ΔK

5.3 Power

a) define and use average power ∆
And instantaneous power, = ⃗⃗⃗ ⃗. ⃗ = ∆

b) verify the law of conservation of energy (Experiment 3 : Energy)

61

5.1 WORK

1. SI unit for work done is

A. Pascal
B. Joules
C. Newton
D. Ohms

2. When force is applied on an object but object does not move, it means that
A. no power is used
B. no work is done
C. work is done
D. power is used

3. Energy is
A. the work needed to create potential or kinetics energy.
B. the ability to do work
C. the work that can be done by an object with Potential Energy and Kinetics Energy
D. All of the above

4. The metric unit of a joule (J) is a unit of
A. Potential energy
B. Work
C. Kinetic energy
D. all of the above

6. A box resting on a horizontal surface is push by a 27N force. The surface exerts a friction
force of 2N as the box moves. If the box moves a total of 3.7 m, how much work done on
the box?
(92.5 J)

7. A lawnmower (25kg) is pushed a horizontal distance of 10m by a 50 N downward
force directed 60 degrees to the horizontal. The coefficient of kinetic frictional force
is 0.017. What is the work done by each of the external forces on the lawn mower?
(Ww=0, WN = 0 , Wf = -49.1 J , WF = +250 J)

62

8. A 4.1 kg box is lifted vertically from rest to a distance of 1.6 m with a constant upward
applied force of 60.0 N. Calculate

a) the work done by the applied force
b) the work done by gravity.
c) final velocity

(96 J , -64 J , 3.9 m s-1)

9. A 500 kg helicopter ascends vertically from the ground with an acceleration of 2 m s-2
over a 5.0 s interval. Calculate the
a) work done by the lifting force.
b) work done by the gravitational force.
c) net work done on the helicopter.
(1.48 x 105J , -1.23 x 105 J , 2.5 x 104 J)

10. a) State the physical quantity represent by the area under the slope of the graph in
Figure 5.1.

Figure 5.1
b) The force F on an object varies with the displacement x as shown in Figure 5.2.

Calculate the work done from x  1.0 m to x  2.5 m.

Figure 5.2 (4 J)
63

11.

Figure 5.3
Figure 5.3 show the force as a function of displacement of a moving object is presented
by the graph.
a) How much work is done when the object moves from 0 m to 6 m?
b) How much work is done when the object moves from 6 m to 9 m?
c) How much work is done when the move from 0 to 9 m?

(30 J , 15 J , 45 J)

5.2 ENERGY AND CONSERVATION OF ENERGY
1. The potential energy of a box on a shelf, relative to the floor , is measure of

A. the energy the box has due to its position above the floor
B. the weight of the box times the distance above the floor
C. the work done putting the box on the shelf from the floor
D. all of the above
2. Which quantity has the greatest influence on he amount of kinetic energy that a large truck has
while moving down the highway
A. mass
B. weight
C. velocity
D. size
3. . Potential energy and kinetic energy are types of
A. Electrical energy
B. Magnetic energy
C. Thermal energy
D. Mechanical energy

64

4. Energy cannot be
A. converted or transferred
B. destroyed or transferred
C. created or transferred
D. created or destroyed

5. When a pendulum bob is displaced to one side, it gains
A. Chemical Potential Energy
B. Elastic Potential Energy
C. Gravitational Potential Energy
D. Kinetic energy

6. Energy used by a moving object is termed as
A. Potential Energy
B. Mechanical Energy
C. Kinetic Energy
D. Thermal Energy

7. A spring with a spring constant 2.5 x 104 N m-1 is compressed by 3.0 cm.
a) Calculate the work done to compress the spring.
b) A 12 g steel ball is placed on the compressed spring. When the spring is released,
calculate the velocity of the ball assuming all energy of the spring is transferred to
it.
(11.25 J , 43.3 m s-1)

8. A 20 kg load is connected through a smooth pulley to a spring with constant force, k = 350
Nm-1 as shown in FIGURE 5.4. Initially the load is held so that it is rest and the spring is not
extended. When the load is released, determine the velocity when the elongation is 0.5 m.

(2.33 m s-1)

65

9.

Figure 5.5
An object P of mass 2.0 kg is connected to one end of a light spring. This spring–mass system is
placed on a smooth horizontal plane, as in Figure 5.5. The spring has a spring constant of 4.5 kN
m-1. An object Q of mass 3.0 kg moves at a speed 5.0 m s-1 towards P. Q collides and sticks to P.
Assuming there is no dissipation of energy to the surroundings. Calculate the
a) speed of P and Q immediately after the collision.
b) speed of P and Q at the instant when the spring is compressed by 5.0 cm.
c) maximum compression experienced by the spring.

(-3 m s-1 , 2.6 m s-1 , 0.10 m)
10.

Figure 5.6
Cyclist rides a bicycle down the slope 3 m high hill at an initial velocity of 2 m s-1 as in Figure
5.6. At the foot of the hill, the velocity is 6 m s-1. The total mass of cyclist and his bicycle is 75
kg. Calculate the
(i) initial kinetic energy and potential energy.

(50 J , 2207.25 J )
(ii) work done to overcome friction along the slope.

(907.25 J)

66

11.
A

3m
BC

6m

Figure 5.7

A 10 kg block is released from point A in Figure 5.7. The track is frictionless except for
the portion between point B and C of length 6 m. The block slides down the track and
hits a spring of force constant 2250 N m-1. The spring compressed 30 cm from its
equilibrium position before the block comes to rest momentarily. Calculate the
coefficient of kinetic friction between the block and the rough surface.

(65.8 hp)

12. Figure 5.8 shows a lorry of mass 2000kg slides down a slope of an angle θ and distance L.
After going down the slope, the lorry passes through a flat road and then climbs a second slope
of an angle 30°.

h h’
θ 30⁰

Figure 5.8
a) Calculate the distance s on the second slope if given θ = 25°, v0 = 25 m s-1 and L=15 m

(76.36 m)

b) If after going down the first slope, the lorry driver steps on the brake and stops at a distance of
12 m on the flat road, what is the work done by the brakes and the average force exerted by the
brake?

(-7.49 x 105 J , -6.24 x 104 N)

67

13. A 0.02 kg bullet is fired horizontally with velocity up using a 0.8 kg gun. The bullet hits a
0.88 kg block that is suspended by a light vertical string 1 m in length. After the collision, the
bullet embeds itself in the block and both rises to a maximum height of 0.4 m as shown in
Figure 5.9.

1.0 θm

0.4

Figure 5.9

a) Calculate the bullet velocity up

b) Calculate the maximum angle θm that the bullet-block makes with the vertical (140 m s -1)
(53.1°)

5.3 POWER

1. One watt is equal to

A. One joule per second
B. One second per joule
C. One meter per joule
D. One joule per meter

2. Power is defined as

A. time taken per joule
B. work done or energy converted per distance
C. work done or energy converted per unit time
D. distance covered per unit work done

68

3. a) A man pulls a 100 N crate up a frictionless 30° slope 5 m high as shown in
Figure 5.10. The crate moves at constant speed. Calculate the work done by the
man.

Figure 5.10

(500J)

4. A 20,000 kg train produces a power of 8.1 MW with a traction force of 398 kN on a
level track.

a) Determine the speed of the train.If the train travels on a uphill track with a slope

of 12° above the horizontal.
(20.34 m s-1)

b) draw a diagram showing all the forces acting on the train and determine the new

speed of the train. (18.45 m s-1)

5. A machine is used to push an object of mass of 2.5 kg up a rough inclined plane. A
constant force of 15 N is needed to move the object up the plane in 6 s as shown in
Figure 5.11.

2.5 kg F
5m 3m

4m
Figure 5.11

Calculate the potential energy of the object,

(73.58 J)

69

6. A 10 kg block is released from point A in Figure 5.12. The track is frictionless except for
the portion between point B and C of length 6 m. The block slides down the track and
hits a spring of force constant 2250 N m-1. The spring compressed 30 cm from its
equilibrium position before the block comes to rest momentarily. Calculate the
coefficient of kinetic friction between the block and the rough surface.

A B C
3m 6m

FIGURE 5.12

(0.328)

70

CHAPTER 6 :Circular Motion (1 Hour) 6.1.1 Linear (tangential) velocity ,

Its direction is always perpendicular to
the radius of circular path.
Is always tangential to the circular path
as shown in Figure 6.2

Learning Outcome: the magnitude of v of an object is
6.1 Uniform circular motion and constant. The SI unit is ms-1.
centripetal force
6.1.2 Angular quantities related to
a) Describe uniform circular motion circular motion
Period, T is defined as the time taken for
b) Convert units between degrees, one complete revolution (cycle/rotation).
radian and revolution or rotation The SI unit is second (s).
Frequency, f is defined as the number of
6.2 Centripetal force revolutions (cycles/rotations) completed
in one second.
a) Define and use centripetal The SI unit is hertz (Hz) or s-1.
Equation :
acceleration,  v2 to centripetal
relarted f 1
ac T

b) Solve problems For one complete revolution, 2r

force the distance travelled is
the time interval is one period, T
(i) Uniform circular motion

(ii) Horizontal circular motion

(iii) Vertical circular motion

(iv) Conical pendulum

6.1 Uniform Circular Motion
is defined as a motion in a circle
(circular arc) at a constant speed.
Consider an object moving with uniform

circular motion as shown isn Figurreθ6.1.

71

6.1.3 Centripetal (radial) acceleration Example 2
ac, ar

Centripetal acceleration is the
acceleration of an object moving in
circular path with the magnitude equal to
the square of the speed divided by the
radius.

Its direction is always towards the centre
of the circular path.

The direction of centripetal (radial)
acceleration is also perpendicular to the
linear (tangential) velocity as shown in
Figure 6.5.

Example 1

72

:
73

74

75

76

77

CHAPTER 6
CIRCULAR MOTION
LEARNING OUTCOMES
6.1 Uniform Circular Motion
a) Describe uniform circular motion
b) Convert units between degrees, radian and revolution or rotation
6.2 Centripetal Force
a) Define centripetal acceleration,
b) Solve problems related to centripetal force for uniform circular motion
(i) Horizontal circular motion
(ii) Vertical circular motion
(iii) Conical pendulum (exclude banked curve)

78

6.1 UNIFORM CIRCULAR MOTION
6.2 CENTRIPETAL FORCE

1. For an object to move in uniform circular motion, it must move with
A. constant speed
B. constant velocity
C. constant direction
D. constant momentum

2. A motorcycle of mass m is moving with constant speed in a vertical circular track as
shown in FIGURE 6.1
Z

Y

X
FIGURE 6.1
The normal reaction of the track is…
A. zero at X
B. maximum at X
C. equal to mg at Y
D. equal to mg at Z

3. In a motorcycle racing competition, there is a circular track with radius 80 m inclined at
20° with the horizontal. Calculate the maximum velocity of a motorcycle which is moving
along this part of the track without overturning.
( 16.90 m s-1)

4. A simple pendulum is suspended freely on the ceiling of a bas moving in a horizontal
circular track of radius 80 m. The angle between the string of the pendulum and the
vertical line is 36°. Calculate the centripetal acceleration of the pendulum.
(7.12 ms-1)

79

5. A ball of mass, m moves with constant velocity v, in a horizontal circular path of radius r.
The ball is attached to a string of length 24 cm and makes a conical pendulum as shown
in FIGURE 6.2.

24 cm θ

r

FIGURE 6.2
(a) Sketch a free body diagram showing the forces acting on the object.
(b) If θ = 30°, calculate the radius, r and the velocity, v. (12 cm, 0.824 m s-1)
(c) If θ is more than 30°, what can you deduce about the velocity of the ball?

6. A Ferris wheel with diameter 18 m and rotates at the rate of 4 rpm. Calculate
(a) the acceleration of a rider on this wheel,
(1.58 m s-2)
(b) the magnitude and direction of the force which the seat exerted on a 40 kg rider at
the lowest and the highest point of the wheel.
(455.6 N, 392.2 N)

7. A 4 kg object is attached to a vertical rod by two strings each of length 80 cm. The object
rotates in a horizontal circle at constant speed 6 m s-1 as shown in FIGURE 6.3. Both
strings are inclined at 60° with the vertical axis.

FIGURE 6.3

Find the tension of the lower and upper string. (80.7 N, 159.2 N)

80

8. a) State the conditions for an object to move in a circle motion.

b) A motorcycle is moving with constant speed 10 m s-1 along a curve of a highway.
The total mass of the rider and motorcycle is 200 kg and the radius of the curve is
80 m.

i) Why does the rider cannot remain in a vertical position when the
motorcycle is taking the curve? Sketch a diagram illustrating the forces
acting on the rider.

ii) What is the force that provides the centripetal force for the motorcycle and

the rider? Calculate the centripetal force. (250 N)

9. a) A student is to swing a bucket of water in a vertical circle without spilling any. If

the distance from his shoulder to the centre of mass of the bucket of water is 1.0

m, what is the minimum speed required to keep the water from coming out of the

bucket at the top of the swing? (3.1 ms-1)

b) Calculate the minimum speed that a roller coaster must be travelling when upside

down at the top of a vertical circular track of radius 12.0 m, so that the passengers

do not fall out. (10.8 ms-1)

81

7.1 Gravitational Force and Field Strength
= 2

d) State and use the Newton’s Law of where;

Gravitation, = =
2

e) Define and use gravitational field = universal gravitational constant

strength, = (6.67x10−11 Nm2kg−2)
2
=

=

Gravitational Force, Fg =

1 2

Newton’s Law of Gravitation states that a Note: Every spherical object with constant
magnitude of an attractive force between two  density can be reduced to a point mass
point masses is directly proportional to the at the centre of the sphere.
product of their masses and inversely  The gravitational forces always
proportional to the square of the distance attractive in nature and the forces
between them. always act along the line joining the
two point masses.
∝ 1 2
and
1

∝ 2

| 21| = | 12| = =
2

21 = gravitational force by particle 2 on
particle 1

12 = gravitational force by particle 1 on
particle 2

82

EXAMPLE Gravitational Field Strength, ag

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=6.001024 since = and =
kg, the mass of the Moon, mM=7.351022 kg 2
and the universal gravitational constant,
G=6.671011 N m2 kg2)

Solutions: therefore; =
2

where ;

= ℎ
= &

= 2 Note:
The gravitational field strength in the small
= 2 region near the Earth’s surface( r  R) are uniform
2 ( − )2 where its strength is 9.81 m s2 .

x2  mE Figure 1(a):
2mM Direction of close to
rEM  x2 the Earth's surface.

6.001024 Figure 1(b):
Direction of when we
2 7.351022 look at the planet from a
   x2 2  distance.

3.84108 Figure 1(c):
x Direction of when
even greater distance

x  3.32108 m

83

Variation of gravitational field strength on

the distance from the centre of the Earth EXAMPLE

Determine the magnitude of the gravitational field
strength at a point
a. 5 000 km above the Earth’s surface.
b. on the Earth’s surface.

Outside the Earth (r>R) (Given G = 6.671011 N m2 kg2, mass of the Earth,
1 M = 6.00  1024 kg and radius of the Earth, R = 6.40
 106 m)
∝ 2
Solutions:

a.

On the Earth (r=R) r Rh
r  6.40 106  5.00 106

= 2
= 9.81 −2

r  1.14 107 m


= 2
(6.67 × 10−11)(6.00 × 1024)
=
(1.14 × 107)2

= 3.08 −2

Inside the Earth (r<R)

′ b. r  R  6.40 106 m
= 2
Every spherical object with constant density

can be reduced to a point mass at the centre = 2

of the sphere (6.67 × 10−11)(6.00 × 1024)

′ ′ = (6.40 × 106)2
=

′ = 3 = 9.77 −2
3

= 3
∝

84

7.2 Gravitational Potential Energy EXAMPLE

a) Define gravitational potential energy The Moon has a mass of 7.351022 kg and a radius
b) Use gravitational potential energy, of 1740 km. A probe of mass 100 kg is dropped
from a height 1 km onto the Moon’s surface.
Calculate its change in gravitational potential
= energy.

Gravitational Potential Energy, U

Gravitational potential energy is defined as the Solutions:
work done by an external force in bringing the 1 = 1.74 × 106 + 1.00 × 103
test mass from infinity to a point. 2 = 1.74 × 106

U   GMm
r

where ; ΔU  Uf Ui

= ℎ ΔU    GMm     GMm   GMm 1  1 
= ℎ r2 r1 r1 r2
= ℎ &
  ΔU 100 1 1 
The Change in Gravitational Potential Energy,  6.67 1011 7.35 10 22 1.74110 6  1.74 10 6 
∆ 

ΔU  1.62 105 J

When the test mass moving from one point 7.3 Satellite Motion in a Circular Orbit

to another point;

ΔU  Uf Ui a) Use escape velocity,

ΔU    GMm    GMm  = √2 = √2
r2 r1

b) Use equations for satellite motion:

ΔU  GMm 1  1  i. velocity, = √
r1 r2


GMm r2  r1  ii. period, = 2 √ 3
r1r2


ΔU  Escape Velocity

If the test mass, m stay close to the Earth, then
the denominator we may replace r1 and r2 by R
(Earth’s radius), so

U  mgh

where ;
ℎ = ℎ

85

At infinity, Uf  0 and Kf  0

 Ei   Ef

Ki Ui  Kf Uf

1 2 + (− = 0
2 )

= √2


where ;

=
= ℎ &

ℎ′

If the test mass on the Earth’s surface (r =R)
hence;

rocket escaping from the atmosphere

= √2


where ; EXAMPLE
= ℎ
Calculate the escape velocity from the Earth’s
surface for a rocket of mass 3000 kg.

Since on the Earth’s surface (Given G = 6.671011 N m2 kg2, mass of the Earth,
M = 6.001024 kg, and radius of the Earth , R =
= where = = 9.81 −2 6.40106 m)
2
Solutions:
Therefore,

= √2 = √2 = √2


    
2 6.67 1011 6.001024
6.40 106

 1.12 104 m s1

86

Orbital/ Linear/ Tangential Velocity

Orbital velocity is the velocity needed to
achieve balance between gravity's pull on the
satellite and the inertia of the satellite's motion
-- the satellite's tendency to keep going.

EXAMPLE

= A satellite of mass 200 kg is placed in Earth orbit at a
height of 200 km above the surface.
GMm  mv2 (a) With a circular orbit, how long does the satellite
r2 r
take to complete one orbit?
(b) What is the satellite's speed?

= √ (Given G = 6.671011 N m2 kg2, mass of the Earth,
M = 6.001024 kg, and radius of the Earth , R =
6.40106 m)

where ; Solutions
= ℎ
= ℎ ℎ (a) r  R  200 103
 6.40 106  200 103  6.6106 m
ℎ′

If the satellite close to the Earth; T  2 r3
GM
rR and = = = 9.81 −2
2

Thus, v  gR     T  2
6.6106 3
Period Of The Satellite Orbiting Earth 6.67 1011 6.00 1024

v  r  2r = 5.32 × 103
T
(b)
2r  GM
Tr   v  GM 
r
T  2 r3 6.67 1011 6.001024
GM 6.60 106

= 7.79 × 103 −1

87

CHAPTER 7
GRAVITATION

LEARNING OUTCOMES

7.1 Gravitational Force and Field Strength

a) State and use the Newton’s Law of Gravitation, =
2

b) Define and use gravitational field strength, =
2

7.2 Gravitational Potential Energy

a) Define gravitational potential energy
b) Use gravitational potential energy, =



7.3 Satellite Motion in a Circular Orbit

a) Use escape velocity, = √2 = √2



b) Use equations for satellite motion:

i. velocity, = √



ii. period, = 2 √ 3



List of Selected Constant Values (Additional)

Gravitational constant G = 6.67×10-11 N m2 kg-2
Mass of Earth ME = 6.0×1024 kg
Radius of Earth R
Acceleration due to gravity g = 6.4106 m
Radius of the sun RS = 9.81 m s-2

= 7.0108 m

88

7.1 GRAVITATIONAL FORCE AND FIELD STRENGTH

1. The gravitational force of a particle X of mass 1on another particle Y of mass 2 is 1.
The gravitational force of Y on X is 2. If 1 > 2, which of the following statements
about the gravitational forces 1and 2 is not correct?
A. 1and 2 act along the line joining 1 and 2.
B. 1and 2are in opposite directions.
C. Magnitude of 1is greater than the magnitude of 2.
D. 1and 2form an action-reaction pair.

2. The gravitational field strength at a point in a gravitational field is
A. The gravitational attraction on a body at that point.
B. The gravitational force per unit mass at that point.
C. The weight of a body at that point.
D. The potential energy per unit mass at that point.

3. Two objects with masses m1 and m2 respectively are a distance r apart. The magnitude of the
gravitational force between them is F. The masses are changed to 2m1 and 2m2 and the
distance is changed to 4r. Find the magnitude of the new gravitational force in terms of F?

( )

4. A 200 kg object and a 500 kg object are separated by 0.40 m.
(i) Determine the net gravitational force exerted on a 50.0 kg object placed midway
between them.
(ii) At what position the 50.0 kg object should be placed so as to experience a net force of
zero?
( . × − , . )

5. The gravitational field strength at the surface of the Earth is 9.8 N kg-1.
(i) Calculate the gravitational field strength at the surface of a planet which has a radius
1.2 times the radius of the Earth and a mass of 1.5 times the mass of the Earth.
(ii) At what altitude from the surface of the earth will the acceleration due to gravity be
reduced by 10 percent ? Radius of the earth = 6.4 x 106 kg
(10.2 N kg-1, 3.37 x m)

89

7.2 GRAVITATIONAL POTENTIAL ENERGY

1. A rocket is moving away from the Earth. Which of the following about the gravitational
force on the rocket and its gravitational potential energy is correct?
A. Gravitational force decreases, gravitational potential energy increases.
B. Gravitational force decreases, gravitational potential energy decreases.
C. Gravitational force increases, gravitational potential energy increases.
D. Gravitational force increases, gravitational potential energy decreases.

2. Gravitational potential is always
A. zero
B. infinite
C. positive
D. negative

3. The Moon takes 27 days to go around the Earth in an orbit of radius 3.8 × 108 . The mass
of the Moon is 7.4 × 1022 . Calculate the gravitational potential energy.
(− . × )

7.3 SATELLITE MOTION IN A CIRCULAR ORBIT

1. Calculate the escape velocity of a helium atom from the surface of the Sun.
(Mass of a helium atom = 6.6 × 10−27 , mass of the Sun = 2.0 × 1030 )
( . × − )

2. (i) Find the speed of a satellite that orbits close to the Earth’s surface. You may assume
that the radius of the orbit of the satellite equals the radius of the Earth.

(ii) What is the period of orbit of the satellite in question (i).
( . × − , . × )

3. The orbits of the Earth and Venus are essentially circular. If the radius of Venus’ orbit is
0.723 times the radius of the Earth’s orbit, calculate the period of orbit of Venus round the
Earth.
( . )

90

8.0 Rotation of Rigid Body 8.1 Rotational Kinematics

Learning outcomes

1. Define and use:

i) angular displacement ()

ii) average angular velocity (av)

iii) instantaneous angular velocity ()

iv) average angular acceleration (av)

v) instantaneous angular acceleration ().

2. Relate parameters in rotational motion with

their corresponding quantities in linear motion:

v2

s = rθ, v = rω, at = rα , ac= rω² =

r

3. Use equations for rotational motion with

constant angular acceleration;

 = ω0 + αt, θ = ω0 t + 1 αt² and
2

ω² = ω0²+2 αθ .

4. Define torque, τ= r x F

5. Solve problems related to equilibrium of a

uniform rigid body

6. Define and use the moment of inertia of a

uniform rigid body (sphere, cylinder, ring, disc

and rod)

7. Determine the moment of inertia of a flywheel

8. State and use torque, τ =Iα.

9. Define and use angular momentum, L=Iω.

10. State and use the principle of conservation of

angular momentum.

91

92

8.2 Equilibrium of a uniform rigid body
93

94

8.3 Rotational Dynamics
95

96

8.4 Conservation of angular momentum
97

98

99