2. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a particle moves with constant angular
velocity.
3. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a particle moves with constant angular
velocity. the magnitude of the linear velocity will also be constant
4. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a particle moves with constant angular
velocity. the magnitude of the linear velocity will also be constant
A
r
O
5. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a particle moves with constant angular
velocity. the magnitude of the linear velocity will also be constant
A particle moves from A to P with constant
A
angular velocity.
r
O
P
6. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a particle moves with constant angular
velocity. the magnitude of the linear velocity will also be constant
A particle moves from A to P with constant
A
angular velocity.
r
O
The acceleration of the particle is the change
in velocity with respect to time.
P
7. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a particle moves with constant angular
velocity. the magnitude of the linear velocity will also be constant
A particle moves from A to P with constant
A
angular velocity.
r v
O
The acceleration of the particle is the change
in velocity with respect to time.
P
8. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a particle moves with constant angular
velocity. the magnitude of the linear velocity will also be constant
A particle moves from A to P with constant
A
angular velocity.
r v
O
The acceleration of the particle is the change
in velocity with respect to time.
P
v’
9. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a particle moves with constant angular
velocity. the magnitude of the linear velocity will also be constant
A particle moves from A to P with constant
A
angular velocity.
r v
O
The acceleration of the particle is the change
in velocity with respect to time.
P
v’ v
10. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a particle moves with constant angular
velocity. the magnitude of the linear velocity will also be constant
A particle moves from A to P with constant
A
angular velocity.
r v
O
The acceleration of the particle is the change
in velocity with respect to time.
P
v’ v
v’
11. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a particle moves with constant angular
velocity. the magnitude of the linear velocity will also be constant
A particle moves from A to P with constant
A
angular velocity.
r v
O
The acceleration of the particle is the change
in velocity with respect to time.
P
v’ v
v’
v
12. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a particle moves with constant angular
velocity. the magnitude of the linear velocity will also be constant
A A particle moves from A to P with constant
angular velocity.
r v
O
The acceleration of the particle is the change
in velocity with respect to time.
P
v’ v’ v
v
13. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a particle moves with constant angular
velocity. the magnitude of the linear velocity will also be constant
A A particle moves from A to P with constant
angular velocity.
r v
O
The acceleration of the particle is the change
in velocity with respect to time.
P
v’ v
v’ This triangle of vectors
is similar to OAP
v
16. A
v’
v r
O
v r
P
v v
AP r
v AP
v
r
17. A
v’
v r
O
v r
P
v v
AP r
v AP
v
r
v
a
t
18. A
v’
v r
O
v r
P
v v
AP r
v AP
v
r
v
a
t
v AP
a
r t
19. A
v’
v r
O
v r
P
v v
AP r
v AP
v
r
v
a
t
v AP
a
r t
But, as Δt 0, AP arcAP
20. A
v’
v r
O
v r
P
v v
AP r
v AP
v
r
v
a
t
v AP
a
r t
But, as Δt 0, AP arcAP
v arcAP
a lim
t 0 r t
21. A
v’
v r
distance speed time
O
arcAP v t
v r
P
v v
AP r
v AP
v
r
v
a
t
v AP
a
r t
But, as Δt 0, AP arcAP
v arcAP
a lim
t 0 r t
22. A
v’ v r
distance speed time
O
arcAP v t
v r
P
v v
v v t
AP r a lim
t 0 r t
v AP
v
r v2
lim
v t 0 r
a
t v2
v AP r
a
r t
But, as Δt 0, AP arcAP
v arcAP
a lim
t 0 r t
23. A
v’ v r
distance speed time
O
arcAP v t
v r
P
v v
v v t
AP r a lim
t 0 r t
v AP
v
r v2
lim
v t 0 r
a
t v2
v AP r
a
r t
the acceleration in uniform circular
But, as Δt 0, AP arcAP
v2
v arcAP motion has magnitude and is
a lim r
t 0 r t
directed towards the centre
26. Acceleration Involved in Forces Involved in
Uniform Circular Motion Uniform Circular Motion
v2
a
r
OR
a r 2
27. Acceleration Involved in Forces Involved in
Uniform Circular Motion Uniform Circular Motion
v2 mv 2
a F
r r
OR
a r 2
28. Acceleration Involved in Forces Involved in
Uniform Circular Motion Uniform Circular Motion
v2 mv 2
a F
r r
OR OR
a r 2 F mr 2
29. Acceleration Involved in Forces Involved in
Uniform Circular Motion Uniform Circular Motion
v2 mv 2
a F
r r
OR OR
a r 2 F mr 2
e.g. (i) (2003)
A particle P of mass m moves with constant angular velocity
on a circle of radius r. Its position at time t is given by;
x r cos
y r sin , where t
30. Acceleration Involved in Forces Involved in
Uniform Circular Motion Uniform Circular Motion
v2 mv 2
a F
r r
OR OR
a r 2 F mr 2
e.g. (i) (2003)
A particle P of mass m moves with constant angular velocity
on a circle of radius r. Its position at time t is given by;
x r cos
y r sin , where t
a) Show that there is an inward radial force of magnitude mr 2 acting
on P.
33. x r cos y y r sin
P
r y r sin
x r cos x
34. x r cos y y r sin
d
x r sin
P
dt
r y r sin
r sin
x r cos x
35. x r cos y y r sin
d d
x r sin
P y r cos
dt dt
r y r sin
r sin r cos
x r cos x
36. x r cos y y r sin
d d
x r sin
P y r cos
dt dt
r y r sin
r sin r cos
d x r cos x
r cos
x
dt
r 2 cos
2 x
37. x r cos y y r sin
d d
x r sin
P y r cos
dt dt
r y r sin
r sin r cos
r cos
x
d x r cos x r sin d
y
dt dt
r 2 cos r 2 sin
x2 2 y
38. x r cos y y r sin
d d
x r sin
P y r cos
dt dt
r y r sin
r sin r cos
r cos
x
d x r cos x r sin d
y
dt dt
r 2 cos r 2 sin
x 2 2 y
y
x
39. x r cos y y r sin
d d
x r sin
P y r cos
dt dt
r y r sin
r sin r cos
r cos
x
d x r cos x r sin d
y
dt dt
r 2 cos r 2 sin
x 2 2 y
a y
x
40. x r cos y y r sin
d d
x r sin
P y r cos
dt dt
r y r sin
r sin r cos
r cos
x
d x r cos x r sin d
y
dt dt
r 2 cos r 2 sin
x 2
a
2
x
2
y
2 2 y
a y
x
41. x r cos y y r sin
d d
x r sin
P y r cos
dt dt
r y r sin
r sin r cos
r cos
x
d x r cos x r sin d
y
dt dt
r 2 cos r 2 sin
x 2
a
2
x
2
y
2 2 y
4 x2 4 y2
4 x 2 y 2
a y
4r 2
x
42. x r cos y y r sin
d d
x r sin
P y r cos
dt dt
r y r sin
r sin r cos
r cos
x
d x r cos x r sin d
y
dt dt
r 2 cos r 2 sin
x 2
a
2
x y
2 2 2 y
4 x2 4 y2
4 x 2 y 2
a y
4r 2
a r 2
x
43. x r cos y y r sin
d d
x r sin
P y r cos
dt dt
r y r sin
r sin r cos
r cos
x
d x r cos x r sin d
y
dt dt
r 2 cos r 2 sin
x 2
a
2
x y
2 2 2 y
4 x2 4 y2
4 x 2 y 2
a y
4r 2
a r 2
x
F ma
F mr 2
44. x r cos y y r sin
d d
x r sin
P y r cos
dt dt
r y r sin
r sin r cos
r cos
x
d x r cos x r sin d
y
dt dt
r 2 cos r 2 sin
y
x 2
a
2
x y
2 2 tan 1 2 y
x
4 x2 4 y2
4 x 2 y 2
a y
4r 2
a r 2
x
F ma
F mr 2
45. x r cos y y r sin
d d
x r sin
P y r cos
dt dt
r y r sin
r sin r cos
r cos
x
d x r cos x r sin d
y
dt dt
r 2 cos r 2 sin
y
x 2
a
2
x y
2 2 tan 1 2 y
x
4 x2 4 y2 2 y
tan 1
4 x 2 y 2 x
2
a y
4r 2 y
tan 1
a r 2 x
F ma
x
F mr 2
46. x r cos y y r sin
d d
x r sin
P y r cos
dt dt
r y r sin
r sin r cos
r cos
x
d x r cos x r sin d
y
dt dt
r 2 cos r 2 sin
y
x 2
a
2
x y
2 2 tan 1 2 y
x
4 x2 4 y2 2 y
tan 1
4 x 2 y 2 x
2
a y
4r 2 y
tan 1
a r 2 x
F ma
x
F mr 2 There is a force , F mr 2 , acting
towards the centre
47. b) A telecommunications satellite, of mass m, orbits Earth with constant
angular velocity at a distance r from the centre of the Earth.
The gravitational force exerted by Earth on the satellite is Am where
r2
A is a constant. By considering all other forces on the satellite to be
negligible, show that;
A
r 3
2
48. b) A telecommunications satellite, of mass m, orbits Earth with constant
angular velocity at a distance r from the centre of the Earth.
The gravitational force exerted by Earth on the satellite is Am where
r2
A is a constant. By considering all other forces on the satellite to be
negligible, show that;
A
r 3
2
49. b) A telecommunications satellite, of mass m, orbits Earth with constant
angular velocity at a distance r from the centre of the Earth.
The gravitational force exerted by Earth on the satellite is Am where
r2
A is a constant. By considering all other forces on the satellite to be
negligible, show that;
A
r 3
2
m mr 2
x
50. b) A telecommunications satellite, of mass m, orbits Earth with constant
angular velocity at a distance r from the centre of the Earth.
The gravitational force exerted by Earth on the satellite is Am where
r2
A is a constant. By considering all other forces on the satellite to be
negligible, show that;
A
r 3
2
Am
m mr 2
x
r2
51. b) A telecommunications satellite, of mass m, orbits Earth with constant
angular velocity at a distance r from the centre of the Earth.
The gravitational force exerted by Earth on the satellite is Am where
r2
A is a constant. By considering all other forces on the satellite to be
negligible, show that;
A
r 3
2
Am
mr 2
Am r2
m mr 2
x
r2
52. b) A telecommunications satellite, of mass m, orbits Earth with constant
angular velocity at a distance r from the centre of the Earth.
The gravitational force exerted by Earth on the satellite is Am where
r2
A is a constant. By considering all other forces on the satellite to be
negligible, show that;
A
r 3
2
Am
mr 2
Am r2
m mr 2
x
r2 Am
r3
m 2
A
2
53. b) A telecommunications satellite, of mass m, orbits Earth with constant
angular velocity at a distance r from the centre of the Earth.
The gravitational force exerted by Earth on the satellite is Am where
r2
A is a constant. By considering all other forces on the satellite to be
negligible, show that;
A
r 3
2
Am
mr 2
Am r2
m mr 2
x
r2 Am
r3
m 2
A
2
A
r3
2
54. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string.
55. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string.
56. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string.
m
x
57. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string.
m
x mg
58. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string.
T
m
x mg
59. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string.
m mg T
x
T
m
x mg
60. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string.
m mg T
x
T 0 mg T
m
x mg
61. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string.
m mg T
x
T 0 mg T
T 40 9.8
392 N
m
x mg
62. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string.
m mg T
x
T 0 mg T
T 40 9.8
392 N
m
x mg
63. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string. m
x
m mg T
x
T 0 mg T
T 40 9.8
392 N
m
x mg
64. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string. m
x
T
m mg T
x
T 0 mg T
T 40 9.8
392 N
m
x mg
65. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string. m
x
T
m mg T
x T mr 2
T 0 mg T
T 40 9.8
392 N
m
x mg
66. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string. m
x
T
m mg T
x T mr 2
0 mg T
T 392 2 0.5 2
T 40 9.8
392 N
m
x mg
67. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string. m
x
T
m mg T
x T mr 2
0 mg T
T 392 2 0.5 2
T 40 9.8
2 392
392 N
mg
392
m
x
2 98rad/s
68. (ii) A string is 50cm long and it will break if a ,mass exceeding 40kg is
hung from it.
A mass of 2kg is attached to one end of the string and it is revolved
in a circle.
Find the greatest angular velocity which may be imparted without
breaking the string. m
x
T
m mg T
x T mr 2
0 mg T
T 392 2 0.5 2
T 40 9.8
2 392
392 N
mg
392
m
x
2 98rad/s
Exercise 9B; all