Acceleration with Uniform
Circular Motion

Acceleration with Uniform
Circular Motion
Uniform circular motion is when a particle moves with constant angular
velocity.

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 

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

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

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

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

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’

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

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’

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

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

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

A
v’ 
v r
O

v r

A
v’ 
v r
O

v r
P
v v

AP r

A
v’ 
v r
O

v r
P
v v

AP r
v  AP
v 
r

A
v’ 
v r
O

v r
P
v v

AP r
v  AP
v 
r
v
a
t

A
v’ 
v r
O

v r
P
v v

AP r
v  AP
v 
r
v
a
t
v  AP
a
r  t

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

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

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

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

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

Acceleration Involved in
Uniform Circular Motion
v2
a
r

Acceleration Involved in
Uniform Circular Motion
v2
a
r
OR
a  r 2

Acceleration Involved in Forces Involved in
Uniform Circular Motion Uniform Circular Motion
v2
a
r
OR
a  r 2

Acceleration Involved in Forces Involved in
Uniform Circular Motion Uniform Circular Motion
v2 mv 2
a F
r r
OR
a  r 2

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

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

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.

y
P
r y  r sin 

x  r cos x

x  r cos y
P
r y  r sin 

x  r cos x

x  r cos y y  r sin 
P
r y  r sin 

x  r cos x

x  r cos y y  r sin 
d
x   r sin  
 P
dt
r y  r sin 
  r sin 

x  r cos x

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

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

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

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


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


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


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


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


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

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

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

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

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

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

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

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

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

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


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
r3
2

(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.

(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.

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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

(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