X2 t06 04 uniform circular motion (2013)

1. Acceleration with Uniform
Circular Motion

2. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a pa rticle moves with constant angular
velocity.

3. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a pa rticle moves with constant angular
velocity.


constantbealso ocit
y
will linear vel theof ma
g
nitude the


4. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a pa rticle moves with constant angular
velocity.
O
r
A


constantbealso ocit
y
will linear vel theof ma
g
nitude the


5. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a pa rticle moves with constant angular
velocity.

O
r
A
P


constantbealso ocit
y
will linear vel theof ma
g
nitude the

A particle moves from A

to P with constant
angular velocity.

6. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a pa rticle moves with constant angular
velocity.

O
r
A
P


constantbealso ocit
y
will linear vel theof ma
g
nitude the

A particle moves from A

to P with constant
angular velocity.
The acceleration of the particle is the change
in velocity with respect to time.

7. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a pa rticle moves with constant angular
velocity.

O
r
A
P
v


constantbealso ocit
y
will linear vel theof ma
g
nitude the

A particle moves from A

to P with constant
angular velocity.
The acceleration of the particle is the change
in velocity with respect to time.

8. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a pa rticle moves with constant angular
velocity.

O
r
A
P
v


constantbealso ocit
y
will linear vel theof ma
g
nitude the

v’
A particle moves from A

to P with constant
angular velocity.
The acceleration of the particle is the change
in velocity with respect to time.

9. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a pa rticle moves with constant angular
velocity.

O
r
A
P
v


constantbealso ocit
y
will linear vel theof ma
g
nitude the

v’
A particle moves from A

to P with constant
angular velocity.
The acceleration of the particle is the change
in velocity with respect to time.
v

10. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a pa rticle moves with constant angular
velocity.

O
r
A
P
v


constantbealso ocit
y
will linear vel theof ma
g
nitude the

v’
A particle moves from A

to P with constant
angular velocity.
The acceleration of the particle is the change
in velocity with respect to time.
v
v’

11. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a pa rticle moves with constant angular
velocity.

O
r
A
P
v


constantbealso ocit
y
will linear vel theof ma
g
nitude the

v’
A particle moves from A

to P with constant
angular velocity.
The acceleration of the particle is the change
in velocity with respect to time.
v
v’
v


12. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a pa rticle moves with constant angular
velocity.

O
r
A
P
v


constantbealso ocit
y
will linear vel theof ma
g
nitude the

v’
A particle moves from
A

to
P
with constant
angular velocity.
The acceleration of the particle is the change
in velocity with respect to time.
v
v’
v


13. Acceleration with Uniform
Circular Motion
Uniform circular motion is when a pa rticle moves with constant angular
velocity.

O
r
A
P
v


constantbealso ocit
y
will linear vel theof ma
g
nitude the

v’
A particle moves from
A

to
P
with constant
angular velocity.
The acceleration of the particle is the change
in velocity with respect to time.
v
v’
v

OAP similar tois
vectorsoftriangle This


14. v
v’
v


O
r
A
r


15. v
v’
v


O
r
A
P
r

r
v
AP
v



16. v
v’
v


O
r
A
P
r

r
v
AP
v


rAP
v
v



17. v
v’
v


O
r
A
P
r

r
v
AP
v


rAP
v
v


tv
a




18. v
v’
v


O
r
A
P
r

r
v
AP
v


rAP
v
v


tv
a



t
r
AP
v
a




19. v
v’
v


O
r
A
P
r

r
v
AP
v


rAP
v
v


tv
a



t
r
AP
v
a



arcAP
AP
Δ
t

,0asBut,

20. v
v’
v


O
r
A
P
r

r
v
AP
v


rAP
v
v


tv
a



t
r
AP
v
a



arcAP
AP
Δ
t

,0asBut,
t
r
arcA
P
v
a
t





0
lim

21. v
v’
v


O
r
A
P
r

r
v
AP
v


rAP
v
v


tv
a



t
r
AP
v
a



arcAP
AP
Δ
t

,0asBut,
t
r
arcA
P
v
a
t





0
lim
t
varcA
P





timespee
d
distance

22. v
v’
v


O
r
A
P
r

r
v
AP
v


rAP
v
v


tv
a



t
r
AP
v
a



arcAP
AP
Δ
t

,0asBut,
t
r
arcA
P
v
a
t





0
lim
t
varcA
P





timespee
d
distance
r
v
r
v
tr
t
vv
a
t
t
2
2
0
0

lim
lim











23. v
v’
v
O
r
A
P
r

r
v
AP
v

rAP
v
v


tv
a



t
r
AP
v
a



arcAP
AP
Δ
t

,0asBut,
t
r
arcA
P
v
a
t





0
lim
t
varcA
P





timespeed distance
r
v
r
v
tr
t
vv
a
t
t
2
2
0
0

lim
lim










centrethetowards directed
is and magnitude hasmotion
circular uniformin on accelerati the
2
r
v


24. r
v
a
2

Acceleration Involved in
Uniform Circular Motion

25. r
v
a
2

2

ra
OR

Acceleration Involved in
Uniform Circular Motion

26. r
v
a
2

2

ra
OR

Acceleration Involved in
Uniform Circular Motion
Forces Involved in
Uniform Circular Motion

27. r
v
a
2

2

ra
OR

Acceleration Involved in
Uniform Circular Motion
r
mv
F
2

Forces Involved in
Uniform Circular Motion

28. r
v
a
2

2

ra
OR

Acceleration Involved in
Uniform Circular Motion
r
mv
F
2

2

mr
F
OR

Forces Involved in
Uniform Circular Motion

29. r
v
a
2

2

ra
OR

Acceleration Involved in
Uniform Circular Motion
r
mv
F
2

2

mr
F
OR

Forces Involved in
Uniform Circular Motion
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;

t ry
r
x
 

 

where,sin
cos

30. r
v
a
2

2

ra
OR

Acceleration Involved in
Uniform Circular Motion
r
mv
F
2

2

mr
F
OR

Forces Involved in
Uniform Circular Motion
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;

t ry
r
x
 

 

where,sin
cos
.on
acting magnitude of force radial inwardan is e that ther Show a)
2
P
mr


31. y
x
P

r

cos
r
x


sin
ry

32. y
x
P

r

cos
r
x


sin
ry

cos
r
x


33. y
x
P

r

cos
r
x


sin
ry

cos
r
x


sin
ry


34. y
x
P

r

cos
r
x


sin
ry

cos
r
x


sin
ry





sin
sin
r
dt
d
rx


 

35. y
x
P

r

cos r
x


sin ry

cos r
x


sin ry





sin
sin
r
dt
d
rx


 




cos
cos
r
dt
d
ry

 

36. y
x
P

r

cos r
x


sin ry

cos r
x


sin ry





sin
sin
r
dt
d
rx


 




cos
cos
r
dt
d
ry

 
x
r
dt
d
rx
2
2
cos
cos






 

37. y
x
P

r

cos r
x


sin ry

cos r
x


sin ry





sin
sin
r
dt
d
rx


 




cos
cos
r
dt
d
ry

 
x
r
dt
d
rx
2
2
cos
cos






 
y
r
dt
d
ry
2
2
sin
sin




 



38. y
x
P

r

cos r
x


sin ry

cos r
x


sin ry





sin
sin
r
dt
d
rx








cos
cos
r
dt
d
ry



x
r
dt
d
rx
2
2
cos
cos








y
r
dt
d
ry
2
2
sin
sin




 


y
x

39. y
x
P

r

cos
r
x


sin
ry


cos
r
x


sin
ry





sin
sin
r
dt
d
rx








cos
cos
r
dt
d
ry



x
r
dt
d
rx
2
2
cos
cos








y
r
dt
d
ry
2
2
sin
sin




 


y
x
a


40. y
x
P

r

cos
r
x


sin
ry


cos
r
x


sin
ry





sin
sin
r
dt
d
rx








cos
cos
r
dt
d
ry



x
r
dt
d
rx
2
2
cos
cos








y
r
dt
d
ry
2
2
sin
sin




 


y
x
a




22 2
yxa 


41. y
x
P

r

cos
r
x


sin
ry

cos
r
x


sin
ry





sin
sin
r
dt
d
rx


 




cos
cos
r
dt
d
ry

 
x
r
dt
d
rx
2
2
cos
cos






 
y
r
dt
d
ry
2
2
sin
sin




 
 
y
x
a




22 2
yxa 

24
224
2424
r
yx
yx









42. y
x
P

r

cos
r
x


sin
ry

cos
r
x


sin
ry





sin
sin
r
dt
d
rx


 




cos
cos
r
dt
d
ry

 
x
r
dt
d
rx
2
2
cos
cos






 
y
r
dt
d
ry
2
2
sin
sin




 
 
y
x
a




22 2
yxa 

24
224
2424
r
yx
yx







2

ra



43. y
x
P

r

cos
r
x


sin
ry

cos
r
x


sin
ry





sin
sin
r
dt
d
rx


 




cos
cos
r
dt
d
ry

 
x
r
dt
d
rx
2
2
cos
cos






 
y
r
dt
d
ry
2
2
sin
sin




 
 
y
x
a




22 2
yxa 

24
224
2424
r
yx
yx







2

ra

2

mr
F
ma
F





44. y
x
P

r

cos
r
x


sin
ry

cos
r
x


sin
ry





sin
sin
r
dt
d
rx


 




cos
cos
r
dt
d
ry

 
x
r
dt
d
rx
2
2
cos
cos






 
y
r
dt
d
ry
2
2
sin
sin




 
 
y
x
a




22 2
yxa 

24
224
2424
r
yx
yx







2

ra

2

mr
F
ma
F




xy


1
tan




45. y
x
P

r

cos
r
x


sin
ry

cos
r
x


sin
ry





sin
sin
r
dt
d
rx


 




cos
cos
r
dt
d
ry

 
x
r
dt
d
rx
2
2
cos
cos






 
y
r
dt
d
ry
2
2
sin
sin




 
 
y
x
a




22 2
yxa 

24
224
2424
r
yx
yx







2

ra

2

mr
F
ma
F




xy


1
tan



















x
y
x
y
1
2
2
1
tan
tan

46. y
x
P

r

cos
r
x


sin
ry

cos
r
x


sin
ry





sin
sin
r
dt
d
rx


 




cos
cos
r
dt
d
ry

 
x
r
dt
d
rx
2
2
cos
cos






 
y
r
dt
d
ry
2
2
sin
sin




 
 
y
x
a




22 2
yxa 

24
224
2424
r
yx
yx







2

ra

2

mr
F
ma
F




xy


1
tan



















x
y
x
y
1
2
2
1
tan
tan
centrethes towar
d
acting,, force a is There
2

mrF



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 where
A

is a constant. By considering all other forces on the satellite to be
negligible, show that;

2
rA
m
3
2

A
r

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 where
A

is a constant. By considering all other forces on the satellite to be
negligible, show that;

2
rA
m
3
2

A
r

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 where
A

is a constant. By considering all other forces on the satellite to be
negligible, show that;

2
rA
m
3
2

A
r
2

mr
x
m



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 where
A

is a constant. By considering all other forces on the satellite to be
negligible, show that;

2
rA
m
3
2

A
r
2

mr
x
m


2
rA
m

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 where
A

is a constant. By considering all other forces on the satellite to be
negligible, show that;

2
rA
m
3
2

A
r
2
2
r
A
m
mr

2

mr
x
m


2
rA
m

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 where
A

is a constant. By considering all other forces on the satellite to be
negligible, show that;

2
rA
m
3
2

A
r
2
2
r
A
m
mr

2
2
3


A
m
A
m
r


2

mr
x
m


2
rA
m

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 where
A

is a constant. By considering all other forces on the satellite to be
negligible, show that;

2
rA
m
3
2

A
r
2
2
r
A
m
mr

2
2
3


A
m
A
m
r


3
2

A
r
2

mr
x
m


2
rA
m

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 veloc ity 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 veloc ity 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 veloc ity which may be imparted without
breaking the string.
x
m

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 veloc ity which may be imparted without
breaking the string.
x
m
m
g

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 veloc ity which may be imparted without
breaking the string.
x
m
m
g
T

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 veloc ity which may be imparted without
breaking the string.
x
m
m
g
T
m
g
x
m



T

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 veloc ity which may be imparted without
breaking the string.
x
m
m
g
T
m
g
x
m



T
T
m
g


0

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 veloc ity which may be imparted without
breaking the string.
x
m
m
g
T
m
g
x
m







N
T
392
8.940

T
T
m
g


0

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 veloc ity which may be imparted without
breaking the string.
x
m
m
g
T
m
g
x
m







N
T
392
8.940

T
T
m
g


0

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 veloc ity which may be imparted without
breaking the string.
x
m
m
g
T
m
g
x
m







N
T
392
8.940

T
T
m
g


0
x
m

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 veloc ity which may be imparted without
breaking the string.
x
m
m
g
T
m
g
x
m







N
T
392
8.940

T
T
m
g


0
x
m T

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 veloc ity which may be imparted without
breaking the string.
x
m
m
g
T
m
g
x
m







N
T
392
8.940

2

mr
T

T
T
m
g


0
x
m T

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 veloc ity which may be imparted without
breaking the string.
x
m
m
g
T
m
g
x
m







N
T
392
8.940

2

mr
T





2
5.02392


T
T
m
g


0
x
m T

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. x
m
m
g
T
m
g
x
m







N
T
392
8.940

2

mr
T





2
5.02392


rad/s982
392
392
2



T
T
m
g


0
x
m T

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. x
m
m
g
T
m
g
x
m







N
T
392
8.940

2

mr
T





2
5.02392


rad/s982
392
392
2



Exercise 9B; all
T
T
m
g


0
x
m T