1. Miscellaneous Dynamics
Questions
e.g. (i) (1992) – variable angular velocity
The diagram shows a model train T that is
moving around a circular track, centre O
and radius a metres.
The train is travelling at a constant speed of
u m/s. The point N is in the same plane as
the track and is x metres from the nearest
point on the track. The line NO produced
meets the track at S.
Let TNS and TOS as in the diagram
4. dθ
a) Express in terms of a and u l a
dt dl d
a
dt dt
5. dθ
a) Express in terms of a and u l a
dt dl d
a
dt dt
d
ua
dt
d u
dt a
6. dθ
a) Express in terms of a and u l a
dt dl d
a
dt dt
d
ua
dt
d u
dt a
b) Show that a sin - x a sin 0 and deduce that;
d u cos
dt x a cos a cos
7. dθ
a) Express in terms of a and u l a
dt dl d
a
dt dt
d
ua
dt
d u
dt a
b) Show that a sin - x a sin 0 and deduce that;
d u cos
dt x a cos a cos
NTO TNO TOS (exterior , OTN )
8. dθ
a) Express in terms of a and u l a
dt dl d
a
dt dt
d
ua
dt
d u
dt a
b) Show that a sin - x a sin 0 and deduce that;
d u cos
dt x a cos a cos
NTO TNO TOS (exterior , OTN )
NTO
NTO
9. dθ
a) Express in terms of a and u l a
dt dl d
a
dt dt
d
ua
dt
d u
dt a
b) Show that a sin - x a sin 0 and deduce that;
d u cos
dt x a cos a cos
NTO TNO TOS (exterior , OTN )
NTO a ax
In NTO;
NTO sin sin
10. dθ
a) Express in terms of a and u l a
dt dl d
a
dt dt
d
ua
dt
d u
dt a
b) Show that a sin - x a sin 0 and deduce that;
d u cos
dt x a cos a cos
NTO TNO TOS (exterior , OTN )
NTO a ax
In NTO;
NTO sin sin
a sin a x sin
a sin a x sin 0
12. differentiate with respect to t
a cos d d a x cos d 0
dt dt dt
d d
a cos a cos a x cos 0
dt dt
13. differentiate with respect to t
a cos d d a x cos d 0
dt dt dt
d d
a cos a cos a x cos 0
dt dt
a cos a x cos d a cos u
dt a
14. differentiate with respect to t
a cos d d a x cos d 0
dt dt dt
d d
a cos a cos a x cos 0
dt dt
a cos a x cos d a cos u
dt a
d u cos
dt a cos a x cos
15. differentiate with respect to t
a cos d d a x cos d 0
dt dt dt
d d
a cos a cos a x cos 0
dt dt
a cos a x cos d a cos u
dt a
d u cos
dt a cos a x cos
d
c) Show that 0 when NT is tangential to the track.
dt
when NT is a tangent;
16. differentiate with respect to t
a cos d d a x cos d 0
dt dt dt
d d
a cos a cos a x cos 0
dt dt
a cos a x cos d a cos u
dt a
d u cos
dt a cos a x cos
d
c) Show that 0 when NT is tangential to the track.
dt T
N O
17. differentiate with respect to t
a cos d d a x cos d 0
dt dt dt
d d
a cos a cos a x cos 0
dt dt
a cos a x cos d a cos u
dt a
d u cos
dt a cos a x cos
d
c) Show that 0 when NT is tangential to the track.
dt T
when NT is a tangent;
N O
18. differentiate with respect to t
a cos d d a x cos d 0
dt dt dt
d d
a cos a cos a x cos 0
dt dt
a cos a x cos d a cos u
dt a
d u cos
dt a cos a x cos
d
c) Show that 0 when NT is tangential to the track.
dt T
when NT is a tangent;
NTO 90 (tangent radius)
N O
19. differentiate with respect to t
a cos d d a x cos d 0
dt dt dt
d d
a cos a cos a x cos 0
dt dt
a cos a x cos d a cos u
dt a
d u cos
dt a cos a x cos
d
c) Show that 0 when NT is tangential to the track.
dt T
when NT is a tangent;
NTO 90 (tangent radius)
N O
90
20. differentiate with respect to t
a cos d d a x cos d 0
dt dt dt
d d
a cos a cos a x cos 0
dt dt
a cos a x cos d a cos u
dt a
d u cos
dt a cos a x cos
d
c) Show that 0 when NT is tangential to the track.
dt T
when NT is a tangent;
NTO 90 (tangent radius)
N O
90
d u cos 90
dt a cos 90 a x cos
21. differentiate with respect to t
a cos d d a x cos d 0
dt dt dt
d d
a cos a cos a x cos 0
dt dt
a cos a x cos d a cos u
dt a
d u cos
dt a cos a x cos
d
c) Show that 0 when NT is tangential to the track.
dt T
when NT is a tangent;
NTO 90 (tangent radius)
N O
90
d u cos 90 d
0
dt a cos 90 a x cos dt
22. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
23. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
24. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T
N O
25. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T
a
N O
26. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T
a
N O
2a
27. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T
5a a
N O
2a
28. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T 2
cos
5a 5
a
N O
2a
29. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T 2
cos
5a 5
a
cos
1
N
2a
O 2 5
30. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T u cos
2 d 2
cos
5 dt
5a a a cos 2a cos
2
cos
1
N
2a
O 2 5
31. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T u cos
2 d 2
cos
5 dt
5a a a cos 2a cos
2
cos
1
1
N
2a
O 2 5 u
5
1 2a 2
a
5 5
32. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T u cos
2 d 2
cos
5 dt
5a a a cos 2a cos
2
cos
1
1
N
2a
O 2 5 u
5
1 2a 2
a
5 5
u
5a
33. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T u cos
2 d 2
cos
5 dt
5a a a cos 2a cos
2
cos
1
1
N
2a
O 2 5 u
5
when 0 1 2a 2
a
5 5
u
5a
34. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T u cos
2 d 2
cos
5 dt
5a a a cos 2a cos
2
cos
1
1
N
2a
O 2 5 u
5
when 0 1 2a 2
a
d u cos 0 5 5
dt a cos 0 2a cos 0 u
5a
35. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T u cos
2 d 2
cos
5 dt
5a a a cos 2a cos
2
cos
1
1
N
2a
O 2 5 u
5
when 0 1 2a 2
a
d u cos 0 5 5
dt a cos 0 2a cos 0 u
u
5a
3a
36. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T u cos
2 d 2
cos
5 dt
5a a a cos 2a cos
2
cos
1
1
N
2a
O 2 5 u
5
when 0 1 2a 2
a
d u cos 0 5 5
dt a cos 0 2a cos 0 u
u
5a
3a
5 u
3 5a
37. d) Suppose that x = a
3
Show that the train’s angular velocity about N when is
times the angular velocity about N when 0 2 5
when
2
T u cos
2 d 2
cos
5 dt
5a a a cos 2a cos
2
cos
1
1
N
2a
O 2 5 u
5
when 0 1 2a 2
a
d u cos 0 5 5
dt a cos 0 2a cos 0 u
u
5a
3a 3
Thus the angular velocity when is times
5 u 2 5
3 5a the angular velocity when 0
38. (ii) (2000) A string of length l is initially vertical and has a mass
P of m kg attached to it. The mass P is given a
horizontal velocity of magnitude V and begins to
move along the arc of a circle in a counterclockwise
direction.
Let O be the centre of this circle and A the initial
ds
position of P. Let s denote the arc length AP, v ,
dt
AOP and let the tension in the string be T. The acceleration due to
gravity is g and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.
39. (ii) (2000) A string of length l is initially vertical and has a mass
P of m kg attached to it. The mass P is given a
horizontal velocity of magnitude V and begins to
move along the arc of a circle in a counterclockwise
direction.
Let O be the centre of this circle and A the initial
ds
position of P. Let s denote the arc length AP, v ,
dt
AOP and let the tension in the string be T. The acceleration due to
gravity is g and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.
d 2s 1 d 1 2
a) Show that the tangential acceleration of P is given by 2 v
dt l d 2
40. (ii) (2000) A string of length l is initially vertical and has a mass
P of m kg attached to it. The mass P is given a
horizontal velocity of magnitude V and begins to
move along the arc of a circle in a counterclockwise
direction.
Let O be the centre of this circle and A the initial
ds
position of P. Let s denote the arc length AP, v ,
dt
AOP and let the tension in the string be T. The acceleration due to
gravity is g and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.
d 2s 1 d 1 2
a) Show that the tangential acceleration of P is given by 2 v
dt l d 2
s l
41. (ii) (2000) A string of length l is initially vertical and has a mass
P of m kg attached to it. The mass P is given a
horizontal velocity of magnitude V and begins to
move along the arc of a circle in a counterclockwise
direction.
Let O be the centre of this circle and A the initial
ds
position of P. Let s denote the arc length AP, v ,
dt
AOP and let the tension in the string be T. The acceleration due to
gravity is g and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.
d 2s 1 d 1 2
a) Show that the tangential acceleration of P is given by 2 v
dt l d 2
s l
ds
v
dt
d
l
dt
42. (ii) (2000) A string of length l is initially vertical and has a mass
P of m kg attached to it. The mass P is given a
horizontal velocity of magnitude V and begins to
move along the arc of a circle in a counterclockwise
direction.
Let O be the centre of this circle and A the initial
ds
position of P. Let s denote the arc length AP, v ,
dt
AOP and let the tension in the string be T. The acceleration due to
gravity is g and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.
d 2s 1 d 1 2
a) Show that the tangential acceleration of P is given by 2 v
dt l d 2
s l d s dv
2
ds dt 2
dt
v
dt
d
l
dt
43. (ii) (2000) A string of length l is initially vertical and has a mass
P of m kg attached to it. The mass P is given a
horizontal velocity of magnitude V and begins to
move along the arc of a circle in a counterclockwise
direction.
Let O be the centre of this circle and A the initial
ds
position of P. Let s denote the arc length AP, v ,
dt
AOP and let the tension in the string be T. The acceleration due to
gravity is g and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.
d 2s 1 d 1 2
a) Show that the tangential acceleration of P is given by 2 v
dt l d 2
s l d s dv
2
ds dt 2
dt
v dv d
dt
d d dt
l
dt
44. (ii) (2000) A string of length l is initially vertical and has a mass
P of m kg attached to it. The mass P is given a
horizontal velocity of magnitude V and begins to
move along the arc of a circle in a counterclockwise
direction.
Let O be the centre of this circle and A the initial
ds
position of P. Let s denote the arc length AP, v ,
dt
AOP and let the tension in the string be T. The acceleration due to
gravity is g and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.
d 2s 1 d 1 2
a) Show that the tangential acceleration of P is given by 2 v
dt l d 2
s l d s dv
2
ds dt 2
dt
v dv d
dt
d d dt
l
dt dv v
d l
45. (ii) (2000) A string of length l is initially vertical and has a mass
P of m kg attached to it. The mass P is given a
horizontal velocity of magnitude V and begins to
move along the arc of a circle in a counterclockwise
direction.
Let O be the centre of this circle and A the initial
ds
position of P. Let s denote the arc length AP, v ,
dt
AOP and let the tension in the string be T. The acceleration due to
gravity is g and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.
d 2s 1 d 1 2
a) Show that the tangential acceleration of P is given by 2 v
dt l d 2
s l d s dv
2 1 dv
v
ds dt 2 dt l d
v dv d
dt
d d dt
l
dt dv v
d l
46. (ii) (2000) A string of length l is initially vertical and has a mass
P of m kg attached to it. The mass P is given a
horizontal velocity of magnitude V and begins to
move along the arc of a circle in a counterclockwise
direction.
Let O be the centre of this circle and A the initial
ds
position of P. Let s denote the arc length AP, v ,
dt
AOP and let the tension in the string be T. The acceleration due to
gravity is g and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.
d 2s 1 d 1 2
a) Show that the tangential acceleration of P is given by 2 v
dt l d 2
s l d s dv
2 1 dv
v
ds dt 2 dt l d
v 1 dv d 1
dt dv d v2
l d dv 2
d d dt
l
dt dv v
d l
47. (ii) (2000) A string of length l is initially vertical and has a mass
P of m kg attached to it. The mass P is given a
horizontal velocity of magnitude V and begins to
move along the arc of a circle in a counterclockwise
direction.
Let O be the centre of this circle and A the initial
ds
position of P. Let s denote the arc length AP, v ,
dt
AOP and let the tension in the string be T. The acceleration due to
gravity is g and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.
d 2s 1 d 1 2
a) Show that the tangential acceleration of P is given by 2 v
dt l d 2
s l d s dv
2 1 dv
v
ds dt 2 dt l d
v 1 dv d 1
dt dv d v2
l d dv 2
d d dt
l 1 d 1 2
dt dv v v
l d 2
d l
48. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
49. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
50. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T
51. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T
mg
52. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T
mg
53. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T
mg
mg sin
54. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T m
s
mg
mg sin
55. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T m
s m mg sin
s
mg
mg sin
56. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T m
s m mg sin
s
mg g sin
s
mg sin
57. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T m
s m mg sin
s
mg g sin
s
mg sin
1 d 1 2
v g sin
l d 2
58. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T m
s m mg sin
s
mg g sin
s
mg sin
1 d 1 2
v g sin
l d 2
c) Deduce that V 2 v 2 2 gl 1 cos
59. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T m
s m mg sin
s
mg g sin
s
mg sin
1 d 1 2
v g sin
l d 2
c) Deduce that V 2 v 2 2 gl 1 cos
1 d 1 2
v g sin
l d 2
60. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T m
s m mg sin
s
mg g sin
s
mg sin
1 d 1 2
v g sin
l d 2
c) Deduce that V 2 v 2 2 gl 1 cos
1 d 1 2
v g sin
l d 2
d 1 2
v gl sin
d 2
61. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T m
s m mg sin
s
mg g sin
s
mg sin
1 d 1 2
v g sin
l d 2
c) Deduce that V 2 v 2 2 gl 1 cos
1 d 1 2
v g sin
l d 2
d 1 2
v gl sin
d 2
1 2
v gl cos c
2
v 2 2 gl cos c
62. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T m
s m mg sin
s
mg g sin
s
mg sin
1 d 1 2
v g sin
l d 2
c) Deduce that V 2 v 2 2 gl 1 cos
when 0, v V
1 d 1 2
v g sin V 2 2 gl c
l d 2
c V 2 2 gl
d 1 2
v gl sin
d 2
1 2
v gl cos c
2
v 2 2 gl cos c
63. 1 d 1 2
b) Show that the equation of motion of P is v g sin
l d 2
T m
s m mg sin
s
mg g sin
s
mg sin
1 d 1 2
v g sin
l d 2
c) Deduce that V 2 v 2 2 gl 1 cos
when 0, v V
1 d 1 2
v g sin V 2 2 gl c
l d 2
c V 2 2 gl
d 1 2
v gl sin v 2 2 gl cos V 2 2 gl
d 2
1 2 V 2 v 2 2 gl 2 gl cos
v gl cos c
2 V 2 v 2 2 gl 1 cos
v 2 2 gl cos c
69. 1 2
d) Explain why T-mg cos θ mv
l
T mg cos m T mg cos
x
m
x
70. 1 2
d) Explain why T-mg cos θ mv
l
T mg cos m T mg cos
x
m
x But, the resultant force towards the centre is
centripetal force.
71. 1 2
d) Explain why T-mg cos θ mv
l
T mg cos m T mg cos
x
m
x But, the resultant force towards the centre is
centripetal force.
mv 2
T mg cos
l
1 2
T mg cos mv
l
72. 1 2
d) Explain why T-mg cos θ mv
l
T mg cos m T mg cos
x
m
x But, the resultant force towards the centre is
centripetal force.
mv 2
T mg cos
l
1 2
T mg cos mv
l
e) Suppose that V 2 3 gl. Find the value of at which T 0
73. 1 2
d) Explain why T-mg cos θ mv
l
T mg cos m T mg cos
x
m
x But, the resultant force towards the centre is
centripetal force.
mv 2
T mg cos
l
1 2
T mg cos mv
l
e) Suppose that V 2 3 gl. Find the value of at which T 0
T mg cos mV 2 2 gl 1 cos
1
l
74. 1 2
d) Explain why T-mg cos θ mv
l
T mg cos m T mg cos
x
m
x But, the resultant force towards the centre is
centripetal force.
mv 2
T mg cos
l
1 2
T mg cos mv
l
e) Suppose that V 2 3 gl. Find the value of at which T 0
T mg cos mV 2 2 gl 1 cos
1
l
1
0 mg cos m3 gl 2 gl 1 cos
l
75. 1 2
d) Explain why T-mg cos θ mv
l
T mg cos m T mg cos
x
m
x But, the resultant force towards the centre is
centripetal force.
mv 2
T mg cos
l
1 2
T mg cos mv
l
e) Suppose that V 2 3 gl. Find the value of at which T 0
T mg cos mV 2 2 gl 1 cos
1
l
1
0 mg cos m3 gl 2 gl 1 cos
l
mg cos m g 2 g cos
76. 1 2
d) Explain why T-mg cos θ mv
l
T mg cos m T mg cos
x
m
x But, the resultant force towards the centre is
centripetal force.
mv 2
T mg cos
l
1 2
T mg cos mv
l
e) Suppose that V 2 3 gl. Find the value of at which T 0
T mg cos mV 2 2 gl 1 cos
1
l
1
0 mg cos m3 gl 2 gl 1 cos
l
mg cos m g 2 g cos
3mg cos mg
77. 1 2
d) Explain why T-mg cos θ mv
l
T mg cos m T mg cos
x
m
x But, the resultant force towards the centre is
centripetal force.
mv 2
T mg cos
l
1 2
T mg cos mv
l
e) Suppose that V 2 3 gl. Find the value of at which T 0
T mg cos mV 2 2 gl 1 cos
1
l cos
1
1
0 mg cos m3 gl 2 gl 1 cos 3
l
mg cos m g 2 g cos
3mg cos mg
78. 1 2
d) Explain why T-mg cos θ mv
l
T mg cos m T mg cos
x
m
x But, the resultant force towards the centre is
centripetal force.
mv 2
T mg cos
l
1 2
T mg cos mv
l
e) Suppose that V 2 3 gl. Find the value of at which T 0
T mg cos mV 2 2 gl 1 cos
1
l cos
1
1
0 mg cos m3 gl 2 gl 1 cos 3
l 1.911radians
mg cos m g 2 g cos
3mg cos mg
79. f) Consider the situation in part e). Briefly describe, in words, the path of
P after the tension T becomes zero.
80. f) Consider the situation in part e). Briefly describe, in words, the path of
P after the tension T becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it
would follow a parabolic arc.
Its initial velocity would be tangential to the circle with magnitude;
81. f) Consider the situation in part e). Briefly describe, in words, the path of
P after the tension T becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it
would follow a parabolic arc.
Its initial velocity would be tangential to the circle with magnitude;
1 2
T mg cos mv
l
82. f) Consider the situation in part e). Briefly describe, in words, the path of
P after the tension T becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it
would follow a parabolic arc.
Its initial velocity would be tangential to the circle with magnitude;
1 2
T mg cos mv
l
1 1 mv 2
mg
3 l
83. f) Consider the situation in part e). Briefly describe, in words, the path of
P after the tension T becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it
would follow a parabolic arc.
Its initial velocity would be tangential to the circle with magnitude;
1 2
T mg cos mv
l
1 1 mv 2
mg
3 l
gl
v2
3
gl
v
3
84. f) Consider the situation in part e). Briefly describe, in words, the path of
P after the tension T becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it
would follow a parabolic arc.
Its initial velocity would be tangential to the circle with magnitude;
1 2
T mg cos mv
l
1 1 mv 2
mg
3 l
gl
v2
3
gl
v
3
85. f) Consider the situation in part e). Briefly describe, in words, the path of
P after the tension T becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it
would follow a parabolic arc.
Its initial velocity would be tangential to the circle with magnitude;
1 2
T mg cos mv
l
1 1 mv 2
mg
3 l
gl
v2
3
gl
v
3
86. f) Consider the situation in part e). Briefly describe, in words, the path of
P after the tension T becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it
would follow a parabolic arc.
Its initial velocity would be tangential to the circle with magnitude;
1 2
T mg cos mv
l
1 1 mv 2
mg
3 l
gl
v2
3
gl
v
3
87. f) Consider the situation in part e). Briefly describe, in words, the path of
P after the tension T becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it
would follow a parabolic arc.
Its initial velocity would be tangential to the circle with magnitude;
1 2
T mg cos mv
l
1 1 mv 2
mg
3 l
gl
v2
3
gl
v
3
88. f) Consider the situation in part e). Briefly describe, in words, the path of
P after the tension T becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it
would follow a parabolic arc.
Its initial velocity would be tangential to the circle with magnitude;
1 2
T mg cos mv
l
1 1 mv 2
mg
3 l
gl
v2
3
gl
v
3
89. (iii) (2003)
A particle of mass m is thrown from the top, O, of a very tall building
with an initial velocity u at an angle of to the horizontal. The
particle experiences the effect of gravity, and a resistance
proportional to its velocity in both directions.
The equations of motion in the horizontal and
vertical directions are given respectively by
kx and ky g
x y
where k is a constant and the acceleration due
to gravity is g.
(You are NOT required to show these)
91. a) Derive the result x ue kt cos
dx
kx
dt
92. a) Derive the result x ue kt cos
dx
kx
dt x
1 dx
t
k u cos x
93. a) Derive the result x ue kt cos
dx
kx
dt x
1 dx
t
k u cos x
1
t log x u cos
x
k
94. a) Derive the result x ue kt cos
dx
kx
dt x
1 dx
t
k u cos x
1
t log x u cos
x
k
1
t log x logu cos
k
95. a) Derive the result x ue kt cos
x
t log
1
dx
kx
k u cos
dt x
1 dx
t
k u cos x
1
t log x u cos
x
k
1
t log x logu cos
k
96. a) Derive the result x ue kt cos
x
t log
1
dx
kx
k u cos
dt x
1 dx x
t kt log
k u cos x u cos
1 x
t log x u cos e kt
x
k u cos
1
t log x logu cos
x ue kt cos
k
97. a) Derive the result x ue kt cos
x
t log
1
dx
kx
k u cos
dt x
1 dx x
t kt log
k u cos x u cos
1 x
t log x u cos e kt
x
k u cos
1
t log x logu cos
x ue kt cos
k
b) Verify that y ku sin g e kt g satisfies the appropriate
1
k
equation of motion and initial condition
98. a) Derive the result x ue kt cos
x
t log
1
dx
kx
k u cos
dt x
1 dx x
t kt log
k u cos x u cos
1 x
t log x u cos e kt
x
k u cos
1
t log x logu cos
x ue kt cos
k
b) Verify that y ku sin g e kt g satisfies the appropriate
1
k
equation of motion and initial condition
dy
ky g
dt
99. a) Derive the result x ue kt cos
x
t log
1
dx
kx k u cos
dt x
1 dx x
t kt log
k u cos x u cos
1 x
t log x u cos e kt
x
k u cos
1
t log x logu cos
x ue kt cos
k
b) Verify that y ku sin g e kt g satisfies the appropriate
1
k
equation of motion and initial condition
dy
ky g
dt y
dy
t
u sin
ky g
100. a) Derive the result x ue kt cos
x
t log
1
dx
kx k u cos
dt x
1 dx x
t kt log
k u cos x u cos
1 x
t log x u cos e kt
x
k u cos
1
t log x logu cos
x ue kt cos
k
b) Verify that y ku sin g e kt g satisfies the appropriate
1
k
equation of motion and initial condition
dy
ky g
dt y
dy
t
u sin
ky g
1
t logky g u sin
y
k
101. a) Derive the result x ue kt cos
x
t log
1
dx
kx k u cos
dt x
1 dx x
t kt log
k u cos x u cos
1 x
t log x u cos e kt
x
k u cos
1
t log x logu cos
x ue kt cos
k
b) Verify that y ku sin g e kt g satisfies the appropriate
1
k
equation of motion and initial condition
dy kt logky g logku sin g
ky g
dt y
dy
t
u sin
ky g
1
t logky g u sin
y
k
102. a) Derive the result x ue kt cos
x
t log
1
dx
kx k u cos
dt x
1 dx x
t kt log
k u cos x u cos
1 x
t log x u cos e kt
x
k u cos
1
t log x logu cos
x ue kt cos
k
b) Verify that y ku sin g e kt g satisfies the appropriate
1
k
equation of motion and initial condition
dy kt logky g logku sin g
ky g
dt y ky g
dy kt log
t ku sin g
u sin
ky g
1
t logky g u sin
y
k
103. a) Derive the result x ue kt cos
x
t log
1
dx
kx k u cos
dt x
1 dx x
t kt log
k u cos x u cos
1 x
t log x u cos e kt
x
k u cos
1
t log x logu cos
x ue kt cos
k
b) Verify that y ku sin g e kt g satisfies the appropriate
1
k
equation of motion and initial condition
dy kt logky g logku sin g
ky g
dt y ky g
dy kt log
t ku sin g
u sin
ky g
ky g
e kt
1
t logky g u sin
y
ku sin g
k
104. a) Derive the result x ue kt cos
x
t log
1
dx
kx k u cos
dt x
1 dx x
t kt log
k u cos x u cos
1 x
t log x u cos e kt
x
k u cos
1
t log x logu cos
x ue kt cos
k
b) Verify that y ku sin g e kt g satisfies the appropriate
1
k
equation of motion and initial condition
dy kt logky g logku sin g
ky g
dt y ky g
dy kt log
t ku sin g
u sin
ky g
ky g
e kt
1
t logky g u sin
y
ku sin g
y ku sin g e kt g
1
k
k
105. c) Find the value of t when the particle reaches its maximum height
106. c) Find the value of t when the particle reaches its maximum height
Maximum height occurs when y 0
107. c) Find the value of t when the particle reaches its maximum height
Maximum height occurs when y 0
1
t logky g u sin
0
k
108. c) Find the value of t when the particle reaches its maximum height
Maximum height occurs when y 0
1
t logky g u sin
0
k
1
t log g logku sin g
k
109. c) Find the value of t when the particle reaches its maximum height
Maximum height occurs when y 0
1
t logky g u sin
0
k
1
t log g logku sin g
k
1 ku sin g
t log
k g
110. c) Find the value of t when the particle reaches its maximum height
Maximum height occurs when y 0
1
t logky g u sin
0
k
1
t log g logku sin g
k
1 ku sin g
t log
k g
d) What is the limiting value of the horizontal displacement of the
particle?
111. c) Find the value of t when the particle reaches its maximum height
Maximum height occurs when y 0
1
t logky g u sin
0
k
1
t log g logku sin g
k
1 ku sin g
t log
k g
d) What is the limiting value of the horizontal displacement of the
particle?
x ue kt cos
dx
ue kt cos
dt
112. c) Find the value of t when the particle reaches its maximum height
Maximum height occurs when y 0
1
t logky g u sin
0
k
1
t log g logku sin g
k
1 ku sin g
t log
k g
d) What is the limiting value of the horizontal displacement of the
particle?
x lim u cos e kt dt
x ue kt cos
t
0
dx
ue kt cos
dt
113. c) Find the value of t when the particle reaches its maximum height
Maximum height occurs when y 0
1
t logky g u sin
0
k
1
t log g logku sin g
k
1 ku sin g
t log
k g
d) What is the limiting value of the horizontal displacement of the
particle?
x lim u cos e kt dt
x ue kt cos
t
0
t
dx x lim u cos e kt
1
ue cos
kt
k 0
dt t
114. c) Find the value of t when the particle reaches its maximum height
Maximum height occurs when y 0
1
t logky g u sin
0
k
1
t log g logku sin g
k
1 ku sin g
t log
k g
d) What is the limiting value of the horizontal displacement of the
particle?
x lim u cos e kt dt
x ue kt cos
t
0
t
dx x lim u cos e kt
1
ue cos
kt
k 0
dt t
u cos
x lim e kt 1
t k
115. c) Find the value of t when the particle reaches its maximum height
Maximum height occurs when y 0
1
t logky g u sin
0
k
1
t log g logku sin g
k
1 ku sin g
t log
k g
d) What is the limiting value of the horizontal displacement of the
particle?
x lim u cos e kt dt
x ue kt cos
t
0
t
dx x lim u cos e kt
1
ue cos
kt
k 0
dt t
u cos
x lim e kt 1
t k
u cos
x
k