1) The document describes a model train T moving in a circular track with radius a meters at a constant speed u m/s. A point N is x meters from the track. 2) Expressions are derived for the angular velocity dθ/dt of the train in terms of x, a, and u. It is shown that when N is a tangent to the track, the angular velocity is 0. 3) Further relationships between the angles and distances in the diagram are deduced, including an expression relating the rate of change of the angle φ to the coordinates x and a.

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.
diagramin theasandLet   TOSTNS

2. ua
dt
dθ
andofin termsExpressa)

3. ua
dt
dθ
andofin termsExpressa) al 

4. ua
dt
dθ
andofin termsExpressa) al 
dt
d
a
dt
dl 


5. ua
dt
dθ
andofin termsExpressa) al 
dt
d
a
dt
dl 

a
u
dt
d
dt
d
au





6. ua
dt
dθ
andofin termsExpressa) al 
dt
d
a
dt
dl 

a
u
dt
d
dt
d
au




   
 
   






coscos
cos
that;deduceand0sinsinthatShowb)
aax
u
dt
d
ax-a

7. ua
dt
dθ
andofin termsExpressa) al 
dt
d
a
dt
dl 

a
u
dt
d
dt
d
au




   
 
   






coscos
cos
that;deduceand0sinsinthatShowb)
aax
u
dt
d
ax-a
),(exterior OTNTOSTNONTO 

8. ua
dt
dθ
andofin termsExpressa) al 
dt
d
a
dt
dl 

a
u
dt
d
dt
d
au




   
 
   






coscos
cos
that;deduceand0sinsinthatShowb)
aax
u
dt
d
ax-a
),(exterior OTNTOSTNONTO 




NTO
NTO

9. ua
dt
dθ
andofin termsExpressa) al 
dt
d
a
dt
dl 

a
u
dt
d
dt
d
au




   
 
   






coscos
cos
that;deduceand0sinsinthatShowb)
aax
u
dt
d
ax-a
),(exterior OTNTOSTNONTO 




NTO
NTO
  


sinsin
;In
xaa
NTO

10. ua
dt
dθ
andofin termsExpressa) al 
dt
d
a
dt
dl 

a
u
dt
d
dt
d
au




   
 
   






coscos
cos
that;deduceand0sinsinthatShowb)
aax
u
dt
d
ax-a
),(exterior OTNTOSTNONTO 




NTO
NTO
  


sinsin
;In
xaa
NTO
   
    0sinsin
sinsin




xaa
xaa

11.     0coscos
respect towithatedifferenti





 
dt
d
xa
dt
d
dt
d
a
t





12.     0coscos
respect towithatedifferenti





 
dt
d
xa
dt
d
dt
d
a
t




       0coscoscos 
dt
d
xaa
dt
d
a





13.     0coscos
respect towithatedifferenti





 
dt
d
xa
dt
d
dt
d
a
t




       0coscoscos 
dt
d
xaa
dt
d
a




      
a
u
a
dt
d
xaa  

 coscoscos

14.     0coscos
respect towithatedifferenti





 
dt
d
xa
dt
d
dt
d
a
t




       0coscoscos 
dt
d
xaa
dt
d
a




      
a
u
a
dt
d
xaa  

 coscoscos
 
    

coscos
cos
xaa
u
dt
d




15.     0coscos
respect towithatedifferenti





 
dt
d
xa
dt
d
dt
d
a
t




       0coscoscos 
dt
d
xaa
dt
d
a




      
a
u
a
dt
d
xaa  

 coscoscos
 
    

coscos
cos
xaa
u
dt
d



track.thetoltangentiaiswhen0thatShowc) NT
dt
d


when NT is a tangent;

16.     0coscos
respect towithatedifferenti





 
dt
d
xa
dt
d
dt
d
a
t




       0coscoscos 
dt
d
xaa
dt
d
a




      
a
u
a
dt
d
xaa  

 coscoscos
 
    

coscos
cos
xaa
u
dt
d



track.thetoltangentiaiswhen0thatShowc) NT
dt
d


N
T
O

17.     0coscos
respect towithatedifferenti





 
dt
d
xa
dt
d
dt
d
a
t




       0coscoscos 
dt
d
xaa
dt
d
a




      
a
u
a
dt
d
xaa  

 coscoscos
 
    

coscos
cos
xaa
u
dt
d



track.thetoltangentiaiswhen0thatShowc) NT
dt
d


N
T
O
when NT is a tangent;

18.     0coscos
respect towithatedifferenti





 
dt
d
xa
dt
d
dt
d
a
t




       0coscoscos 
dt
d
xaa
dt
d
a




      
a
u
a
dt
d
xaa  

 coscoscos
 
    

coscos
cos
xaa
u
dt
d



track.thetoltangentiaiswhen0thatShowc) NT
dt
d


N
T
O
when NT is a tangent;
radius)(tangent90  
NTO

19.     0coscos
respect towithatedifferenti





 
dt
d
xa
dt
d
dt
d
a
t




       0coscoscos 
dt
d
xaa
dt
d
a




      
a
u
a
dt
d
xaa  

 coscoscos
 
    

coscos
cos
xaa
u
dt
d



track.thetoltangentiaiswhen0thatShowc) NT
dt
d


N
T
O
when NT is a tangent;
radius)(tangent90  
NTO

90 

20.     0coscos
respect towithatedifferenti





 
dt
d
xa
dt
d
dt
d
a
t




       0coscoscos 
dt
d
xaa
dt
d
a




      
a
u
a
dt
d
xaa  

 coscoscos
 
    

coscos
cos
xaa
u
dt
d



track.thetoltangentiaiswhen0thatShowc) NT
dt
d


N
T
O
when NT is a tangent;
radius)(tangent90  
NTO

90 
  

cos90cos
90cos
xaa
u
dt
d

 


21.     0coscos
respect towithatedifferenti





 
dt
d
xa
dt
d
dt
d
a
t




       0coscoscos 
dt
d
xaa
dt
d
a




      
a
u
a
dt
d
xaa  

 coscoscos
 
    

coscos
cos
xaa
u
dt
d



track.thetoltangentiaiswhen0thatShowc) NT
dt
d


N
T
O
when NT is a tangent;
radius)(tangent90  
NTO

90 
  

cos90cos
90cos
xaa
u
dt
d

 

0
dt
d

22. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
5
3
0

23. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0

24. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON


25. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON

a

26. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON

a
2a

27. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON

a
2a
a5

28. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON

a
2a
a5 5
2
cos 

29. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON

a
2a
a5 5
2
cos 
5
1
2
cos 




 


30. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON

a
2a
a5 5
2
cos 
5
1
2
cos 




 






cos2
2
cos
2
cos
aa
u
dt
d












 


31. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON

a
2a
a5 5
2
cos 
5
1
2
cos 




 






cos2
2
cos
2
cos
aa
u
dt
d












 

  

















5
2
2
5
1
5
1
aa
u

32. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON

a
2a
a5 5
2
cos 
5
1
2
cos 




 






cos2
2
cos
2
cos
aa
u
dt
d












 

  

















5
2
2
5
1
5
1
aa
u
a
u
5


33. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON

a
2a
a5 5
2
cos 
5
1
2
cos 




 






cos2
2
cos
2
cos
aa
u
dt
d












 

  

















5
2
2
5
1
5
1
aa
u
a
u
5

0when 

34. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON

a
2a
a5 5
2
cos 
5
1
2
cos 




 






cos2
2
cos
2
cos
aa
u
dt
d












 

  

















5
2
2
5
1
5
1
aa
u
a
u
5

0when 
0cos20cos
0cos
aa
u
dt
d




35. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON

a
2a
a5 5
2
cos 
5
1
2
cos 




 






cos2
2
cos
2
cos
aa
u
dt
d












 

  

















5
2
2
5
1
5
1
aa
u
a
u
5

0when 
0cos20cos
0cos
aa
u
dt
d



a
u
3


36. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON

a
2a
a5 5
2
cos 
5
1
2
cos 




 






cos2
2
cos
2
cos
aa
u
dt
d












 

  

















5
2
2
5
1
5
1
aa
u
a
u
5

0when 
0cos20cos
0cos
aa
u
dt
d



a
u
3

a
u
53
5


37. d) Suppose that x = a
Show that the train’s angular velocity about N when is
times the angular velocity about N when 2

 
2
when

 
5
3
0
T
ON

a
2a
a5 5
2
cos 
5
1
2
cos 




 






cos2
2
cos
2
cos
aa
u
dt
d












 

  

















5
2
2
5
1
5
1
aa
u
a
u
5

0when 
0cos20cos
0cos
aa
u
dt
d



a
u
3

a
u
53
5

0nlocity wheangular vethe
times
5
3
is
2
nlocity wheangular vetheThus






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
position of P. Let s denote the arc length AP, ,
dt
ds
v 
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
position of P. Let s denote the arc length AP, ,
dt
ds
v 
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.
a) Show that the tangential acceleration of P is given by 




 2
2
2
2
11
v
d
d
ldt
sd


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
position of P. Let s denote the arc length AP, ,
dt
ds
v 
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.
a) Show that the tangential acceleration of P is given by 




 2
2
2
2
11
v
d
d
ldt
sd

ls 

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
position of P. Let s denote the arc length AP, ,
dt
ds
v 
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.
a) Show that the tangential acceleration of P is given by 




 2
2
2
2
11
v
d
d
ldt
sd

ls 
dt
d
l
dt
ds
v




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
position of P. Let s denote the arc length AP, ,
dt
ds
v 
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.
a) Show that the tangential acceleration of P is given by 




 2
2
2
2
11
v
d
d
ldt
sd

dt
dv
dt
sd
2
2
ls 
dt
d
l
dt
ds
v




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
position of P. Let s denote the arc length AP, ,
dt
ds
v 
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.
a) Show that the tangential acceleration of P is given by 




 2
2
2
2
11
v
d
d
ldt
sd

dt
dv
dt
sd
2
2
ls 
dt
d
l
dt
ds
v



dt
d
d
dv 



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
position of P. Let s denote the arc length AP, ,
dt
ds
v 
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.
a) Show that the tangential acceleration of P is given by 




 2
2
2
2
11
v
d
d
ldt
sd

dt
dv
dt
sd
2
2
ls 
dt
d
l
dt
ds
v



dt
d
d
dv 


l
v
d
dv



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
position of P. Let s denote the arc length AP, ,
dt
ds
v 
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.
a) Show that the tangential acceleration of P is given by 




 2
2
2
2
11
v
d
d
ldt
sd

dt
dv
dt
sd
2
2
ls 
dt
d
l
dt
ds
v



dt
d
d
dv 


l
v
d
dv


v
d
dv
l


1

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
position of P. Let s denote the arc length AP, ,
dt
ds
v 
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.
a) Show that the tangential acceleration of P is given by 




 2
2
2
2
11
v
d
d
ldt
sd

dt
dv
dt
sd
2
2
ls 
dt
d
l
dt
ds
v



dt
d
d
dv 


l
v
d
dv


v
d
dv
l


1





 2
2
11
v
dv
d
d
dv
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
position of P. Let s denote the arc length AP, ,
dt
ds
v 
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.
a) Show that the tangential acceleration of P is given by 




 2
2
2
2
11
v
d
d
ldt
sd

dt
dv
dt
sd
2
2
ls 
dt
d
l
dt
ds
v



dt
d
d
dv 


l
v
d
dv


v
d
dv
l


1





 2
2
11
v
dv
d
d
dv
l 





 2
2
11
v
d
d
l 

48. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l







49. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l







50. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T

51. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T
mg

52. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T
mg


53. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T
mg

sinmg

54. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T
mg

sm 
sinmg

55. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T
mg

sm 
sinmg
sinmgsm 

56. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T
mg

sm 
sinmg
sinmgsm 
sings 

57. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T
mg

sm 
sinmg
sinmgsm 
sings 


sin
2
11 2
gv
d
d
l







58. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T
mg

sm 
sinmg
sinmgsm 
sings 


sin
2
11 2
gv
d
d
l






 cos12thatDeducec) 22
 glvV

59. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T
mg

sm 
sinmg
sinmgsm 
sings 


sin
2
11 2
gv
d
d
l






 cos12thatDeducec) 22
 glvV


sin
2
11 2
gv
d
d
l







60. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T
mg

sm 
sinmg
sinmgsm 
sings 


sin
2
11 2
gv
d
d
l






 cos12thatDeducec) 22
 glvV


sin
2
11 2
gv
d
d
l








sin
2
1 2
glv
d
d







61. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T
mg

sm 
sinmg
sinmgsm 
sings 


sin
2
11 2
gv
d
d
l






 cos12thatDeducec) 22
 glvV


sin
2
11 2
gv
d
d
l








sin
2
1 2
glv
d
d






cglv
cglv




cos2
cos
2
1
2
2

62. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T
mg

sm 
sinmg
sinmgsm 
sings 


sin
2
11 2
gv
d
d
l






 cos12thatDeducec) 22
 glvV


sin
2
11 2
gv
d
d
l








sin
2
1 2
glv
d
d






cglv
cglv




cos2
cos
2
1
2
2
glVc
cglV
Vv
2
2
0,when
2
2




63. b) Show that the equation of motion of P is 

sin
2
11 2
gv
d
d
l






T
mg

sm 
sinmg
sinmgsm 
sings 


sin
2
11 2
gv
d
d
l






 cos12thatDeducec) 22
 glvV


sin
2
11 2
gv
d
d
l








sin
2
1 2
glv
d
d






cglv
cglv




cos2
cos
2
1
2
2
glVc
cglV
Vv
2
2
0,when
2
2



 


cos12
cos22
2cos2
22
22
22



glvV
glglvV
glVglv

64. 21
cosyExplain whd) mv
l
θT-mg 

65. 21
cosyExplain whd) mv
l
θT-mg 

66. 21
cosyExplain whd) mv
l
θT-mg 
T

67. 21
cosyExplain whd) mv
l
θT-mg 
T cosmg

68. 21
cosyExplain whd) mv
l
θT-mg 
T cosmg
xm 

69. 21
cosyExplain whd) mv
l
θT-mg 
T cosmg
xm 
cosmgTxm 

70. 21
cosyExplain whd) mv
l
θT-mg 
T cosmg
xm 
cosmgTxm 
But, the resultant force towards the centre is
centripetal force.

71. 21
cosyExplain whd) mv
l
θT-mg 
T cosmg
xm 
cosmgTxm 
2
2
1
cos
cos
mv
l
mgT
mgT
l
mv




But, the resultant force towards the centre is
centripetal force.

72. 21
cosyExplain whd) mv
l
θT-mg 
T cosmg
xm 
cosmgTxm 
2
2
1
cos
cos
mv
l
mgT
mgT
l
mv




But, the resultant force towards the centre is
centripetal force.
0at whichofvaluetheFind.3thatSupposee) 2
 TglV 

73. 21
cosyExplain whd) mv
l
θT-mg 
T cosmg
xm 
cosmgTxm 
2
2
1
cos
cos
mv
l
mgT
mgT
l
mv




But, the resultant force towards the centre is
centripetal force.
0at whichofvaluetheFind.3thatSupposee) 2
 TglV 
   cos12
1
cos 2
 glVm
l
mgT

74. 21
cosyExplain whd) mv
l
θT-mg 
T cosmg
xm 
cosmgTxm 
2
2
1
cos
cos
mv
l
mgT
mgT
l
mv




But, the resultant force towards the centre is
centripetal force.
0at whichofvaluetheFind.3thatSupposee) 2
 TglV 
   cos12
1
cos 2
 glVm
l
mgT
   cos123
1
cos0  glglm
l
mg

75. 21
cosyExplain whd) mv
l
θT-mg 
T cosmg
xm 
cosmgTxm 
2
2
1
cos
cos
mv
l
mgT
mgT
l
mv




But, the resultant force towards the centre is
centripetal force.
0at whichofvaluetheFind.3thatSupposee) 2
 TglV 
   cos12
1
cos 2
 glVm
l
mgT
   cos123
1
cos0  glglm
l
mg
  cos2cos ggmmg 

76. 21
cosyExplain whd) mv
l
θT-mg 
T cosmg
xm 
cosmgTxm 
2
2
1
cos
cos
mv
l
mgT
mgT
l
mv




But, the resultant force towards the centre is
centripetal force.
0at whichofvaluetheFind.3thatSupposee) 2
 TglV 
   cos12
1
cos 2
 glVm
l
mgT
   cos123
1
cos0  glglm
l
mg
  cos2cos ggmmg 
mgmg cos3

77. 21
cosyExplain whd) mv
l
θT-mg 
T cosmg
xm 
cosmgTxm 
2
2
1
cos
cos
mv
l
mgT
mgT
l
mv




But, the resultant force towards the centre is
centripetal force.
0at whichofvaluetheFind.3thatSupposee) 2
 TglV 
   cos12
1
cos 2
 glVm
l
mgT
   cos123
1
cos0  glglm
l
mg
  cos2cos ggmmg 
mgmg cos3
3
1
cos 

78. 21
cosyExplain whd) mv
l
θT-mg 
T cosmg
xm 
cosmgTxm 
2
2
1
cos
cos
mv
l
mgT
mgT
l
mv




But, the resultant force towards the centre is
centripetal force.
0at whichofvaluetheFind.3thatSupposee) 2
 TglV 
   cos12
1
cos 2
 glVm
l
mgT
   cos123
1
cos0  glglm
l
mg
  cos2cos ggmmg 
mgmg cos3
3
1
cos 
radians911.1

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;
21
cos mv
l
mgT  

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;
21
cos mv
l
mgT  
21
3
1
mv
l
mg 






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;
21
cos mv
l
mgT  
21
3
1
mv
l
mg 





3
3
2
gl
v
gl
v



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;
21
cos mv
l
mgT  
21
3
1
mv
l
mg 





3
3
2
gl
v
gl
v



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;
21
cos mv
l
mgT  
21
3
1
mv
l
mg 





3
3
2
gl
v
gl
v



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;
21
cos mv
l
mgT  
21
3
1
mv
l
mg 





3
3
2
gl
v
gl
v



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;
21
cos mv
l
mgT  
21
3
1
mv
l
mg 





3
3
2
gl
v
gl
v



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;
21
cos mv
l
mgT  
21
3
1
mv
l
mg 





3
3
2
gl
v
gl
v



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
gykyxkx   and
where k is a constant and the acceleration due
to gravity is g.
(You are NOT required to show these)

90. cosresulttheDerivea) kt
uex 


91. cosresulttheDerivea) kt
uex 

xk
dt
xd




92. cosresulttheDerivea) kt
uex 

xk
dt
xd




x
u
x
xd
k
t



cos
1

93. cosresulttheDerivea) kt
uex 

xk
dt
xd




x
u
x
xd
k
t



cos
1
 x
ux
k
t

 coslog
1


94. cosresulttheDerivea) kt
uex 

xk
dt
xd




x
u
x
xd
k
t



cos
1
 x
ux
k
t

 coslog
1

  cosloglog
1
ux
k
t  

95. cosresulttheDerivea) kt
uex 

xk
dt
xd




x
u
x
xd
k
t



cos
1
 x
ux
k
t

 coslog
1

  cosloglog
1
ux
k
t  






cos
log
1
u
x
k
t


96. cosresulttheDerivea) kt
uex 

xk
dt
xd




x
u
x
xd
k
t



cos
1
 x
ux
k
t

 coslog
1

  cosloglog
1
ux
k
t  






cos
log
1
u
x
k
t




cos
cos
cos
log
kt
kt
uex
e
u
x
u
x
kt














97. cosresulttheDerivea) kt
uex 

xk
dt
xd




x
u
x
xd
k
t



cos
1
 x
ux
k
t

 coslog
1

  cosloglog
1
ux
k
t  






cos
log
1
u
x
k
t




cos
cos
cos
log
kt
kt
uex
e
u
x
u
x
kt













  
conditioninitialandmotionofequation
eappropriatthesatisfiessin
1
tVerify thab) gegku
k
y kt
 


98. cosresulttheDerivea) kt
uex 

xk
dt
xd




x
u
x
xd
k
t



cos
1
 x
ux
k
t

 coslog
1

  cosloglog
1
ux
k
t  






cos
log
1
u
x
k
t




cos
cos
cos
log
kt
kt
uex
e
u
x
u
x
kt













  
conditioninitialandmotionofequation
eappropriatthesatisfiessin
1
tVerify thab) gegku
k
y kt
 

gyk
dt
yd
 


99. cosresulttheDerivea) kt
uex 

xk
dt
xd




x
u
x
xd
k
t



cos
1
 x
ux
k
t

 coslog
1

  cosloglog
1
ux
k
t  






cos
log
1
u
x
k
t




cos
cos
cos
log
kt
kt
uex
e
u
x
u
x
kt













  
conditioninitialandmotionofequation
eappropriatthesatisfiessin
1
tVerify thab) gegku
k
y kt
 

gyk
dt
yd
 

 

y
u
gyk
yd
t



sin

100. cosresulttheDerivea) kt
uex 

xk
dt
xd




x
u
x
xd
k
t



cos
1
 x
ux
k
t

 coslog
1

  cosloglog
1
ux
k
t  






cos
log
1
u
x
k
t




cos
cos
cos
log
kt
kt
uex
e
u
x
u
x
kt













  
conditioninitialandmotionofequation
eappropriatthesatisfiessin
1
tVerify thab) gegku
k
y kt
 

gyk
dt
yd
 

 

y
u
gyk
yd
t



sin
  y
ugyk
k
t

 sinlog
1


101. cosresulttheDerivea) kt
uex 

xk
dt
xd




x
u
x
xd
k
t



cos
1
 x
ux
k
t

 coslog
1

  cosloglog
1
ux
k
t  






cos
log
1
u
x
k
t




cos
cos
cos
log
kt
kt
uex
e
u
x
u
x
kt













  
conditioninitialandmotionofequation
eappropriatthesatisfiessin
1
tVerify thab) gegku
k
y kt
 

gyk
dt
yd
 

 

y
u
gyk
yd
t



sin
  y
ugyk
k
t

 sinlog
1

   gkugykkt  sinloglog 

102. cosresulttheDerivea) kt
uex 

xk
dt
xd




x
u
x
xd
k
t



cos
1
 x
ux
k
t

 coslog
1

  cosloglog
1
ux
k
t  






cos
log
1
u
x
k
t




cos
cos
cos
log
kt
kt
uex
e
u
x
u
x
kt













  
conditioninitialandmotionofequation
eappropriatthesatisfiessin
1
tVerify thab) gegku
k
y kt
 

gyk
dt
yd
 

 

y
u
gyk
yd
t



sin
  y
ugyk
k
t

 sinlog
1

   gkugykkt  sinloglog 









gku
gyk
kt
sin
log


103. cosresulttheDerivea) kt
uex 

xk
dt
xd




x
u
x
xd
k
t



cos
1
 x
ux
k
t

 coslog
1

  cosloglog
1
ux
k
t  






cos
log
1
u
x
k
t




cos
cos
cos
log
kt
kt
uex
e
u
x
u
x
kt













  
conditioninitialandmotionofequation
eappropriatthesatisfiessin
1
tVerify thab) gegku
k
y kt
 

gyk
dt
yd
 

 

y
u
gyk
yd
t



sin
  y
ugyk
k
t

 sinlog
1

   gkugykkt  sinloglog 









gku
gyk
kt
sin
log

kt
e
gku
gyk 



sin


104. cosresulttheDerivea) kt
uex 

xk
dt
xd




x
u
x
xd
k
t



cos
1
 x
ux
k
t

 coslog
1

  cosloglog
1
ux
k
t  






cos
log
1
u
x
k
t




cos
cos
cos
log
kt
kt
uex
e
u
x
u
x
kt













  
conditioninitialandmotionofequation
eappropriatthesatisfiessin
1
tVerify thab) gegku
k
y kt
 

gyk
dt
yd
 

 

y
u
gyk
yd
t



sin
  y
ugyk
k
t

 sinlog
1

   gkugykkt  sinloglog 









gku
gyk
kt
sin
log

kt
e
gku
gyk 



sin

  gegku
k
y kt
 
sin
1


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
0whenoccursheightMaximum y

107. c) Find the value of t when the particle reaches its maximum height
0whenoccursheightMaximum y
  0
sinlog
1
ugyk
k
t  

108. c) Find the value of t when the particle reaches its maximum height
0whenoccursheightMaximum y
  0
sinlog
1
ugyk
k
t  
    gkug
k
t  sinloglog
1

109. c) Find the value of t when the particle reaches its maximum height
0whenoccursheightMaximum y
  0
sinlog
1
ugyk
k
t  
    gkug
k
t  sinloglog
1





 

g
gku
k
t
sin
log
1

110. c) Find the value of t when the particle reaches its maximum height
0whenoccursheightMaximum y
  0
sinlog
1
ugyk
k
t  
    gkug
k
t  sinloglog
1





 

g
gku
k
t
sin
log
1
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
0whenoccursheightMaximum y
  0
sinlog
1
ugyk
k
t  
    gkug
k
t  sinloglog
1





 

g
gku
k
t
sin
log
1
d) What is the limiting value of the horizontal displacement of the
particle?


cos
cos
kt
kt
ue
dt
dx
uex





112. c) Find the value of t when the particle reaches its maximum height
0whenoccursheightMaximum y
  0
sinlog
1
ugyk
k
t  
    gkug
k
t  sinloglog
1





 

g
gku
k
t
sin
log
1
d) What is the limiting value of the horizontal displacement of the
particle?


cos
cos
kt
kt
ue
dt
dx
uex



 




0
coslim dteux kt
t


113. c) Find the value of t when the particle reaches its maximum height
0whenoccursheightMaximum y
  0
sinlog
1
ugyk
k
t  
    gkug
k
t  sinloglog
1





 

g
gku
k
t
sin
log
1
d) What is the limiting value of the horizontal displacement of the
particle?


cos
cos
kt
kt
ue
dt
dx
uex



 




0
coslim dteux kt
t

t
kt
t
e
k
ux
0
1
coslim 


 



114. c) Find the value of t when the particle reaches its maximum height
0whenoccursheightMaximum y
  0
sinlog
1
ugyk
k
t  
    gkug
k
t  sinloglog
1





 

g
gku
k
t
sin
log
1
d) What is the limiting value of the horizontal displacement of the
particle?


cos
cos
kt
kt
ue
dt
dx
uex



 




0
coslim dteux kt
t

t
kt
t
e
k
ux
0
1
coslim 


 


 1
cos
lim  

kt
t
e
k
u
x


115. c) Find the value of t when the particle reaches its maximum height
0whenoccursheightMaximum y
  0
sinlog
1
ugyk
k
t  
    gkug
k
t  sinloglog
1





 

g
gku
k
t
sin
log
1
d) What is the limiting value of the horizontal displacement of the
particle?


cos
cos
kt
kt
ue
dt
dx
uex



 




0
coslim dteux kt
t

t
kt
t
e
k
ux
0
1
coslim 


 


 1
cos
lim  

kt
t
e
k
u
x

k
u
x
cos


116. Exercise 9E; 1 to 4, 7
Exercise 9F; 1, 2, 4, 7, 9, 12, 14, 16, 20, 22, 25