1. Dynamics

2. Dynamics
Newton’s Laws Of Motion

3. Dynamics
Newton’s Laws Of Motion
Law 1
“ Corpus omne perseverare in statu suo quiescendi vel movendi
uniformiter, nisi quatenus a viribus impressis cogitur statum illum
mutare.”

4. Dynamics
Newton’s Laws Of Motion
Law 1
“ Corpus omne perseverare in statu suo quiescendi vel movendi
uniformiter, nisi quatenus a viribus impressis cogitur statum illum
mutare.”
Everybody continues in its state of rest, or of uniform motion in a
right line, unless it is compelled to change that state by forces
impressed upon it.

5. Dynamics
Newton’s Laws Of Motion
Law 1
“ Corpus omne perseverare in statu suo quiescendi vel movendi
uniformiter, nisi quatenus a viribus impressis cogitur statum illum
mutare.”
Everybody continues in its state of rest, or of uniform motion in a
right line, unless it is compelled to change that state by forces
impressed upon it.
Law 2
“ Mutationem motus proportionalem esse vi motrici impressae, et
fieri secundum rectam qua vis illa impritmitur.”

6. Dynamics
Newton’s Laws Of Motion
Law 1
“ Corpus omne perseverare in statu suo quiescendi vel movendi
uniformiter, nisi quatenus a viribus impressis cogitur statum illum
mutare.”
Everybody continues in its state of rest, or of uniform motion in a
right line, unless it is compelled to change that state by forces
impressed upon it.
Law 2
“ Mutationem motus proportionalem esse vi motrici impressae, et
fieri secundum rectam qua vis illa impritmitur.”
The change of motion is proportional to the motive force
impressed, and it is made in the direction of the right line in which
that force is impressed.

7. Motive force  change in motion

8. Motive force  change in motion
F acceleration

9. Motive force  change in motion
F acceleration
F  constant  acceleration

10. Motive force  change in motion
F acceleration
F  constant  acceleration
 F  ma (constant = mass)

11. Motive force  change in motion
F acceleration
F  constant  acceleration
 F  ma (constant = mass)
Law 3
“ Actioni contrariam semper et aequalem esse reactionem: sive
corporum duorum actions in se mutuo semper esse aequales et in
partes contrarias dirigi.”

12. Motive force  change in motion
F acceleration
F  constant  acceleration
 F  ma (constant = mass)
Law 3
“ Actioni contrariam semper et aequalem esse reactionem: sive
corporum duorum actions in se mutuo semper esse aequales et in
partes contrarias dirigi.”
To every action there is always opposed an equal reaction: or, the
mutual actions of two bodies upon each other are always equal, and
directed in contrary parts.

13. Acceleration

14. Acceleration
d 2x
  2
x
dt
dv

dt

15. Acceleration
d 2x
  2
x (acceleration is a function of t)
dt
dv

dt

16. Acceleration
d 2x
  2
x (acceleration is a function of t)
dt
dv

dt
d 1 2 
  v  (acceleration is a function of x)
dx  2 

17. Acceleration
d 2x
  2
x (acceleration is a function of t)
dt
dv

dt
d 1 2 
  v  (acceleration is a function of x)
dx  2 
dv
v (acceleration is a function of v)
dx

18. Acceleration
d 2x
  2
x (acceleration is a function of t)
dt
dv

dt
d 1 2 
  v  (acceleration is a function of x)
dx  2 
dv
v (acceleration is a function of v)
dx
e.g. (1981)
Assume that the earth is a sphere of radius R and that, at a point r  R 
from the centre of the earth, the acceleration due to gravity is
proportional to 1 and is directed towards the earth’s centre.
r2
A body is projected vertically upwards from the surface of the earth with
initial speed V.

19. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.

20. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.
O

21. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.
O

22. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.
r  R, v  V ,    g
r
O

23. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.
v0
r  R, v  V ,    g
r
O

24. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.
v0
r  R, v  V ,    g
r
O

25. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.
v0 resultant
m
r
force
r  R, v  V ,    g
r
O

26. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.
v0 resultant
m
r
force
m
r2
r  R, v  V ,    g
r
O

27. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.
m
v0 resultant m   2
r
m
r r
force
m
r2
r  R, v  V ,    g
r
O

28. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.
m
v0 resultant m   2
r
m
r r
force 
  2
r
m r
r2
r  R, v  V ,    g
r
O

29. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.
m
v0 resultant m   2
r
m
r r
force 
  2
r
m r
2 when r  R,    g
r
r
r  R, v  V ,    g
r
O

30. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.
m
v0 resultant m   2
r
m
r r
force 
  2
r
m r
2 when r  R,    g
r
r 
r  R, v  V ,    g
r i.e.  g  2
R
  gR 2
O

31. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.
m
v0 resultant m   2
r
m
r r
force 
  2
r
m r
2 when r  R,    g
r
r 
r  R, v  V ,    g
r i.e.  g  2
R
  gR 2
O  gR 2
  2
r
r

32. (i) Prove that it will escape the earth if and only if V  2 gR where g
is the magnitude of the acceleration due to gravity at the earth’s
surface.
m
v0 resultant m   2
r
m
r r
force 
  2
r
m r
2 when r  R,    g
r
r 
r  R, v  V ,    g
r i.e.  g  2
R
  gR 2
O  gR 2
  2
r
r
d  1 2   gR 2
 v  2
dr  2  r

33. 1 2 gR 2
v  c
2 r
2 gR 2
v2  c
r

34. 1 2 gR 2
v  c
2 r
2 gR 2
v2  c
r
when r  R, v  V

35. 1 2 gR 2
v  c
2 r
2 gR 2
v2  c
r
when r  R, v  V
2 gR 2
i.e. V 2  c
R
c  V 2  2 gR

36. 1 2 gR 2
v  c
2 r
2 gR 2
v2  c
r
when r  R, v  V
2 gR 2
i.e. V 2  c
R
c  V 2  2 gR
2 gR 2
v 
2
 V 2  2 gR
r

37. 1 2 gR 2
v  c
2 r
2 gR 2 If particle never stops then r  
v2  c
r
when r  R, v  V
2 gR 2
i.e. V 2  c
R
c  V 2  2 gR
2 gR 2
v 
2
 V 2  2 gR
r

38. 1 2 gR 2
v  c
2 r
2 gR 2 If particle never stops then r  
v2  c
r  2 gR 2 
lim v  lim
2
 V  2 gR 
2
when r  R, v  V r  r 
 r 
2 gR 2  V 2  2 gR
i.e. V 2  c
R
c  V 2  2 gR
2 gR 2
v 
2
 V 2  2 gR
r

39. 1 2 gR 2
v  c
2 r
2 gR 2 If particle never stops then r  
v2  c
r  2 gR 2 
lim v  lim
2
 V  2 gR 
2
when r  R, v  V r  r 
 r 
2 gR 2  V 2  2 gR
i.e. V 2  c
R
Now v 2  0
c  V 2  2 gR
2 gR 2
v 
2
 V 2  2 gR
r

40. 1 2 gR 2
v  c
2 r
2 gR 2 If particle never stops then r  
v2  c
r  2 gR 2 
lim v  lim
2
 V  2 gR 
2
when r  R, v  V r  r 
 r 
2 gR 2  V 2  2 gR
i.e. V 2  c
R
Now v 2  0
c  V 2  2 gR
V 2  2 gR  0
2 gR 2
v 
2
 V 2  2 gR
r

41. 1 2 gR 2
v  c
2 r
2 gR 2 If particle never stops then r  
v2  c
r  2 gR 2 
lim v  lim
2
 V  2 gR 
2
when r  R, v  V r  r 
 r 
2 gR 2  V 2  2 gR
i.e. V 2  c
R
Now v 2  0
c  V 2  2 gR
V 2  2 gR  0
2 gR 2 V 2  2 gR
v 
2
 V 2  2 gR
r

42. 1 2 gR 2
v  c
2 r
2 gR 2 If particle never stops then r  
v2  c
r  2 gR 2 
lim v  lim
2
 V  2 gR 
2
when r  R, v  V r  r 
 r 
2 gR 2  V 2  2 gR
i.e. V 2  c
R
Now v 2  0
c  V 2  2 gR
V 2  2 gR  0
2 gR 2 V 2  2 gR
v 
2
 V 2  2 gR
r
Thus if V  2 gR the particle never stops,
i.e. the particle escapes the earth.

43. (ii) If V  2 gR ,prove that the time taken to rise to a height R above
the earth’s surface is;
1
4  2  R
3 g

44. (ii) If V  2 gR ,prove that the time taken to rise to a height R above
the earth’s surface is;
1
4  2  R
3 g
2 gR 2
v 
2
 V 2  2 gR
r

45. (ii) If V  2 gR ,prove that the time taken to rise to a height R above
the earth’s surface is;
1
4  2  R
3 g
2 gR 2
v 
2
 V 2  2 gR
r
If V  2 gR ,

46. (ii) If V  2 gR ,prove that the time taken to rise to a height R above
the earth’s surface is;
1
4  2  R
3 g
2 gR 2
v 
2
 V 2  2 gR
r
2 gR 2
If V  2 gR , v   2 gR  2 gR
2
r
2 gR 2
v2 
r

47. (ii) If V  2 gR ,prove that the time taken to rise to a height R above
the earth’s surface is;
1
4  2  R
3 g
2 gR 2
v 
2
 V 2  2 gR
r
2 gR 2
If V  2 gR , v   2 gR  2 gR
2
r
2 gR 2
v2 
r
2 gR 2
v (cannot be –ve, as it does not return)
r

48. (ii) If V  2 gR ,prove that the time taken to rise to a height R above
the earth’s surface is;
1
4  2  R
3 g
2 gR 2
v 
2
 V 2  2 gR
r
2 gR 2
If V  2 gR , v   2 gR  2 gR
2
r
2 gR 2
v2 
r
2 gR 2
v (cannot be –ve, as it does not return)
r
dr 2 gR 2

dt r

49. (ii) If V  2 gR ,prove that the time taken to rise to a height R above
the earth’s surface is;
1
4  2  R
3 g
2 gR 2
v 
2
 V 2  2 gR
r
2 gR 2
If V  2 gR , v   2 gR  2 gR
2
r
2 gR 2
v2 
r
2 gR 2
v (cannot be –ve, as it does not return)
r
dr 2 gR 2

dt r
dt r

dr 2gR 2

50. 2R
1
t
2 gR 2 
R
r dr

51. 2R
1 2 R 
t
2 gR 2  r dr   as travelling from R to 2 R 
R R 

52. 2R
1 2 R 
t
2 gR 2  r dr   as travelling from R to 2 R 
R R 
2R
1 2  3
 3 r 
2
2 gR 2  R

53. 2R
1 2 R 
t
2 gR 2  r dr   as travelling from R to 2 R 
R R 
2R
1 2  3
 3 r 
2
2 gR 2  R

2
2
2 R 2 R  R R 
3 gR

54. 2R
1 2 R 
t
2 gR 2  r dr   as travelling from R to 2 R 
R R 
2R
1 2  3
 3 r 
2
2 gR 2  R

2
2
2 R 2 R  R R 
3 gR

2
2 2  1R R
3R g

55. 2R
1 2 R 
t
2 gR 2  r dr   as travelling from R to 2 R 
R R 
2R
1 2  3
 3 r 
2
2 gR 2  R

2
2
2 R 2 R  R R 
3 gR

2
2 2  1R R
3R g

R
4  2 
3 g

56. 2R
1 2 R 
t
2 gR 2  r dr   as travelling from R to 2 R 
R R 
2R
1 2  3
 3 r 
2
2 gR 2  R

2
2
2 R 2 R  R R 
3 gR

2
2 2  1R R
3R g

R
4  2 
3 g
 4  2 
1 R
3 g

57. 2R
1 2 R 
t
2 gR 2  r dr   as travelling from R to 2 R 
R R 
2R
1 2  3
 3 r 
2
2 gR 2  R

2
2
2 R 2 R  R R 
3 gR

2
2 2  1R R
3R g

R
4  2 
3 g
 time taken to rise to a height R above
 4  2 
1 R
earth' s surface is 4  2 
1 R
3 g seconds
3 g

58. Exercise 8C; 3, 15, 19