In the previous section, we saw that the velocity of the 'bead executing vertical circular motion along a loop’ will be different at different points along the loop. We saw the method to find the velocity at any point.
• When the motion takes place along the loop, if there is sufficient velocity, the bead will be trying to escape from the loop
• So the loop will be in contact with the inner surface of the bead. This is shown in figs.6.49(a) and (b) below:
■ However, at the top most point C, two types of contacts can occur
• If the velocity is large, the contact will be as shown in fig.6.49(c)
♦ The inner surface will be in contact with the loop
• If the velocity is small, the contact will be as shown in fig.6.49(d)
♦ The outer surface will be in contact with the loop
■ Whenever the bead is in contact with the loop, the bead will experience a normal reaction FN from the loop
• This FN will vary at different points along the path
• Our next task is to find the reason for such a variation
• We will also see the method to find the magnitude and direction of this FN at various points
1. In fig.6.50(a) below, the bead is shown at the four quadrant points and also the 'any point P'
• ‘O’ is the center of the loop and ‘r’ is the radius.
2. The FBD of the bead when it is at A is shown in fig.b
• In this fig.b, the bead is trying to escape from the loop. (Note that, the inner surface of the bead is in contact with the loop)
• So it will exert a force on the loop.
♦ This is a force ‘exerted by the bead’. So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(A) on the bead.
♦ This is a force ‘exerted on the bead’. So we must show it in the FBD
3. The force ‘exerted by the bead’ is directed away from the center ‘O’
• So the reaction FN(A) will be towards the center ‘O’
4. The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.b
5. Let us considered the other possibility:
• In fig.c, the bead is trying to move towards the center of the loop.
♦ It is not trying to escape from the loop
♦ Note that, now the outer surface of the bead is in contact with the loop
• So it will exert a force on the loop. This is a force ‘exerted by the bead’
♦ So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(A) on the bead. This is a force ‘exerted on the bead’
♦ So we must show it in the FBD
• The force ‘exerted by the bead’ is directed towards the center ‘O’
• So the reaction FN(A) will be directed away from the center ‘O’
• The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.c
■ However such a situation generally does not occur.
• Because, then it would mean that, the bead is trying to move vertically upwards when it is at A
• So we do not need the fig.c
6. Let us consider fig.b:
• Both the forces in fig.b act along the same line but opposite in direction
• So the resultant force is (FN(A)-mg)
7. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_A^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_A^2}{r}}$
• This is the centripetal force at point A.
• So we can write: $\mathbf\small{F_{N(A)}-mg=\frac{mv_A^2}{r}}$
$\mathbf\small{\Longrightarrow F_{N(A)}=m \left(\frac{v_A^2}{r}+g \right)}$
• The centripetal force is always directed towards the center of the circle. So it is considered as positive
■ Thus, if we know the velocity at A, we can easily calculate the normal reaction exerted by the loop at A
8. Next we will calculate FN(B), the normal reaction at B
• The FBD of the bead when it is at B is shown in fig.d
• In this fig.d, the bead is trying to escape from the loop. (Note that, the inner surface of the bead is in contact with the loop)
• So it will exert a force on the loop.
♦ This is a force ‘exerted by the bead’. So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(B) on the bead.
♦ This is a force ‘exerted on the bead’. So we must show it in the FBD
9. The force ‘exerted by the bead’ is directed away from the center ‘O’
• So the reaction FN(B) will be towards the center ‘O’
10. The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.d
11. Let us considered the other possibility:
• In fig.e, the bead is trying to move towards the center of the loop.
♦ It is not trying to escape from the loop
♦ Note that, now the outer surface of the bead is in contact with the loop
• So it will exert a force on the loop. This is a force ‘exerted by the bead’
♦ So we cannot show it in the FBD
• The loop will exert a normal reaction FN(B) on the bead. This is a force ‘exerted on the bead’
♦ So we must show it in the FBD
• The force ‘exerted by the bead’ is directed towards the center ‘O’
• So the reaction FN(B) will be directed away from the center ‘O’
• The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.e
■ However such a situation generally does not occur.
• Because, then it would mean that, the bead is trying to move towards left when it is at B
• So we do not need the fig.e
12. Let us consider fig.d:
• The only force in the radial direction is FN(B)
• This is because, here 'mg' is perpendicular to FN(B) and so do not have any component along the radial direction
• So the resultant force is FN(B) itself
13. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_B^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_B^2}{r}}$
• This is the centripetal force at point B
• So we can write: $\mathbf\small{F_{N(B)}=\frac{mv_B^2}{r}}$
■ Thus, if we know the velocity vB at B, we can easily calculate FN(B)
14. But how do we find vB?
• If we know the velocity vA at the lowest point A, we can calculate the velocity vB at B
• For that, we use the equation obtained in (10) in the previous section
♦ Here we will write it again: $\mathbf\small{v_B=\sqrt{v_A^2-2gr}}$
■ Once we calculate vB, we can easily calculate FN(B) using the equation in (13) above
15. Next we will calculate FN(C), the normal reaction at C
• The FBD of the bead when it is at C is shown in fig.f
• In this fig.f, the bead is trying to escape from the loop. (Note that, the inner surface of the bead is in contact with the loop)
• So it will exert a force on the loop.
♦ This is a force ‘exerted by the bead’. So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(C) on the bead.
♦ This is a force ‘exerted on the bead’. So we must show it in the FBD
16. The force ‘exerted by the bead’ is directed away from the center ‘O’
• So the reaction FN(A) will be towards the center ‘O’
17. The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.f
18. Let us considered the other possibility:
• In fig.c, the bead is trying to move towards the center of the loop.
♦ It is not trying to escape from the loop
♦ Note that, now the outer surface of the bead is in contact with the loop
• So it will exert a force on the loop. This is a force ‘exerted by the bead’
♦ So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(C) on the bead. This is a force ‘exerted on the bead’
♦ So we must show it in the FBD
• The force ‘exerted by the bead’ is directed towards the center ‘O’
• So the reaction FN(C) will be directed away from the center ‘O’
• The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.g
■ Such a situation can indeed occur if the velocity is not sufficient
• So we need both the figs.(f) and (g)
19. First, let us consider fig.f:
• Both the forces in fig.f act along the same line and in same direction
• So the resultant force is (FN(C)+mg)
20. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_C^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_C^2}{r}}$
• This is the centripetal force at point C
• So we can write: $\mathbf\small{F_{N(C)}+mg=\frac{mv_C^2}{r}}$
$\mathbf\small{\Longrightarrow F_{N(C)}=m \left(\frac{v_C^2}{r}-g \right)}$
■ Thus, if we know the velocity at C, we can easily calculate the normal reaction exerted by the loop at C
21. Now, let us consider fig.g:
• Both the forces in fig.f act along the same line but in opposite direction
• So the resultant force is (-FN(C)+mg)
22. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_C^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_C^2}{r}}$
• This is the centripetal force at point C
• So we can write: $\mathbf\small{-F_{N(C)}+mg=\frac{mv_C^2}{r}}$
$\mathbf\small{\Longrightarrow F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
■ Thus, if we know the velocity vC at C, we can easily calculate FN(C)
23. But how do we find vC?
• If we know the velocity vA at the lowest point A, we can calculate the velocity vC at C
• For that, we use the equation obtained in (13) in the previous section
♦ Here we will write it again: $\mathbf\small{v_C=\sqrt{v_A^2-4gr}}$
■ Once we calculate vC, we can easily calculate FN(C) using the equation in (20) or (22) above
24. So we have two equations at C. They are:
(i) From step (20), we have: $\mathbf\small{F_{N(C)}=m \left(\frac{v_C^2}{r}-g \right)}$
(ii) From step (22), we have: $\mathbf\small{F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
• We will see their applications later in this section
25. Next we will calculate FN(D), the normal reaction D
• The FBD of the bead when it is at D is shown in fig.h
• In this fig.h, the bead is trying to escape from the loop. (Note that, the inner surface of the bead is in contact with the loop)
• So it will exert a force on the loop.
♦ This is a force ‘exerted by the bead’. So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(D) on the bead.
♦ This is a force ‘exerted on the bead’. So we must show it in the FBD
26. The force ‘exerted by the bead’ is directed away from the center ‘O’
• So the reaction FN(D) will be towards the center ‘O’
27. The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.h
28. Let us considered the other possibility:
• In fig.i, the bead is trying to move towards the center of the loop.
♦ It is not trying to escape from the loop
♦ Note that, now the outer surface of the bead is in contact with the loop
• So it will exert a force on the loop. This is a force ‘exerted by the bead’
♦ So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(D) on the bead. This is a force ‘exerted on the bead’
♦ So we must show it in the FBD
• The force ‘exerted by the bead’ is directed towards the center ‘O’
• So the reaction FN(D) will be directed away from the center ‘O’
• The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.i
■ However such a situation generally does not occur.
• Because, then it would mean that, the bead is trying to move towards right when it is at B
• So we do not need the fig.i
29. Let us consider fig.h:
• The only force in the radial direction is FN(D)
• This is because, here 'mg' is perpendicular to FN(D) and so do not have any component along the radial direction
• So the resultant force is FN(D) itself
30. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_D^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_D^2}{r}}$
• This is the centripetal force at point D
• So we can write: $\mathbf\small{F_{N(D)}=\frac{mv_D^2}{r}}$
■ Thus, if we know the velocity vD at D, we can easily calculate FN(D)
31. But how do we find vD?
• If we know the velocity vA at the lowest point A, we can calculate the velocity vD at D
• For that, we use the equation obtained in (16) in the previous section
♦ Here we will write it again: $\mathbf\small{v_D=\sqrt{v_A^2-2gr}}$
■ Once we calculate vD, we can easily calculate FN(D) using the equation in (30)
■ It is interesting to note that, at B and D, just like the 'magnitudes of velocities', the 'magnitudes of reactions' are also the same
• So we successfully calculated the tensions at all the four quadrant points. But what about the intermediate points?
• For that, we will have to apply 'resolution of forces' (Details here). Let us see how it is done:
32. In fig.6.51(a) below, the bead is at P
• At that instant, the 'imaginary line OP connecting the bead to the center O' makes an angle θ with the vertical. This is shown in fig.b
33. In fig.6.65(b), the portion of the bead alone is shown
• In this fig.b, a vertical is drawn through the bead
• The following two verticals will be parallel:
♦ Vertical through O
♦ Vertical through P
• So, if we extend OP downwards along the same line, that extension will make the same angle θ with the vertical
34. This 'same angle' is our clue
• We know that, the 'component which is adjacent to the angle' will get the cosine. And the other component will get the sine
• So the component of the weight 'mg' which acts along the line of the string is mg cosθ
• This is shown in the FBD in fig.c
• In this fig.c, the bead is trying to escape from the loop. (Note that, the inner surface of the bead is in contact with the loop)
• So it will exert a force on the loop.
♦ This is a force ‘exerted by the bead’. So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(P) on the bead.
♦ This is a force ‘exerted on the bead’. So we must show it in the FBD
35. The force ‘exerted by the bead’ is directed away from the center ‘O’
• So the reaction FN(P) will be towards the center ‘O’
36. The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.c
37. Let us considered the other possibility:
• In fig.d, the bead is trying to move towards the center of the loop.
♦ It is not trying to escape from the loop
♦ Note that, now the outer surface of the bead is in contact with the loop
• So it will exert a force on the loop. This is a force ‘exerted by the bead’
♦ So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(P) on the bead. This is a force ‘exerted on the bead’
♦ So we must show it in the FBD
• The force ‘exerted by the bead’ is directed towards the center ‘O’
• So the reaction FN(P) will be directed away from the center ‘O’
• The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.d
■ Such a situation can indeed occur at the highest point 'C', if the velocity is not sufficient
• So we need both the figs.(c) and (d)
38. First let us consider fig.c:
• We see that the net force in the radial direction is (FN(P) -mg cosθ)
39. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_P^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_P^2}{r}}$
• This is the centripetal force at point P.
• So we can write: $\mathbf\small{F_{N(P)}-mg \cos \theta =\frac{mv_P^2}{r}}$
• $\mathbf\small{\Longrightarrow F_{N(P)}=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
■ Thus, if we know the velocity vP at P, we can easily calculate FN(P)
40. Now, let us consider fig.d:
• We see that the net force in the radial direction is (-FN(P) - mg cosθ)
41. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_P^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_P^2}{r}}$
• This is the centripetal force at point P.
• So we can write: $\mathbf\small{-F_{N(P)}-mg \cos \theta =\frac{mv_P^2}{r}}$
• $\mathbf\small{\Longrightarrow F_{N(P)}=-m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
■ Thus, if we know the velocity vP at P, we can easily calculate FN(P)
42. But how do we find vP?
• If we know the velocity vA at the lowest point A, we can calculate the velocity vP at P
• For that, we use the equation obtained in (23) in the previous section
♦ Here we will write it again: $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$
■ Once we calculate vP, we can easily calculate FN(P) using the equation in (39) or (41) above
• If we can use the equation derived in (39) or (41) above for 'any point', it must be applicable to points A, B, C and D also. Let us check:
43. First we will check the equation at point 'A'
• We have: $\mathbf\small{F_{N(P)}=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at A, θ = 0o
♦ So cos θ = cos 0 = 1
• Substituting the known values, we get:
$\mathbf\small{F_{N(A)}=m \left(\frac{v_A^2}{r}+g \right)}$
■ This is the same equation that we obtained in (7) above
44. Next we will check the equation at point 'B'
• We have: $\mathbf\small{F_{N(P)}=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at B, θ = 90o
♦ So cos θ = cos 90 = 0
• Substituting the known values, we get: $\mathbf\small{F_{N(B)}=m \left(\frac{v_B^2}{r}+g \times 0 \right)}$
$\mathbf\small{\Longrightarrow F_{N(B)}=\frac{mv_B^2}{r}}$
■ This is the same equation that we obtained in (13) above
45. Next we will check the equation at point 'C'
• We have: $\mathbf\small{F_{N(P)}=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at C, θ = 180o
♦ So cos θ = cos 180 = -1
• Substituting the known values, we get:
$\mathbf\small{F_{N(C)}=m \left(\frac{v_C^2}{r}-g \right)}$
■ This is the same equation that we obtained in (20) above
46. When the bead tries to move inwards we have to use the equation in (41)
We have: $\mathbf\small{F_{N(P)}=-m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at C, θ = 180o
♦ So cos θ = cos 180 = -1
• Substituting the known values, we get:
$\mathbf\small{F_{N(C)}=-m \left(\frac{v_C^2}{r}+g \times (-1) \right)}$
So we get: $\mathbf\small{F_{N(C)}=-m \left(\frac{v_C^2}{r}-g \right)}$
$\mathbf\small{\Longrightarrow F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
■ This is the same equation that we obtained in (22) above
47. Finally, we will check the equation at point 'D'
• We have: $\mathbf\small{F_{N(P)}=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at D, θ = 270o
♦ So cos θ = cos 270 = 0
• Substituting the known values, we get: $\mathbf\small{F_{N(D)}=m \left(\frac{v_D^2}{r}+g \times 0 \right)}$
$\mathbf\small{\Longrightarrow F_{N(D)}=\frac{mv_D^2}{r}}$
■ This is the same equation that we obtained in (30) above
• We can use the appropriate one from the two equations:
♦ $\mathbf\small{F_{N(P)}=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
♦ $\mathbf\small{F_{N(P)}=-m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
to find the reaction at any point along the vertical circular loop
• But of course, we need to find the velocity at that point first
♦ For that, we use the equation $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$ which we derived in (23) of the previous section
♦ Where vA is the velocity at the lowest point A
1. In the fig.6.52(a) below, the bead is at the highest point C
• We know that if the bead tries to escape at C, the normal reaction is given by:
$\mathbf\small{F_{N(C)}=m \left(\frac{v_C^2}{r}-g \right)}$
[From step (20) above]
2. If $\mathbf\small{\frac{v_C^2}{r}}$ becomes equal to g, the normal reaction will become zero
• r and g are constants
• So $\mathbf\small{\frac{v_C^2}{r}}$ becomes equal to g when vC has a low value
3. Then the position of the bead will be as shown in fig.b above
• But, if the normal reaction is zero, it means that action is also zero. Because there will be reaction only if there is action
• That means, the bead is not exerting any force on the loop
• In such a situation, it will appear that, the bead is touching the loop as in fig.6.52(b) above
• But the bead is 'just touching' with out exerting any force on the loop
4. Now, if the value of vC becomes still lesser, the bead will fall down and actually touch the loop
• This is shown in fig.c
5. Also because of the low velocity, the bead will try to move inwards
• We can prove this mathematically by the following five steps:
(i) If vC is still lower than in (2), $\mathbf\small{\left(\frac{v_C^2}{r}-g \right)}$ will become a negative quantity
(ii) That means $\mathbf\small{F_{N(C)}}$ becomes a negative quantity
(iii) A negative $\mathbf\small{F_{N(C)}}$ means that, the reaction is in the upward direction
(iv) Upward reaction indicates that the bead is applying force in the downward direction
(v) That is., the bead is trying to move inwards
6. Now we will see the condition for the 'reaction not becoming zero'
• For that, $\mathbf\small{\frac{v_C^2}{r}}$ must be greater than g
• But r and g are constants. $\mathbf\small{v_C}$ is the only variable
■ So we can write: $\mathbf\small{v_C^2}$ must be greater than $\mathbf\small{rg}$
That is., $\mathbf\small{v_C}$ must be greater than $\mathbf\small{\sqrt{rg}}$
• Let a bead of mass 'm' move along a vertical circular loop of radius 1.5 m
• The required velocity at the top most point C = $\mathbf\small{\sqrt{rg}=\sqrt{1.5 \times 9.8}=3.834\,\,ms^{-1}}$
• If vC is exactly equal to 3.834 ms-1, the reaction will become zero as shown below:
$\mathbf\small{F_{N(C)}=m \left(\frac{3.834^2}{1.5}-9.8 \right)=(9.8-9.8)=0}$
• Let us now see what happens if vC is less than 3.834 ms-1
(i) Let vC = 3.5 ms-1
(ii) Then we get: $\mathbf\small{F_{N(C)}=m \left(\frac{3.5^2}{1.5}-9.8 \right)=(8.167-9.8)m=-1.633m\,\,N}$
(iii) The negative value indicates that, the reaction is in the opposite direction
• This is indeed true because, if the velocity is less than the 'required value', the bead will be trying to move inwards
• So we have to use the other equation obtained in step (22):
$\mathbf\small{F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
(i) Substituting the values, we get: $\mathbf\small{F_{N(C)}=m \left(9.8-\frac{3.5^2}{1.5} \right)=1.633m\,\,N}$
• The relation is: $\mathbf\small{v_C=\sqrt{v_A^2-4gr}}$
$\mathbf\small{\Longrightarrow v_C^2={v_A^2-4gr}}$
• Substituting this result in (3), we get:
$\mathbf\small{v_A^2-4gr}$ must be greater than $\mathbf\small{rg}$
• That is., $\mathbf\small{v_A^2}$ must be greater than $\mathbf\small{rg+4rg}$
• That is., $\mathbf\small{v_A^2}$ must be greater than $\mathbf\small{5rg}$
8. Note that we cannot blame vC if the reaction becomes zero at C
• This is because, vC depends on vA
• We have to give the sufficient velocity at the lowest point A. Only then will the object have the required velocity at the highest point C
9. We saw that if the reaction is not to brecome zero at C, the velocity vC at C must be greater than $\mathbf\small{\sqrt{rg}}$
■ What if vC is exactly equal to $\mathbf\small{\sqrt{rg}}$?
Ans: Then at that instant the reaction FN(C) will become zero
• But vC is not equal to zero. It is equal to $\mathbf\small{\sqrt{rg}}$
• Since vC is not equal to zero, the bead will pass the point C
• Once it passes the point C, it's velocity goes on increasing. So it will complete the circular path and will return to the point A
The velocity vC is must be greater less than $\mathbf\small{\sqrt{rg}}$
■ In some practical situations, vC must indeed be greater than $\mathbf\small{\sqrt{rg}}$
■ But in some other practical situations, vC must be less than $\mathbf\small{\sqrt{rg}}$
• We will see both cases in the next section
• When the motion takes place along the loop, if there is sufficient velocity, the bead will be trying to escape from the loop
• So the loop will be in contact with the inner surface of the bead. This is shown in figs.6.49(a) and (b) below:
 |
| Fig.6.49 |
• If the velocity is large, the contact will be as shown in fig.6.49(c)
♦ The inner surface will be in contact with the loop
• If the velocity is small, the contact will be as shown in fig.6.49(d)
♦ The outer surface will be in contact with the loop
■ Whenever the bead is in contact with the loop, the bead will experience a normal reaction FN from the loop
• This FN will vary at different points along the path
• Our next task is to find the reason for such a variation
• We will also see the method to find the magnitude and direction of this FN at various points
1. In fig.6.50(a) below, the bead is shown at the four quadrant points and also the 'any point P'
 |
| Fig.6.50 |
2. The FBD of the bead when it is at A is shown in fig.b
• In this fig.b, the bead is trying to escape from the loop. (Note that, the inner surface of the bead is in contact with the loop)
• So it will exert a force on the loop.
♦ This is a force ‘exerted by the bead’. So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(A) on the bead.
♦ This is a force ‘exerted on the bead’. So we must show it in the FBD
3. The force ‘exerted by the bead’ is directed away from the center ‘O’
• So the reaction FN(A) will be towards the center ‘O’
4. The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.b
5. Let us considered the other possibility:
• In fig.c, the bead is trying to move towards the center of the loop.
♦ It is not trying to escape from the loop
♦ Note that, now the outer surface of the bead is in contact with the loop
• So it will exert a force on the loop. This is a force ‘exerted by the bead’
♦ So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(A) on the bead. This is a force ‘exerted on the bead’
♦ So we must show it in the FBD
• The force ‘exerted by the bead’ is directed towards the center ‘O’
• So the reaction FN(A) will be directed away from the center ‘O’
• The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.c
■ However such a situation generally does not occur.
• Because, then it would mean that, the bead is trying to move vertically upwards when it is at A
• So we do not need the fig.c
6. Let us consider fig.b:
• Both the forces in fig.b act along the same line but opposite in direction
• So the resultant force is (FN(A)-mg)
7. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_A^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_A^2}{r}}$
• This is the centripetal force at point A.
• So we can write: $\mathbf\small{F_{N(A)}-mg=\frac{mv_A^2}{r}}$
$\mathbf\small{\Longrightarrow F_{N(A)}=m \left(\frac{v_A^2}{r}+g \right)}$
• The centripetal force is always directed towards the center of the circle. So it is considered as positive
■ Thus, if we know the velocity at A, we can easily calculate the normal reaction exerted by the loop at A
8. Next we will calculate FN(B), the normal reaction at B
• The FBD of the bead when it is at B is shown in fig.d
• In this fig.d, the bead is trying to escape from the loop. (Note that, the inner surface of the bead is in contact with the loop)
• So it will exert a force on the loop.
♦ This is a force ‘exerted by the bead’. So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(B) on the bead.
♦ This is a force ‘exerted on the bead’. So we must show it in the FBD
9. The force ‘exerted by the bead’ is directed away from the center ‘O’
• So the reaction FN(B) will be towards the center ‘O’
10. The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.d
11. Let us considered the other possibility:
• In fig.e, the bead is trying to move towards the center of the loop.
♦ It is not trying to escape from the loop
♦ Note that, now the outer surface of the bead is in contact with the loop
• So it will exert a force on the loop. This is a force ‘exerted by the bead’
♦ So we cannot show it in the FBD
• The loop will exert a normal reaction FN(B) on the bead. This is a force ‘exerted on the bead’
♦ So we must show it in the FBD
• The force ‘exerted by the bead’ is directed towards the center ‘O’
• So the reaction FN(B) will be directed away from the center ‘O’
• The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.e
■ However such a situation generally does not occur.
• Because, then it would mean that, the bead is trying to move towards left when it is at B
• So we do not need the fig.e
12. Let us consider fig.d:
• The only force in the radial direction is FN(B)
• This is because, here 'mg' is perpendicular to FN(B) and so do not have any component along the radial direction
• So the resultant force is FN(B) itself
13. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_B^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_B^2}{r}}$
• This is the centripetal force at point B
• So we can write: $\mathbf\small{F_{N(B)}=\frac{mv_B^2}{r}}$
■ Thus, if we know the velocity vB at B, we can easily calculate FN(B)
14. But how do we find vB?
• If we know the velocity vA at the lowest point A, we can calculate the velocity vB at B
• For that, we use the equation obtained in (10) in the previous section
♦ Here we will write it again: $\mathbf\small{v_B=\sqrt{v_A^2-2gr}}$
■ Once we calculate vB, we can easily calculate FN(B) using the equation in (13) above
15. Next we will calculate FN(C), the normal reaction at C
• The FBD of the bead when it is at C is shown in fig.f
• In this fig.f, the bead is trying to escape from the loop. (Note that, the inner surface of the bead is in contact with the loop)
• So it will exert a force on the loop.
♦ This is a force ‘exerted by the bead’. So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(C) on the bead.
♦ This is a force ‘exerted on the bead’. So we must show it in the FBD
16. The force ‘exerted by the bead’ is directed away from the center ‘O’
• So the reaction FN(A) will be towards the center ‘O’
17. The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.f
18. Let us considered the other possibility:
• In fig.c, the bead is trying to move towards the center of the loop.
♦ It is not trying to escape from the loop
♦ Note that, now the outer surface of the bead is in contact with the loop
• So it will exert a force on the loop. This is a force ‘exerted by the bead’
♦ So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(C) on the bead. This is a force ‘exerted on the bead’
♦ So we must show it in the FBD
• The force ‘exerted by the bead’ is directed towards the center ‘O’
• So the reaction FN(C) will be directed away from the center ‘O’
• The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.g
■ Such a situation can indeed occur if the velocity is not sufficient
• So we need both the figs.(f) and (g)
19. First, let us consider fig.f:
• Both the forces in fig.f act along the same line and in same direction
• So the resultant force is (FN(C)+mg)
20. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_C^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_C^2}{r}}$
• This is the centripetal force at point C
• So we can write: $\mathbf\small{F_{N(C)}+mg=\frac{mv_C^2}{r}}$
$\mathbf\small{\Longrightarrow F_{N(C)}=m \left(\frac{v_C^2}{r}-g \right)}$
■ Thus, if we know the velocity at C, we can easily calculate the normal reaction exerted by the loop at C
21. Now, let us consider fig.g:
• Both the forces in fig.f act along the same line but in opposite direction
• So the resultant force is (-FN(C)+mg)
22. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_C^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_C^2}{r}}$
• This is the centripetal force at point C
• So we can write: $\mathbf\small{-F_{N(C)}+mg=\frac{mv_C^2}{r}}$
$\mathbf\small{\Longrightarrow F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
■ Thus, if we know the velocity vC at C, we can easily calculate FN(C)
23. But how do we find vC?
• If we know the velocity vA at the lowest point A, we can calculate the velocity vC at C
• For that, we use the equation obtained in (13) in the previous section
♦ Here we will write it again: $\mathbf\small{v_C=\sqrt{v_A^2-4gr}}$
■ Once we calculate vC, we can easily calculate FN(C) using the equation in (20) or (22) above
24. So we have two equations at C. They are:
(i) From step (20), we have: $\mathbf\small{F_{N(C)}=m \left(\frac{v_C^2}{r}-g \right)}$
(ii) From step (22), we have: $\mathbf\small{F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
• We will see their applications later in this section
25. Next we will calculate FN(D), the normal reaction D
• The FBD of the bead when it is at D is shown in fig.h
• In this fig.h, the bead is trying to escape from the loop. (Note that, the inner surface of the bead is in contact with the loop)
• So it will exert a force on the loop.
♦ This is a force ‘exerted by the bead’. So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(D) on the bead.
♦ This is a force ‘exerted on the bead’. So we must show it in the FBD
26. The force ‘exerted by the bead’ is directed away from the center ‘O’
• So the reaction FN(D) will be towards the center ‘O’
27. The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.h
28. Let us considered the other possibility:
• In fig.i, the bead is trying to move towards the center of the loop.
♦ It is not trying to escape from the loop
♦ Note that, now the outer surface of the bead is in contact with the loop
• So it will exert a force on the loop. This is a force ‘exerted by the bead’
♦ So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(D) on the bead. This is a force ‘exerted on the bead’
♦ So we must show it in the FBD
• The force ‘exerted by the bead’ is directed towards the center ‘O’
• So the reaction FN(D) will be directed away from the center ‘O’
• The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.i
■ However such a situation generally does not occur.
• Because, then it would mean that, the bead is trying to move towards right when it is at B
• So we do not need the fig.i
29. Let us consider fig.h:
• The only force in the radial direction is FN(D)
• This is because, here 'mg' is perpendicular to FN(D) and so do not have any component along the radial direction
• So the resultant force is FN(D) itself
30. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_D^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_D^2}{r}}$
• This is the centripetal force at point D
• So we can write: $\mathbf\small{F_{N(D)}=\frac{mv_D^2}{r}}$
■ Thus, if we know the velocity vD at D, we can easily calculate FN(D)
31. But how do we find vD?
• If we know the velocity vA at the lowest point A, we can calculate the velocity vD at D
• For that, we use the equation obtained in (16) in the previous section
♦ Here we will write it again: $\mathbf\small{v_D=\sqrt{v_A^2-2gr}}$
■ Once we calculate vD, we can easily calculate FN(D) using the equation in (30)
■ It is interesting to note that, at B and D, just like the 'magnitudes of velocities', the 'magnitudes of reactions' are also the same
• So we successfully calculated the tensions at all the four quadrant points. But what about the intermediate points?
• For that, we will have to apply 'resolution of forces' (Details here). Let us see how it is done:
32. In fig.6.51(a) below, the bead is at P
 |
| Fig.6.51 |
33. In fig.6.65(b), the portion of the bead alone is shown
• In this fig.b, a vertical is drawn through the bead
• The following two verticals will be parallel:
♦ Vertical through O
♦ Vertical through P
• So, if we extend OP downwards along the same line, that extension will make the same angle θ with the vertical
34. This 'same angle' is our clue
• We know that, the 'component which is adjacent to the angle' will get the cosine. And the other component will get the sine
• So the component of the weight 'mg' which acts along the line of the string is mg cosθ
• This is shown in the FBD in fig.c
• In this fig.c, the bead is trying to escape from the loop. (Note that, the inner surface of the bead is in contact with the loop)
• So it will exert a force on the loop.
♦ This is a force ‘exerted by the bead’. So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(P) on the bead.
♦ This is a force ‘exerted on the bead’. So we must show it in the FBD
35. The force ‘exerted by the bead’ is directed away from the center ‘O’
• So the reaction FN(P) will be towards the center ‘O’
36. The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.c
37. Let us considered the other possibility:
• In fig.d, the bead is trying to move towards the center of the loop.
♦ It is not trying to escape from the loop
♦ Note that, now the outer surface of the bead is in contact with the loop
• So it will exert a force on the loop. This is a force ‘exerted by the bead’
♦ So we cannot show it in the FBD.
• The loop will exert a normal reaction FN(P) on the bead. This is a force ‘exerted on the bead’
♦ So we must show it in the FBD
• The force ‘exerted by the bead’ is directed towards the center ‘O’
• So the reaction FN(P) will be directed away from the center ‘O’
• The weight ‘mg’ is always acting vertically downwards
• Thus we get the FBD shown in fig.d
■ Such a situation can indeed occur at the highest point 'C', if the velocity is not sufficient
• So we need both the figs.(c) and (d)
38. First let us consider fig.c:
• We see that the net force in the radial direction is (FN(P) -mg cosθ)
39. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_P^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_P^2}{r}}$
• This is the centripetal force at point P.
• So we can write: $\mathbf\small{F_{N(P)}-mg \cos \theta =\frac{mv_P^2}{r}}$
• $\mathbf\small{\Longrightarrow F_{N(P)}=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
■ Thus, if we know the velocity vP at P, we can easily calculate FN(P)
40. Now, let us consider fig.d:
• We see that the net force in the radial direction is (-FN(P) - mg cosθ)
41. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_P^2}{r}}$
• So (mass × acceleration) = $\mathbf\small{\frac{mv_P^2}{r}}$
• This is the centripetal force at point P.
• So we can write: $\mathbf\small{-F_{N(P)}-mg \cos \theta =\frac{mv_P^2}{r}}$
• $\mathbf\small{\Longrightarrow F_{N(P)}=-m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
■ Thus, if we know the velocity vP at P, we can easily calculate FN(P)
42. But how do we find vP?
• If we know the velocity vA at the lowest point A, we can calculate the velocity vP at P
• For that, we use the equation obtained in (23) in the previous section
♦ Here we will write it again: $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$
■ Once we calculate vP, we can easily calculate FN(P) using the equation in (39) or (41) above
• If we can use the equation derived in (39) or (41) above for 'any point', it must be applicable to points A, B, C and D also. Let us check:
43. First we will check the equation at point 'A'
• We have: $\mathbf\small{F_{N(P)}=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at A, θ = 0o
♦ So cos θ = cos 0 = 1
• Substituting the known values, we get:
$\mathbf\small{F_{N(A)}=m \left(\frac{v_A^2}{r}+g \right)}$
■ This is the same equation that we obtained in (7) above
44. Next we will check the equation at point 'B'
• We have: $\mathbf\small{F_{N(P)}=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at B, θ = 90o
♦ So cos θ = cos 90 = 0
• Substituting the known values, we get: $\mathbf\small{F_{N(B)}=m \left(\frac{v_B^2}{r}+g \times 0 \right)}$
$\mathbf\small{\Longrightarrow F_{N(B)}=\frac{mv_B^2}{r}}$
■ This is the same equation that we obtained in (13) above
45. Next we will check the equation at point 'C'
• We have: $\mathbf\small{F_{N(P)}=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at C, θ = 180o
♦ So cos θ = cos 180 = -1
• Substituting the known values, we get:
$\mathbf\small{F_{N(C)}=m \left(\frac{v_C^2}{r}-g \right)}$
■ This is the same equation that we obtained in (20) above
46. When the bead tries to move inwards we have to use the equation in (41)
We have: $\mathbf\small{F_{N(P)}=-m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at C, θ = 180o
♦ So cos θ = cos 180 = -1
• Substituting the known values, we get:
$\mathbf\small{F_{N(C)}=-m \left(\frac{v_C^2}{r}+g \times (-1) \right)}$
So we get: $\mathbf\small{F_{N(C)}=-m \left(\frac{v_C^2}{r}-g \right)}$
$\mathbf\small{\Longrightarrow F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
■ This is the same equation that we obtained in (22) above
47. Finally, we will check the equation at point 'D'
• We have: $\mathbf\small{F_{N(P)}=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at D, θ = 270o
♦ So cos θ = cos 270 = 0
• Substituting the known values, we get: $\mathbf\small{F_{N(D)}=m \left(\frac{v_D^2}{r}+g \times 0 \right)}$
$\mathbf\small{\Longrightarrow F_{N(D)}=\frac{mv_D^2}{r}}$
■ This is the same equation that we obtained in (30) above
■ So it is confirmed:
♦ $\mathbf\small{F_{N(P)}=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
♦ $\mathbf\small{F_{N(P)}=-m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
to find the reaction at any point along the vertical circular loop
• But of course, we need to find the velocity at that point first
♦ For that, we use the equation $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$ which we derived in (23) of the previous section
♦ Where vA is the velocity at the lowest point A
Minimum velocity required at the lowest point A
 |
| Fig.6.52 |
$\mathbf\small{F_{N(C)}=m \left(\frac{v_C^2}{r}-g \right)}$
[From step (20) above]
2. If $\mathbf\small{\frac{v_C^2}{r}}$ becomes equal to g, the normal reaction will become zero
• r and g are constants
• So $\mathbf\small{\frac{v_C^2}{r}}$ becomes equal to g when vC has a low value
3. Then the position of the bead will be as shown in fig.b above
• But, if the normal reaction is zero, it means that action is also zero. Because there will be reaction only if there is action
• That means, the bead is not exerting any force on the loop
• In such a situation, it will appear that, the bead is touching the loop as in fig.6.52(b) above
• But the bead is 'just touching' with out exerting any force on the loop
4. Now, if the value of vC becomes still lesser, the bead will fall down and actually touch the loop
• This is shown in fig.c
5. Also because of the low velocity, the bead will try to move inwards
• We can prove this mathematically by the following five steps:
(i) If vC is still lower than in (2), $\mathbf\small{\left(\frac{v_C^2}{r}-g \right)}$ will become a negative quantity
(ii) That means $\mathbf\small{F_{N(C)}}$ becomes a negative quantity
(iii) A negative $\mathbf\small{F_{N(C)}}$ means that, the reaction is in the upward direction
(iv) Upward reaction indicates that the bead is applying force in the downward direction
(v) That is., the bead is trying to move inwards
6. Now we will see the condition for the 'reaction not becoming zero'
• For that, $\mathbf\small{\frac{v_C^2}{r}}$ must be greater than g
• But r and g are constants. $\mathbf\small{v_C}$ is the only variable
■ So we can write: $\mathbf\small{v_C^2}$ must be greater than $\mathbf\small{rg}$
That is., $\mathbf\small{v_C}$ must be greater than $\mathbf\small{\sqrt{rg}}$
An example:
• The required velocity at the top most point C = $\mathbf\small{\sqrt{rg}=\sqrt{1.5 \times 9.8}=3.834\,\,ms^{-1}}$
• If vC is exactly equal to 3.834 ms-1, the reaction will become zero as shown below:
$\mathbf\small{F_{N(C)}=m \left(\frac{3.834^2}{1.5}-9.8 \right)=(9.8-9.8)=0}$
• Let us now see what happens if vC is less than 3.834 ms-1
(i) Let vC = 3.5 ms-1
(ii) Then we get: $\mathbf\small{F_{N(C)}=m \left(\frac{3.5^2}{1.5}-9.8 \right)=(8.167-9.8)m=-1.633m\,\,N}$
(iii) The negative value indicates that, the reaction is in the opposite direction
• This is indeed true because, if the velocity is less than the 'required value', the bead will be trying to move inwards
• So we have to use the other equation obtained in step (22):
$\mathbf\small{F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
(i) Substituting the values, we get: $\mathbf\small{F_{N(C)}=m \left(9.8-\frac{3.5^2}{1.5} \right)=1.633m\,\,N}$
Now we will continue the discussion:
7. But $\mathbf\small{v_C}$ depends on the velocity $\mathbf\small{v_A}$ at the lowest point A
7. But $\mathbf\small{v_C}$ depends on the velocity $\mathbf\small{v_A}$ at the lowest point A
$\mathbf\small{\Longrightarrow v_C^2={v_A^2-4gr}}$
• Substituting this result in (3), we get:
$\mathbf\small{v_A^2-4gr}$ must be greater than $\mathbf\small{rg}$
• That is., $\mathbf\small{v_A^2}$ must be greater than $\mathbf\small{rg+4rg}$
• That is., $\mathbf\small{v_A^2}$ must be greater than $\mathbf\small{5rg}$
8. Note that we cannot blame vC if the reaction becomes zero at C
• This is because, vC depends on vA
• We have to give the sufficient velocity at the lowest point A. Only then will the object have the required velocity at the highest point C
9. We saw that if the reaction is not to brecome zero at C, the velocity vC at C must be greater than $\mathbf\small{\sqrt{rg}}$
■ What if vC is exactly equal to $\mathbf\small{\sqrt{rg}}$?
Ans: Then at that instant the reaction FN(C) will become zero
• But vC is not equal to zero. It is equal to $\mathbf\small{\sqrt{rg}}$
• Since vC is not equal to zero, the bead will pass the point C
• Once it passes the point C, it's velocity goes on increasing. So it will complete the circular path and will return to the point A
■ In the above discussion we saw that:
■ In some practical situations, vC must indeed be greater than $\mathbf\small{\sqrt{rg}}$
■ But in some other practical situations, vC must be less than $\mathbf\small{\sqrt{rg}}$
• We will see both cases in the next section
No comments:
Post a Comment