In the previous section, we saw the potential energy when a spring is stretched/extended vertically. In this section we will see motion of objects in a vertical circle.
1. In fig.6.43(a) below, a small object of mass ‘m’ is moving along a vertical circle
• Four points are marked on the circular path.
♦ A and C are the lowest and highest points
♦ B and D are the quadrant points
2. Let us analyze the travel of the object along the circular path ABCD
• Wherever be the position of the object along ABCD, it will be experiencing the downward acceleration due to gravity ‘g’
• When the object travels upwards from A to C along B, this g will be negative (opposite to the direction of motion)
♦ So the linear speed ‘v’ of the object goes on decreasing
• When the object travels downwards from C to A along D, this g will be positive
♦ So the ‘v’ of the object goes on increasing
■ We can write:
When an object is in a vertical circular motion, it’s linear speed ‘v’ changes continuously
■ Also we can write:
When an object is in a vertical circular motion, since it’s linear speed is not uniform, it is not a uniform circular motion (UCM).
• We have seen the details about UCM in a previous chapter (Details here)
3. When we discussed about UCM, we considered the motion in a horizontal circle.
• When the circle is horizontal, there is no 'change in height' for the object
♦ So the effect due to g will be same at every point along the path
• Thus there is no change in linear speed ‘v’
♦ As a consequence, the centripetal acceleration $\mathbf\small{\frac{v^2}{r}}$ will be uniform
♦ The centripetal force $\mathbf\small{\frac{mv^2}{r}}$ also will be uniform
4. But in our present case, we saw that the linear speed ‘v’ changes continuously.
♦ So the centripetal acceleration $\mathbf\small{\frac{v^2}{r}}$ changes continuously
♦ The centripetal force $\mathbf\small{\frac{mv^2}{r}}$ will also change continuously
5. Another difference between UCM and ‘motion in vertical circle’:
• In UCM, the linear speed ‘v’ remains constant. So there is no linear acceleration
• In the case of ‘motion in vertical circle’ the ‘v’ changes continuously. So there will be linear acceleration also in addition to centripetal acceleration
■ UCM:
• Centripetal acceleration only
♦ Magnitude of this centripetal acceleration is a constant
• No linear acceleration
■ Motion in vertical circle:
• Centripetal acceleration is present
♦ Magnitude of this centripetal acceleration is not constant
• Linear acceleration is also present
♦ If there is no force other than gravity, magnitude of this linear acceleration is constant
6. If a person ties an object of mass ‘m’ to the end of a string and whirl it in a vertical circle (fig.6.43.b), is he doing any work on the object? Let us analyze:
(i) Fig.6.43(c) above shows the force and velocity when the object is at any point ‘P’
• The only contact that the person have with the object is ‘through the string’
• If he wants to do any work on the object, it will be through the string
(ii) The tension ‘T’ in the string will always be in a radial direction
• The linear speed ‘v’ will always be tangential to the path
(iii) Any tangent is perpendicular to the radius
• So the force ‘T’ is perpendicular to the direction of motion
(v) That means, ‘T’ cannot do any work on the object. (Details here)
7. The only external force that is capable of doing any work in this situation is the ‘force of gravity’
• ‘Force of gravity’ is a conservative force
♦ It does not depend on the path
♦ The work done by it can be stored up for later use
• So the law of conservation of energy can be applied to motion in a vertical circle
• That is., Ei = Ef
8. Let us apply it to the points A and B
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EB
9. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = $\mathbf\small{\frac{m\,v_A^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_A}$
♦ Where ‘hA’ is the height of the point ‘A’ from the datum
♦ In the previous section we saw that, the datum can be drawn through any convenient point
♦ Let us assume that the datum passes through ‘A’ itself. Then hA = 0
♦ So potential energy at A = 0
(Remember that, once the datum is fixed, we must follow it for the whole problem)
■ So total energy EA = $\mathbf\small{\frac{m\,v_A^2}{2}}$
(ii) Total energy EB
• Kinetic energy = $\mathbf\small{\frac{m\,v_B^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_B}$
♦ Where ‘hB’ is the height of the point ‘B’ from the datum
♦ From the fig., we can see that hB = r, the radius of the circle
■ So total energy EB = $\mathbf\small{\frac{m\,v_B^2}{2}+mgr}$
10. Equating the two energies we get:
$\mathbf\small{\frac{m\,v_A^2}{2}=\frac{m\,v_B^2}{2}+mgr}$
Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{v_A^2=v_B^2+2gr}$
$\mathbf\small{\Longrightarrow v_B=\sqrt{v_A^2-2gr}}$
■ So, if we know the velocity vA at the lower most point A, we can easily obtain the velocity vB at the quadrant point B on the right side
11. Next, let us apply the law of conservation of energy to the points A and C
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EC
12. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = $\mathbf\small{\frac{m\,v_A^2}{2}}$
• Gravitational potential energy = 0
■ So total energy EA = $\mathbf\small{\frac{m\,v_A^2}{2}}$
(ii) Total energy EC
• Kinetic energy = $\mathbf\small{\frac{m\,v_C^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_C}$
♦ Where ‘hC’ is the height of the point ‘C’ from the datum
♦ From the fig., we can see that hC = 2r, the diameter of the circle
■ So total energy EB = $\mathbf\small{\frac{m\,v_C^2}{2}+2mgr}$
13. Equating the two energies we get:
$\mathbf\small{\frac{m\,v_A^2}{2}=\frac{m\,v_B^2}{2}+2mgr}$
Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{v_A^2=v_C^2+4gr}$
$\mathbf\small{\Longrightarrow v_C=\sqrt{v_A^2-4gr}}$
■ So, if we know the velocity vA at the lower most point A, we can easily obtain the velocity vC at the top most point C
14. Next, let us apply the law of conservation of energy to the points A and D
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = ED
15. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = $\mathbf\small{\frac{m\,v_A^2}{2}}$
• Gravitational potential energy = 0
■ So total energy EA = $\mathbf\small{\frac{m\,v_A^2}{2}}$
(ii) Total energy ED
• Kinetic energy = $\mathbf\small{\frac{m\,v_D^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_D}$
♦ Where ‘hD’ is the height of the point ‘D’ from the datum
♦ From the fig., we can see that hD = r, the radius of the circle
■ So total energy ED = $\mathbf\small{\frac{m\,v_D^2}{2}+mgr}$
16. Equating the two energies we get:
$\mathbf\small{\frac{m\,v_A^2}{2}=\frac{m\,v_D^2}{2}+mgr}$
• Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{v_A^2=v_D^2+2gr}$
$\mathbf\small{\Longrightarrow v_D=\sqrt{v_A^2-2gr}}$
■ So, if we know the velocity vA at the lower most point A, we can easily obtain the velocity vD at the quadrant point D on the left side
■ Also it is interesting to note that magnitudes of vB and vD are the same
• However, they are opposite in directions
♦ vB is upwards
♦ vD is downwards
17. So we successfully calculated the velocities at the two quadrant points and the top most point.
• What about the velocities at the intermediate points?
• For that, we will have to use trigonometry. Let us see how it is done:
18. Consider the instant when the object is at a point 'P' shown in fig.6.44(a) below:
• At that instant, the string makes an angle θ with the vertical
19. We want the vertical height (hP) of 'P' above the datum
• The datum passes through 'A'
• So we want the vertical height of 'P' above 'A'
20. Drop a perpendicular from P onto OA. This is shown in fig.6.44(b)
• Let 'Q' be the foot of this perpendicular
• Then AQ = hP. In the right triangle OPQ, cos θ = OQ⁄OP
• So OQ = OP cos θ = r cos θ
• Thus we get: hP = AQ = (OA - OQ) = (r - r cos θ) = r(1- cos θ)
• Now that we know the height of 'P' above the datum, we can apply the law of conservation of energy to the two points A and P
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EP
22. Let us write the various energies:
(i) Total energy EA:
We have already calculated it as: $\mathbf\small{\frac{m\,v_A^2}{2}}$
(ii) Total energy EP
• Kinetic energy = $\mathbf\small{\frac{m\,v_P^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_P}$
♦ Where ‘hP’ is the height of the point ‘P’ from the datum
♦ From the fig., we have calculated it as: r(1- cos θ)
■ So total energy EP = $\mathbf\small{\frac{m\,v_P^2}{2}+mgr(1-\cos \theta)}$
23. Equating the two energies we get:
$\mathbf\small{\frac{m\,v_A^2}{2}=\frac{m\,v_P^2}{2}+mgr(1-\cos \theta)}$
• Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{v_A^2=v_P^2+2gr(1-\cos \theta)}$
$\mathbf\small{\Longrightarrow v_P^2=v_A^2-2gr(1-\cos \theta)}$
$\mathbf\small{\Longrightarrow v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$
■ So, if we know the velocity vA at the lower most point A, we can easily obtain the velocity vP at any point P on the circle. The only additional information required is the angle which OP makes with the vertical
• If we can use the expression derived in (23) above for 'any point', it must be applicable to points B, C and D also. Let us check:
24. First we will check the equation at point 'B'
• We have: $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$
• When the object is at B, θ = 90o
♦ So cos θ = cos 90 = 0
• Substituting the known values, we get:
$\mathbf\small{v_B=\sqrt{v_A^2-2gr(1-0)}}$
$\mathbf\small{\Longrightarrow v_B=\sqrt{v_A^2-2gr}}$
■ This is the same expression that we obtained in (10)
25. Next we will check the equation at point 'C'
• We have: $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$
• When the object is at C, θ = 180o
♦ So cos θ = cos 180 = -1
• Substituting the known values, we get:
$\mathbf\small{v_C=\sqrt{v_A^2-2gr(1-(-1))}}$
$\mathbf\small{\Longrightarrow v_C=\sqrt{v_A^2-2gr \times 2}}$
$\mathbf\small{\Longrightarrow v_C=\sqrt{v_A^2-4gr}}$
■ This is the same expression that we obtained in (14)
26. Finally, we will check the equation at point 'D'
• We have: $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$
• When the object is at C, θ = 270o
♦ So cos θ = cos 270 = 0
• Substituting the known values, we get:
$\mathbf\small{v_D=\sqrt{v_A^2-2gr(1-0)}}$
$\mathbf\small{\Longrightarrow v_D=\sqrt{v_A^2-2gr}}$
■ This is the same expression that we obtained in (16)
1. In fig.6.43(a) below, a small object of mass ‘m’ is moving along a vertical circle
 |
| Fig.6.43 |
♦ A and C are the lowest and highest points
♦ B and D are the quadrant points
2. Let us analyze the travel of the object along the circular path ABCD
• Wherever be the position of the object along ABCD, it will be experiencing the downward acceleration due to gravity ‘g’
• When the object travels upwards from A to C along B, this g will be negative (opposite to the direction of motion)
♦ So the linear speed ‘v’ of the object goes on decreasing
• When the object travels downwards from C to A along D, this g will be positive
♦ So the ‘v’ of the object goes on increasing
■ We can write:
When an object is in a vertical circular motion, it’s linear speed ‘v’ changes continuously
■ Also we can write:
When an object is in a vertical circular motion, since it’s linear speed is not uniform, it is not a uniform circular motion (UCM).
• We have seen the details about UCM in a previous chapter (Details here)
3. When we discussed about UCM, we considered the motion in a horizontal circle.
• When the circle is horizontal, there is no 'change in height' for the object
♦ So the effect due to g will be same at every point along the path
• Thus there is no change in linear speed ‘v’
♦ As a consequence, the centripetal acceleration $\mathbf\small{\frac{v^2}{r}}$ will be uniform
♦ The centripetal force $\mathbf\small{\frac{mv^2}{r}}$ also will be uniform
4. But in our present case, we saw that the linear speed ‘v’ changes continuously.
♦ So the centripetal acceleration $\mathbf\small{\frac{v^2}{r}}$ changes continuously
♦ The centripetal force $\mathbf\small{\frac{mv^2}{r}}$ will also change continuously
5. Another difference between UCM and ‘motion in vertical circle’:
• In UCM, the linear speed ‘v’ remains constant. So there is no linear acceleration
• In the case of ‘motion in vertical circle’ the ‘v’ changes continuously. So there will be linear acceleration also in addition to centripetal acceleration
We can write a summary:
• Centripetal acceleration only
♦ Magnitude of this centripetal acceleration is a constant
• No linear acceleration
■ Motion in vertical circle:
• Centripetal acceleration is present
♦ Magnitude of this centripetal acceleration is not constant
• Linear acceleration is also present
♦ If there is no force other than gravity, magnitude of this linear acceleration is constant
Now we will continue the discussion:
(i) Fig.6.43(c) above shows the force and velocity when the object is at any point ‘P’
• The only contact that the person have with the object is ‘through the string’
• If he wants to do any work on the object, it will be through the string
(ii) The tension ‘T’ in the string will always be in a radial direction
• The linear speed ‘v’ will always be tangential to the path
(iii) Any tangent is perpendicular to the radius
• So the force ‘T’ is perpendicular to the direction of motion
(v) That means, ‘T’ cannot do any work on the object. (Details here)
7. The only external force that is capable of doing any work in this situation is the ‘force of gravity’
• ‘Force of gravity’ is a conservative force
♦ It does not depend on the path
♦ The work done by it can be stored up for later use
• So the law of conservation of energy can be applied to motion in a vertical circle
• That is., Ei = Ef
8. Let us apply it to the points A and B
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EB
9. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = $\mathbf\small{\frac{m\,v_A^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_A}$
♦ Where ‘hA’ is the height of the point ‘A’ from the datum
♦ In the previous section we saw that, the datum can be drawn through any convenient point
♦ Let us assume that the datum passes through ‘A’ itself. Then hA = 0
♦ So potential energy at A = 0
(Remember that, once the datum is fixed, we must follow it for the whole problem)
■ So total energy EA = $\mathbf\small{\frac{m\,v_A^2}{2}}$
(ii) Total energy EB
• Kinetic energy = $\mathbf\small{\frac{m\,v_B^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_B}$
♦ Where ‘hB’ is the height of the point ‘B’ from the datum
♦ From the fig., we can see that hB = r, the radius of the circle
■ So total energy EB = $\mathbf\small{\frac{m\,v_B^2}{2}+mgr}$
10. Equating the two energies we get:
$\mathbf\small{\frac{m\,v_A^2}{2}=\frac{m\,v_B^2}{2}+mgr}$
Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{v_A^2=v_B^2+2gr}$
$\mathbf\small{\Longrightarrow v_B=\sqrt{v_A^2-2gr}}$
■ So, if we know the velocity vA at the lower most point A, we can easily obtain the velocity vB at the quadrant point B on the right side
11. Next, let us apply the law of conservation of energy to the points A and C
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EC
12. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = $\mathbf\small{\frac{m\,v_A^2}{2}}$
• Gravitational potential energy = 0
■ So total energy EA = $\mathbf\small{\frac{m\,v_A^2}{2}}$
(ii) Total energy EC
• Kinetic energy = $\mathbf\small{\frac{m\,v_C^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_C}$
♦ Where ‘hC’ is the height of the point ‘C’ from the datum
♦ From the fig., we can see that hC = 2r, the diameter of the circle
■ So total energy EB = $\mathbf\small{\frac{m\,v_C^2}{2}+2mgr}$
13. Equating the two energies we get:
$\mathbf\small{\frac{m\,v_A^2}{2}=\frac{m\,v_B^2}{2}+2mgr}$
Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{v_A^2=v_C^2+4gr}$
$\mathbf\small{\Longrightarrow v_C=\sqrt{v_A^2-4gr}}$
■ So, if we know the velocity vA at the lower most point A, we can easily obtain the velocity vC at the top most point C
14. Next, let us apply the law of conservation of energy to the points A and D
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = ED
15. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = $\mathbf\small{\frac{m\,v_A^2}{2}}$
• Gravitational potential energy = 0
■ So total energy EA = $\mathbf\small{\frac{m\,v_A^2}{2}}$
(ii) Total energy ED
• Kinetic energy = $\mathbf\small{\frac{m\,v_D^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_D}$
♦ Where ‘hD’ is the height of the point ‘D’ from the datum
♦ From the fig., we can see that hD = r, the radius of the circle
■ So total energy ED = $\mathbf\small{\frac{m\,v_D^2}{2}+mgr}$
16. Equating the two energies we get:
$\mathbf\small{\frac{m\,v_A^2}{2}=\frac{m\,v_D^2}{2}+mgr}$
• Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{v_A^2=v_D^2+2gr}$
$\mathbf\small{\Longrightarrow v_D=\sqrt{v_A^2-2gr}}$
■ So, if we know the velocity vA at the lower most point A, we can easily obtain the velocity vD at the quadrant point D on the left side
■ Also it is interesting to note that magnitudes of vB and vD are the same
• However, they are opposite in directions
♦ vB is upwards
♦ vD is downwards
17. So we successfully calculated the velocities at the two quadrant points and the top most point.
• What about the velocities at the intermediate points?
• For that, we will have to use trigonometry. Let us see how it is done:
18. Consider the instant when the object is at a point 'P' shown in fig.6.44(a) below:
 |
| Fig.6.44 |
19. We want the vertical height (hP) of 'P' above the datum
• The datum passes through 'A'
• So we want the vertical height of 'P' above 'A'
20. Drop a perpendicular from P onto OA. This is shown in fig.6.44(b)
• Let 'Q' be the foot of this perpendicular
• Then AQ = hP. In the right triangle OPQ, cos θ = OQ⁄OP
• So OQ = OP cos θ = r cos θ
• Thus we get: hP = AQ = (OA - OQ) = (r - r cos θ) = r(1- cos θ)
• Now that we know the height of 'P' above the datum, we can apply the law of conservation of energy to the two points A and P
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EP
22. Let us write the various energies:
(i) Total energy EA:
We have already calculated it as: $\mathbf\small{\frac{m\,v_A^2}{2}}$
(ii) Total energy EP
• Kinetic energy = $\mathbf\small{\frac{m\,v_P^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mgh_P}$
♦ Where ‘hP’ is the height of the point ‘P’ from the datum
♦ From the fig., we have calculated it as: r(1- cos θ)
■ So total energy EP = $\mathbf\small{\frac{m\,v_P^2}{2}+mgr(1-\cos \theta)}$
23. Equating the two energies we get:
$\mathbf\small{\frac{m\,v_A^2}{2}=\frac{m\,v_P^2}{2}+mgr(1-\cos \theta)}$
• Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{v_A^2=v_P^2+2gr(1-\cos \theta)}$
$\mathbf\small{\Longrightarrow v_P^2=v_A^2-2gr(1-\cos \theta)}$
$\mathbf\small{\Longrightarrow v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$
■ So, if we know the velocity vA at the lower most point A, we can easily obtain the velocity vP at any point P on the circle. The only additional information required is the angle which OP makes with the vertical
• If we can use the expression derived in (23) above for 'any point', it must be applicable to points B, C and D also. Let us check:
24. First we will check the equation at point 'B'
• We have: $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$
• When the object is at B, θ = 90o
♦ So cos θ = cos 90 = 0
• Substituting the known values, we get:
$\mathbf\small{v_B=\sqrt{v_A^2-2gr(1-0)}}$
$\mathbf\small{\Longrightarrow v_B=\sqrt{v_A^2-2gr}}$
■ This is the same expression that we obtained in (10)
25. Next we will check the equation at point 'C'
• We have: $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$
• When the object is at C, θ = 180o
♦ So cos θ = cos 180 = -1
• Substituting the known values, we get:
$\mathbf\small{v_C=\sqrt{v_A^2-2gr(1-(-1))}}$
$\mathbf\small{\Longrightarrow v_C=\sqrt{v_A^2-2gr \times 2}}$
$\mathbf\small{\Longrightarrow v_C=\sqrt{v_A^2-4gr}}$
■ This is the same expression that we obtained in (14)
26. Finally, we will check the equation at point 'D'
• We have: $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$
• When the object is at C, θ = 270o
♦ So cos θ = cos 270 = 0
• Substituting the known values, we get:
$\mathbf\small{v_D=\sqrt{v_A^2-2gr(1-0)}}$
$\mathbf\small{\Longrightarrow v_D=\sqrt{v_A^2-2gr}}$
■ This is the same expression that we obtained in (16)
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