In the previous section, we saw that the velocity of an ‘object in vertical circular motion’ will be different at different points along the path. We saw the method to find the velocity at any point.
• When an object is tied to a string and whirled in a vertical circle, not only the velocity, but also the ‘tension in the string’ will vary.
• In this section we will see the reason for such a variation. We will also see the method to find the tension at any point
1. In fig.6.45(a) below, an object of mass ‘m’ is tied to a string and whirled in vertical circle.
• ‘O’ is the center of the circle and ‘r’ is the radius
2. The FBD of the object when it is at A is shown in fig.b
• We know that, the force in a string will always be tensile in nature (Details here). This tension will be pulling at both the ends of the strings
• So the object will experience a pulling through the string. This pulling force by the string is shown in fig.b as ‘TA’
3. Also the object will experience a pull towards the center of the earth
• This force by the earth is shown in fig.b as ‘mg’
4. When the object is at A,
♦ the force ‘TA’ will be vertically upwards
♦ the force ‘mg’ will be vertically downwards
• So both the forces act along the same line but opposite in direction
• The resultant force is (TA-mg)
5. For circular motion (both horizontal and vertical), centripetal force is required.
• The person who whirls the object has to provide this force
• For the person, the only contact with the object is through the string. So the centripetal force should be provided through the string
• We see that, the net force in the direction of the string is (TA-mg)
6. 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: $\mathbf\small{T_A-mg=\frac{mv_A^2}{r}}$
$\mathbf\small{\Longrightarrow T_A=m \left(\frac{v_A^2}{r}+g \right)}$
• Centripetal force is always directed towards the center. So it is positive
■ Thus, if we know the velocity at A, we can easily calculate the tension at A
7. Next we will calculate TB, the tension at B
• The FBD of the object when it is at B is shown in fig.c
• We see that the net force in the direction of the string is TB itself
• This is because, 'mg' is perpendicular to the string and hence have no force component along the string
8. 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: $\mathbf\small{T_B=\frac{mv_B^2}{r}}$
■ Thus, if we know the velocity (vB) at B, we can easily calculate the tension at B
9. How do we calculate vB?
• If we know the velocity vA at the lowest point A, we can calculate vB
• 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 the tension at B using the equation in (8) above
10. Next we will calculate TC, the tension at C
• The FBD of the object when it is at C is shown in fig.d
• We see that the net force in the direction of the string is (TC+mg)
11. 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: $\mathbf\small{T_C+mg=\frac{mv_C^2}{r}}$
$\mathbf\small{\Longrightarrow T_C=m \left(\frac{v_C^2}{r}-g \right)}$
■ Thus, if we know the velocity at C, we can easily calculate the tension at C
12. If we know the velocity vA at the lowest point A, we can calculate the velocity vC at C
• How do we calculate vC?
• If we know the velocity vA at the lowest point A, we can calculate vC
• 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 the tension at C using the equation in (11) above
13. Next we will calculate TD, the tension at D
• The FBD of the object when it is at D is shown in fig.e
• We see that the net force in the direction of the string is TD itself
• This is because, 'mg' is perpendicular to the string and hence have no force component along the string
14. 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: $\mathbf\small{T_D=\frac{mv_D^2}{r}}$
■ Thus, if we know the velocity at D, we can easily calculate the tension at D
15. If we know the velocity vA at the lowest point A, we can calculate the velocity vD at D
• How do we calculate vD?
• If we know the velocity vA at the lowest point A, we can calculate vD
• 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 the tension at D using the equation in (14) above
■ It is interesting to note that, at B and D, just like the 'magnitudes of velocities', the 'magnitudes of tensions' 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:
16. In fig.6.46(a) below, the object is at 'P'
• At that instant, the string makes an angle θ with the vertical
17. In fig.b, the portion of the object alone is shown
• In this fig.b, a vertical is drawn through the object.
• The following two verticals will be parallel:
♦ Vertical through O
♦ Vertical through P
• So, if we extend the string downwards along the same line, that extension will make the same angle θ with the vertical
18. 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
• We see that the net force in the direction of the string is (TP -mg cosθ)
19. 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: $\mathbf\small{T_P-mg \cos \theta =\frac{mv_P^2}{r}}$
• $\mathbf\small{\Longrightarrow T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
20. 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 the tension at P using the equation in (19) above
■ So, if we know the velocity vP at any point P, we can easily obtain the tension TP at that point
For that, we use the equation in (19) above
• If we can use the equation derived in (19) above for 'any point', it must be applicable to points A, B, C and D also. Let us check:
21. First we will check the equation at point 'A'
• We have: $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at A, θ = 0o
♦ So cos θ = cos 0 = 1
• Substituting the known values, we get:
$\mathbf\small{T_A=m \left(\frac{v_A^2}{r}+g \right)}$
■ This is the same equation that we obtained in (7) above
22. Next we will check the equation at point 'B'
• We have: $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at B, θ = 90o
♦ So cos θ = cos 90 = 0
• Substituting the known values, we get: $\mathbf\small{T_B=m \left(\frac{v_B^2}{r}+g \times 0 \right)}$
$\mathbf\small{\Longrightarrow T_B=\frac{mv_B^2}{r}}$
■ This is the same equation that we obtained in (9) above
23. Next we will check the equation at point 'C'
• We have: $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at C, θ = 180o
♦ So cos θ = cos 180 = -1
• Substituting the known values, we get:
$\mathbf\small{T_C=m \left(\frac{v_C^2}{r}-g \right)}$
■ This is the same equation that we obtained in (11) above
24. Finally, we will check the equation at point 'D'
• We have: $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at D, θ = 270o
♦ So cos θ = cos 270 = 0
• Substituting the known values, we get: $\mathbf\small{T_D=m \left(\frac{v_D^2}{r}+g \times 0 \right)}$
$\mathbf\small{\Longrightarrow T_D=\frac{mv_D^2}{r}}$
■ This is the same equation that we obtained in (14) above
• We can use the equation $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$ to find the tension at any point along the vertical circle
• 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.6.47(a) below, the object is at the highest point C
• We know that the tension in the string at the instant when the object is at C is $\mathbf\small{T_C=m \left(\frac{v_C^2}{r}-g \right)}$
2. If $\mathbf\small{\frac{v_C^2}{r}}$ becomes equal to g, the tension will become zero
• We know that, when the object is being whirled, the string will be in a ‘pulled tight’ condition
♦ If the string is in a ‘pulled tight’ condition, it means that it is experiencing tension
• Conversely, if there is tension in the string, then it would be in a 'pulled tight' condition
• So if there is no tension, the string will not be in a ‘pulled tight’ condition
♦ In other words, if there is no tension, the string will slacken. This is shown in fig.6.47(b)
3. So we can write: $\mathbf\small{\frac{v_C^2}{r}}$ 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 $\mathbf\small{\sqrt{rg}}$
4. But $\mathbf\small{v_C}$ depends on the velocity $\mathbf\small{v_A}$ at the lowest point A
• 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 $\mathbf\small{5rg}$
5. Note that we cannot blame vC if the string slackens 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
6. We saw that if the string is not to slacken 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 tension TC 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 object 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
• When an object is tied to a string and whirled in a vertical circle, not only the velocity, but also the ‘tension in the string’ will vary.
• In this section we will see the reason for such a variation. We will also see the method to find the tension at any point
1. In fig.6.45(a) below, an object of mass ‘m’ is tied to a string and whirled in vertical circle.
 |
| Fig.6.45 |
2. The FBD of the object when it is at A is shown in fig.b
• We know that, the force in a string will always be tensile in nature (Details here). This tension will be pulling at both the ends of the strings
• So the object will experience a pulling through the string. This pulling force by the string is shown in fig.b as ‘TA’
3. Also the object will experience a pull towards the center of the earth
• This force by the earth is shown in fig.b as ‘mg’
4. When the object is at A,
♦ the force ‘TA’ will be vertically upwards
♦ the force ‘mg’ will be vertically downwards
• So both the forces act along the same line but opposite in direction
• The resultant force is (TA-mg)
5. For circular motion (both horizontal and vertical), centripetal force is required.
• The person who whirls the object has to provide this force
• For the person, the only contact with the object is through the string. So the centripetal force should be provided through the string
• We see that, the net force in the direction of the string is (TA-mg)
6. 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: $\mathbf\small{T_A-mg=\frac{mv_A^2}{r}}$
$\mathbf\small{\Longrightarrow T_A=m \left(\frac{v_A^2}{r}+g \right)}$
• Centripetal force is always directed towards the center. So it is positive
■ Thus, if we know the velocity at A, we can easily calculate the tension at A
7. Next we will calculate TB, the tension at B
• The FBD of the object when it is at B is shown in fig.c
• We see that the net force in the direction of the string is TB itself
• This is because, 'mg' is perpendicular to the string and hence have no force component along the string
8. 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: $\mathbf\small{T_B=\frac{mv_B^2}{r}}$
■ Thus, if we know the velocity (vB) at B, we can easily calculate the tension at B
9. How do we calculate vB?
• If we know the velocity vA at the lowest point A, we can calculate vB
• 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 the tension at B using the equation in (8) above
10. Next we will calculate TC, the tension at C
• The FBD of the object when it is at C is shown in fig.d
• We see that the net force in the direction of the string is (TC+mg)
11. 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: $\mathbf\small{T_C+mg=\frac{mv_C^2}{r}}$
$\mathbf\small{\Longrightarrow T_C=m \left(\frac{v_C^2}{r}-g \right)}$
■ Thus, if we know the velocity at C, we can easily calculate the tension at C
12. If we know the velocity vA at the lowest point A, we can calculate the velocity vC at C
• How do we calculate vC?
• If we know the velocity vA at the lowest point A, we can calculate vC
• 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 the tension at C using the equation in (11) above
13. Next we will calculate TD, the tension at D
• The FBD of the object when it is at D is shown in fig.e
• We see that the net force in the direction of the string is TD itself
• This is because, 'mg' is perpendicular to the string and hence have no force component along the string
14. 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: $\mathbf\small{T_D=\frac{mv_D^2}{r}}$
■ Thus, if we know the velocity at D, we can easily calculate the tension at D
15. If we know the velocity vA at the lowest point A, we can calculate the velocity vD at D
• How do we calculate vD?
• If we know the velocity vA at the lowest point A, we can calculate vD
• 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 the tension at D using the equation in (14) above
■ It is interesting to note that, at B and D, just like the 'magnitudes of velocities', the 'magnitudes of tensions' 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:
16. In fig.6.46(a) below, the object is at 'P'
 |
| Fig.6.46 |
17. In fig.b, the portion of the object alone is shown
• In this fig.b, a vertical is drawn through the object.
• The following two verticals will be parallel:
♦ Vertical through O
♦ Vertical through P
• So, if we extend the string downwards along the same line, that extension will make the same angle θ with the vertical
18. 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
• We see that the net force in the direction of the string is (TP -mg cosθ)
19. 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: $\mathbf\small{T_P-mg \cos \theta =\frac{mv_P^2}{r}}$
• $\mathbf\small{\Longrightarrow T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
20. 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 the tension at P using the equation in (19) above
■ So, if we know the velocity vP at any point P, we can easily obtain the tension TP at that point
For that, we use the equation in (19) above
• If we can use the equation derived in (19) above for 'any point', it must be applicable to points A, B, C and D also. Let us check:
21. First we will check the equation at point 'A'
• We have: $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at A, θ = 0o
♦ So cos θ = cos 0 = 1
• Substituting the known values, we get:
$\mathbf\small{T_A=m \left(\frac{v_A^2}{r}+g \right)}$
■ This is the same equation that we obtained in (7) above
22. Next we will check the equation at point 'B'
• We have: $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at B, θ = 90o
♦ So cos θ = cos 90 = 0
• Substituting the known values, we get: $\mathbf\small{T_B=m \left(\frac{v_B^2}{r}+g \times 0 \right)}$
$\mathbf\small{\Longrightarrow T_B=\frac{mv_B^2}{r}}$
■ This is the same equation that we obtained in (9) above
23. Next we will check the equation at point 'C'
• We have: $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at C, θ = 180o
♦ So cos θ = cos 180 = -1
• Substituting the known values, we get:
$\mathbf\small{T_C=m \left(\frac{v_C^2}{r}-g \right)}$
■ This is the same equation that we obtained in (11) above
24. Finally, we will check the equation at point 'D'
• We have: $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at D, θ = 270o
♦ So cos θ = cos 270 = 0
• Substituting the known values, we get: $\mathbf\small{T_D=m \left(\frac{v_D^2}{r}+g \times 0 \right)}$
$\mathbf\small{\Longrightarrow T_D=\frac{mv_D^2}{r}}$
■ This is the same equation that we obtained in (14) above
■ So it is confirmed:
• 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.6.47 |
2. If $\mathbf\small{\frac{v_C^2}{r}}$ becomes equal to g, the tension will become zero
• We know that, when the object is being whirled, the string will be in a ‘pulled tight’ condition
♦ If the string is in a ‘pulled tight’ condition, it means that it is experiencing tension
• Conversely, if there is tension in the string, then it would be in a 'pulled tight' condition
• So if there is no tension, the string will not be in a ‘pulled tight’ condition
♦ In other words, if there is no tension, the string will slacken. This is shown in fig.6.47(b)
3. So we can write: $\mathbf\small{\frac{v_C^2}{r}}$ 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 $\mathbf\small{\sqrt{rg}}$
4. But $\mathbf\small{v_C}$ depends on the velocity $\mathbf\small{v_A}$ at the lowest point A
• 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 $\mathbf\small{5rg}$
5. Note that we cannot blame vC if the string slackens 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
6. We saw that if the string is not to slacken 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 tension TC 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 object 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
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