Vertical Circular Motion

Practice Questions:

Question 1: moderate

A particle of mass $m$ is tied to a string of length $L$ and rotated in vertical circle about other end with critical speed so that it is just able to complete the vertical loop. Then tension in string, when string is at horizontal position will be:

1. $2mg$

2. $3mg$

3. $4mg$

4. $5mg$

View Answer

To just complete the vertical loop, the velocity at the bottom is $\sqrt{5gL}$. By energy conservation, the velocity at the horizontal position is $v = \sqrt{3gL}$. The tension at this point is $T = \frac{mv^2}{L} = 3mg$.

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Question 2: moderate

A point mass ‘$m$’ is moved in a vertical circle of radius ‘$r$’ with the help of a string. The velocity of the mass is $\sqrt{7gr}$ at the lowest point. The tension in the string at the lowest point is

(2020-Covid)

1. $7 \text{ mg}$

2. $8 \text{ mg}$

3. $1 \text{ mg}$

4. $6 \text{ mg}$

View Answer

Tension at the lowest point is given by $T = mg + \frac{mv^2}{r} = mg + \frac{m(\sqrt{7gr})^2}{r} = mg + 7mg = 8mg$.

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Question 3: moderate

A mass $m$ is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:

(2019)

1. The mass is at the highest point

2. The wire is horizontal

3. The mass is at the lowest point

4. Inclined at an angle of $60^\circ$ from vertical

View Answer

Tension in the string/wire is maximum at the lowest point of the vertical circle ($T = mg + \frac{mv^2}{r}$), making it most likely to break there.

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Question 4: moderate

What is the minimum velocity with which a body of mass $m$ must enter a vertical loop of radius $R$ so that it can complete the loop?

(2016-I)

1. $\sqrt{gR}$

2. $\sqrt{2gR}$

3. $\sqrt{3gR}$

4. $\sqrt{5gR}$

View Answer

To complete a vertical loop, the minimum velocity required at the lowest point is $\sqrt{5gR}$ to ensure zero tension at the highest point.

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Question 5: moderate

A stone is tied to a string of length ‘$\ell$’ and is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed ‘$u$’. The magnitude of the change in velocity as it reaches a position where the string is horizontal ($g$ being acceleration due to gravity) is:

(2004)

1. $\sqrt{u^2 - g\ell}$

2. $u - \sqrt{u^2 - 2g\ell}$

3. $\sqrt{2g\ell}$

4. $\sqrt{2(u^2 - g\ell)}$

View Answer

Velocity at horizontal position is $v_2 = \sqrt{u^2 - 2g\ell}$. Since velocities are perpendicular, magnitude of change in velocity is $\Delta v = \sqrt{u^2 + v_2^2} = \sqrt{2(u^2 - g\ell)}$.

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