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 \(T = \frac{mv^2}{L} = 3mg\).
Assertion (A): A bob of mass \( m \) is freely suspended from a light rod of length \( L \). The minimum speed given to bob at lowest position to complete vertical circle is \( 2\sqrt{gL} \).
Reason (R): A bob of mass \( m \) is freely suspended from a light string of length \( L \). If bob is given speed \( \sqrt{6gL} \) at the lower position then bob will be complete vertical circle.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
For a mass attached to a rod, the minimum speed at the lowest point to complete a vertical circle is \( v_{min} = 2\sqrt{gL} \). So (A) is true. For a mass on a string, if \( v_{bottom} = \sqrt{6gL} \), the speed at the top will be \( v_{top} = \sqrt{2gL} \). Since \( v_{top} > \sqrt{gL} \), the circle will be completed. So (R) is true. However, they describe different conditions, so (R) is not a correct explanation of (A).
Assertion (A): A body tied to an end of a string is whirled along a vertical circle by giving some velocity at the lowest position. If the velocity becomes zero before the tension in the string is zero, the body will leave the circular path at the position of its zero velocity and then fall vertically downward.
Reason (R): In vertical circular motion, tension in the string at the highest position is maximum.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
If velocity becomes zero before tension, the object leaves the circular path and follows a parabolic trajectory, not vertically downward. So (A) is false.
Tension is maximum at the lowest point and minimum at the highest point in vertical circular motion. So (R) is false.
Assertion (A): A body tied to an end of a string is whirled along a vertical circle with such a velocity at the lowest point that, at some position, tension in the string is zero but the speed at the position is non-zero. The body will leave the circular path at the position of zero tension.
Reason (R): In vertical circular motion, so as to cross the highest point along the circle, speed at the highest point, \( v_H geq 0 \).
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
An object leaves a vertical circular path when tension becomes zero and speed is non-zero. So (A) is true. To complete a vertical circle, the minimum speed at the highest point is \( \sqrt{gR} \), not just \( 0 \). So (R) is false.
A roller coaster is designed such that riders experience “weightlessness” as they go round the top of a hill whose radius of curvature is \(20\text{ m}\). The speed of the car at the top of the hill is between:
(2008)
1. \(13\text{ m/s}\) and \(14\text{ m/s}\)
2. \(14\text{ m/s}\) and \(15\text{ m/s}\)
3. \(15\text{ m/s}) and \(16\text{ m/s}\)
4. \(16\text{ m/s}\) and \(17\text{ m/s}\)
View Answer
At the top of a hill, "weightlessness" implies the normal force is zero. The centripetal force is provided solely by gravity: (mg = \frac{mv^2}{R}). Thus, (v = \sqrt{gR}). Using (g = 9.8\text{ m/s}^2) and (R = 20\text{ m}), (v = \sqrt{9.8 \times 20} = \sqrt{196} = 14\text{ m/s}). This speed lies between \(14\text{ m/s}\) and \(15\text{ m/s}\).
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.
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 $\Delta v = \sqrt{u^2 + v_2^2} = \sqrt{2(u^2 - g\ell)}$.