Question

A merry-go-round, made of a ring-like platform of radius R and mass M. Is revolving with angular speed \[\omega \]? A person of mass \[M\] is standing on it. At one instant, the person jumps off the round, radially away from the center of the round (as seen from the round). The speed of the round afterwards is:

 A. \[2\omega \]

 B. \[\omega \]

 C. \[\omega {\text{ }}/2\]

 D. \[0\]

Answer 1

Hint: This problem can be solved by drawing a proper diagram and applying the formula for understanding the situation. You should have knowledge about angular motion. Torque is a product of force and displacement from the axis of rotation. It is a vector physical quantity.

Formula used: An object is moving in a circular path(merry-go-round) of R radius and applying different kinds of internal forces. Mass of merry round and the person who is standing at the ring part is \[M\]. Both are moving with angular velocity \[\omega \] and angular momentum \[L\]. Before collision and after collision angular momentum is ${L_i},{L_f}$ respectively.

Then angular momentum $L = I\omega $

Here \[I = \] Moment of inertia of object, \[\omega = \] Angular velocity of an Object

Moment of inertia of object $I = M{R^2}$

Here \[M = \] Mass of an object

\[R = \] Radius of circular path

Conservation of Angular momentum:

Before the collision Angular momentum = After the collision Angular momentum

${L_i} = {L_f}$

Torque due circular motion $\tau = r \times F$

Here r is displacement from the axis of rotation

\[F = \] Force apply at an object

Complete step by step answer:

First let us draw a diagram to understand the problem better.

When we observe the situation then,

Make Free Body Diagram

Number of forces work on the object

- Force due to gravity \[\left( W \right)\] ( perpendicular Downward)

- Force due surface normal resultant \[\left( R \right)\] ( perpendicular Upward)

- Centripetal Force (Towards the centre of merry round)

- Centrifugal Force (opposite the centripetal force)

Concept: At the movement according to the question a person jumps from the merry round then if person and merry round is only the part of the system so total external force will become Zero. Torque due circular motion ${\tau _{ext}} = 0$. If in the system total external force will become zero then angular momentum remains conserved

Applying conservation of Angular momentum,

Before the collision Angular momentum = After the collision Angular momentum

${L_i} = {L_f}$

Assume moment of inertia of merry round and person is $I$, ${I'}$ respectively and moving with same angular velocity ω.

Moment of inertia of person = $M{R^2}$

Moment of inertia of merry round = $M{R^2}$

 By the Conservation of Angular momentum

${L_i} = {L_f}$

$\Rightarrow I\omega = {I'}\omega '$

$\Rightarrow \left( {M{R^2} + M{R^2}} \right)\omega = M{R^2}\omega '$

$\Rightarrow \left( {2M{R^2}} \right)\omega = M{R^2}\omega '$

$\therefore 2\omega = \omega '$

So, the correct answer is “Option A”.

Note:

This problem can be solved by law of motion as well as circular motion. Most important is how you will visualise the problem easily and apply concepts as per the requirement.