Q 14CQ - College Physics (Urone)

Question

Question: Suppose a child walks from the outer edge of a rotating merry-go round to the inside. Does the angular velocity of the merry-go-round increase, decrease, or remain the same? Explain your answer.

Step-by-Step Solution

Answer

There will be increase in the angular velocity of the merry-go-round.

Step 1: Concept of law of conservation of angular momentum.

The law of conservation of angular momentum states that there will be no change in the angular momentum unless an external torque acts on the object. Angular momentum of a system is conserved as long as there is no external torque acting on the system.

Step 2: Explanation for increase in the angular velocity of the merry-go-round.

Let, I₁ be the moment of inertia of merry-go-round, I₂ be the initial moment of inertia of the child, I_2’ be the final angular velocity of child,\({\omega _1}\) is the initial angular velocity of the merry-go-round with child, and \({\omega _2}\) is the final angular velocity of the merry-go-round with child.

 

Write the expression for determining the moment of inertia of child.

\(I = m{r^2}\)

 

When the child is moving towards the inside of the merry-go-round, the radius is decreasing thereby, the moment of inertia of the child also decreases.

Therefore,

\({I_{2'}} < {I_2}\)

 

According to law for conservation of angular momentum.

\(\left( {{I_1} + {I_2}} \right){\omega _1} = \left( {{I_1} + {I_{2'}}} \right){\omega _2}\)

 

Clearly, when the value of \({I_{2'}} < {I_2}\), the angular velocity \({\omega _2}\) will be greater than \({\omega _1}\).

 

Hence, the angular velocity of the merry-go-round will increase.