A playground merry-go-round has a radius R and a rotational inertia I.When the merry-go-round is at rest, a child with mass m runs with speed v along a line tangent to the rim and jumps on.The angular velocity of the merry-go-round is then: A)mv/I B)v/R C)mRv/I D)2mRv/I E)mRv/(mR² + I)

Question

Quizplus · Accepted Answer

The correct answer is derived from the conservation of angular momentum. Initially, the angular momentum of the system (merry-go-round plus child) is solely due to the child, since the merry-go-round is at rest. The child's angular momentum before jumping on is given by $L = mvr$, where $r = R$ is the radius of the merry-go-round. After the child jumps on, the total angular momentum of the system must remain constant, so $L = I_{\text{total}}\omega$, where $I_{\text{total}} = I + mR^2$ (the sum of the merry-go-round's rotational inertia and the rotational inertia of the child treated as a point mass at radius $R$), and $\omega$ is the angular velocity of the system. Solving for $\omega$ gives $\omega = \frac{mRv}{I + mR^2}$, which matches the formula in choice E.

A Playground Merry-Go-Round Has a Radius R and a Rotational