Question

A child with mass $m$ is standing at the edge of a playground merry-go-round (A large uniform disc which rotates in horizontal plane about a fixed vertical axis in parks) with moment of inertia $I$, radius $R$, and the initial angular velocity $\omega$ as shown in the figure. The child jumps off the edge of the merry-go-round with a velocity $v$ with respect to the ground in direction tangent to periphery of the disc as shown. The new angular velocity of the merry-go-round is:

• A. $\sqrt{\frac{I{\omega }^2-m{v}^2}{I}}$
• B. $\sqrt{\frac{(I+m{R}^2){\omega }^2-m{v}^2}{I}}$
• C. $\frac{I\omega -mvR}{I}$
• D. $\frac{(I+m{R}^2)\omega -mvR}{I}$

Answer 1

Answer: (D)

$\frac{(I+m{R}^2)\omega -mvR}{I}$

Let the angular velocity of the merry-go-round be $w^{\prime }$.

As there is no external torque, thus $L$ is conserved.

Initial angular momentum, $L_i=Iw+m{R}^{2}w$

Final angular momentum, $L_f=I{w}^{\prime }+mvR$

$L_i=L_f⟹Iw+mw{R}^2=I{w}^{\prime }+mvR$

$⟹w^{\prime }=\frac{(I+m{R}^2)w-mvR}{I}$