## College Physics (4th Edition)

The merry-go-round is rotating at an angular speed of $0.61~rev/s$ The rotational kinetic energy decreased by $663~J$
We can find the initial rotational inertia of the merry-go-round: $I_0 = \frac{1}{2}MR^2$ $I_0 = \frac{1}{2}(160~kg)(2.0~m)^2$ $I_0 = 320~kg~m^2$ We can find the final rotational inertia of the merry-go-round and child: $I_f = \frac{1}{2}MR^2+mR^2$ $I_f = \frac{1}{2}(160~kg)(2.0~m)^2+(180~N/9.80~m/s^2)(2.0~m)^2$ $I_f = 393.5~kg~m^2$ We can use conservation of angular momentum to find the new angular speed: $L_f = L_0$ $I_f~\omega_f = I_0~\omega_0$ $\omega_f = \frac{I_0~\omega_0}{I_f}$ $\omega_f = \frac{(320~kg~m^2)(0.75~rev/s)}{393.5~kg~m^2}$ $\omega_f = 0.61~rev/s$ The merry-go-round is rotating at an angular speed of $0.61~rev/s$ We can find the change in rotational kinetic energy: $\Delta KE_{rot} = \frac{1}{2}I_f~\omega_f^2 - \frac{1}{2}I_0~\omega_0^2$ $\Delta KE_{rot} = \frac{1}{2}(393.5~kg~m^2)[(0.61~rev/s)(2\pi~rad/rev)]^2 - \frac{1}{2}(320~kg~kg~m^2)[(0.75~rev/s)(2\pi~rad/rev)]^2$ $\Delta KE_{rot} = -663~J$ The rotational kinetic energy decreased by $663~J$.