## College Physics (4th Edition)

(a) The required torque is $~48.1~N \cdot m$ (b) Each child must push with a force of $19.2~N$
(a) Let $M_m$ be the mass of the merry-go-round and let $M_c$ be the mass of each child. We can find the rotational inertia of the system: $I = \frac{1}{2}M_mR^2+2M_cR^2$ $I = \frac{1}{2}(350.0~kg)(1.25~m)^2+(2)(30.0~kg)(1.25~m)^2$ $I = 367.2~kg~m^2$ We can express the final angular speed in units of rad/s: $25~rpm \times \frac{2\pi~rad}{1~rev} \times \frac{1~min}{60~s} = 2.618~rad/s$ We can find the angular acceleration: $\omega_f = \omega_0+\alpha~t$ $\alpha = \frac{\omega_f - \omega_0}{t}$ $\alpha = \frac{2.618~rad/s-0}{20.0~s}$ $\alpha = 0.1309~rad/s^2$ We can find the required torque: $\tau = I~\alpha$ $\tau = (367.2~kg~m^2)(0.1309~rad/s^2)$ $\tau = 48.1~N \cdot m$ The required torque is $~48.1~N \cdot m$ (b) We can find the force $F$ with which each child must push: $2~(R \times F) = \tau$ $F = \frac{\tau}{2R}$ $F = \frac{48.1~N \cdot m}{(2)(1.25~m)}$ $F = 19.2~N$ Each child must push with a force of $19.2~N$.