Answer
The torque $\tau$ required to produce the acceleration is $\tau = 320~m\cdot N$.
The force required at the edge is 130 N.
Work Step by Step
(a) $\omega = (15~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s}) = (\pi/2)~rad/s$
We can find the angular acceleration $\alpha$:
$\alpha = \frac{\omega}{t} = \frac{(\pi/2)~rad/s}{10.0~s}$
$\alpha = \frac{\pi}{20.0}~rad/s^2$
Let $m$ be the mass of a child and let $M$ be the mass of the merry-go-round. We can then find the moment of inertia of the system:
$I = \frac{1}{2}MR^2 + mr^2+mr^2$
$I = \frac{1}{2}(560~kg)(2.5~m)^2 + (25~kg)(2.5~m)^2+(25~kg)(2.5~m)^2$
$I = 2062.5~kg\cdot m^2$
Now, we find the torque required to produce the acceleration:
$\tau = I \omega = ( 2062.5~kg\cdot m^2)( \frac{\pi}{20.0}~rad/s^2)$
$\tau = 320~m\cdot N$
The torque $\tau$ required to produce the acceleration is $\tau = 320~m\cdot N$.
(b) We can use the torque to find the required force at the edge:
$r\cdot F = \tau$
$F = \frac{\tau}{r} = \frac{320~m\cdot N}{2.5~m}$
$F = 130~N$
The force required at the edge is 130 N.
