Answer
(a) $\tau = 7.8~m \cdot N$
(b) The force required by the muscle is 310 N.
Work Step by Step
(a) We can think of the ball as a point mass a distance of 0.31 m from the axis of rotation. We can find the moment of inertia of the ball relative to the axis of rotation.
$I = MR^2 =(3.6~kg)(0.31~m)^2 = 0.346~kg\cdot m^2$
We can find the torque required to accelerate the ball.
$\tau = I \alpha = I ~\frac{a}{r}= ( 0.346~kg\cdot m^2)(\frac{7.0~m/s^2}{0.31~m})$
$\tau = 7.8~m \cdot N$
(b) The distance $r$ from the muscle to the axis of rotation is 0.025 m.
$r \cdot F = \tau$
$F = \frac{\tau}{r} = \frac{7.8~m \cdot N}{0.025~m} = 310~N$
The force required by the muscle is 310 N.