Answer
(a) For no slipping to occur, the minimum coefficient of static friction is 0.613.
(b) The coefficient of static friction found in part (a) would not be enough to prevent a hollow sphere from slipping.
(c) When an object is rolling without slipping, then there is no relative motion at the contact points between the two surfaces. Therefore, the friction between the two surfaces is static friction.
Work Step by Step
(a) If the ball rolls without slipping, then $\alpha = \frac{a}{R}$. We can use a torque equation to find an expression for the acceleration $a$.
$\tau = I\alpha$
$F_f~R = (\frac{2}{5}mR^2)(\frac{a}{R})$
$a = \frac{5F_f}{2m}$
We can use this expression in the force equation for the ball as it moves down the slope.
$\sum F = ma$
$mg~sin(\theta) - F_f = m(\frac{5F_f}{2m})$
$mg~sin(\theta) = (\frac{7F_f}{2})$
$mg~cos(\theta)~\mu_s = \frac{2mg~sin(\theta)}{7}$
$\mu_s = \frac{2~tan(\theta)}{7}$
$\mu_s = \frac{2~tan(65.0^{\circ})}{7}$
$\mu_s = 0.613$
For no slipping to occur, the minimum coefficient of static friction is 0.613.
(b) The coefficient of static friction found in part (a) would not be enough to prevent a hollow sphere from slipping. The acceleration of the hollow ball down the slope would be the same as in part (a), and so the same angular acceleration as in part (a) would be required. However, the moment of inertia of a hollow sphere is $\frac{2}{3}mR^2$, which is larger than the moment of inertia of a solid sphere. Therefore, a larger torque is required to give the hollow sphere the same angular acceleration as in part (a), which would require a larger force of static friction, which would require a larger coefficient of static friction.
(c) When an object is rolling without slipping, then there is no relative motion at the contact points between the two surfaces. Therefore, the friction between the two surfaces is static friction.
