How are the magnitudes of the two forces related if the angular accelerations of the two hoops are the same?
Solution
Use Newton's second law in its rotational motion form:
$$\tau = I \alpha$$
Solve for $\alpha$ :
$$\alpha = \frac{\tau}{I}$$
For case 1.
$$\tau_1 = F_1 R$$
$$I_1 = M R^2$$
and Case 2
$$\tau_2 = F_2 (2R)$$
$$I_2 = M (2R)^2 = 4MR^2$$
The masses and $\alpha$s are the same in both cases.
$$\alpha_1 = \alpha_2$$
$$\frac{ F_1 \cancel{ R}}{\cancel {M R^2}}= \frac{ 2F_2 \cancel{ R}}{ 4\cancel{ M R^2}}$$
$$\frac{ F_1 }{1}= \frac{ F_2 }{ 2}$$
So the correct answer is
$$F_2 = 2 F_1$$.
N. Alberding, 2013