I'm almost done with a problem on rotational mechanics.
I've got a golf ball rolling down a hill and I'm to find the value of the angle of the hill so that the ball can roll without slipping if the constant of friction $\mu = 0.29$.
Now I've got $$ a_t = \frac{0.29 \cdot 9.8 \cdot \cos( \theta ) \cdot 5}{2}$$ and this has to equal $ \alpha$ for it to not slip. I'm just having problems finishing this problem! Note that the torque is applied by $F_f$.
$a_t$ is derived from $$ \alpha = \frac{F_f \cdot R}{I}$$
(As usual I want to remind that I prefer mathematics SE over physics SE and there is a physics tag on maths, so this question is perfectly fine).
Let $\theta$ be the angle between the Hill surface and the ground.
$$ mgcos\theta - N = 0 \cdots(1)$$
$$ mgsin\theta - f_s = ma \cdots (2) $$
$$ \tau = Rf_s$$
$$Rf_s = I\alpha$$
$$ a= R\alpha$$
$$f_{s} = \frac{Ia}{R^2}$$ $$mgsin\theta - \frac{Ia}{R^2}= ma$$
$$ a = \dfrac {gsin\theta}{1+\frac{I}{R^2}}$$
$$ I = \frac{2}{5} MR_2$$
Substituting the value of I and a $$f_s = \frac{2}{7}mgsin\theta$$
For an object rolling on an inclined plane without slipping
$$f_s \leq \mu mgcos\theta$$
$$\frac{2}{7}mgsin\theta \leq \mu mgcos\theta$$
$$tan\theta \leq \frac{7\mu}{2}$$
$$\theta \leq tan^{-1}(\frac{7\mu}{2})$$