Question

A grindstone, initially at rest, is given a constant angular acceleration so that it makes 20.0 rev in the first 8.00 s. what is its angular acceleration?

Answer 1

The angle covered by the grindstone during this time is, converting into radians,
$\theta = 20 rev = 20 rev \cdot \frac{2 \pi rad}{rev} =125. 6rad$

This is a rotational motion with constant angular acceleration $\alpha$, and with initial angular speed $\omega _0 =0$. The angle covered in a time t is given by
$\theta (t)= \omega_0 t+ \frac{1}{2} \alpha t^2 = \frac{1}{2} \alpha t^2$
because the initial angular speed is zero.
Using t=8.00 s, we can find the value of angular acceleration:
$\alpha = \frac{2 \theta}{t^2}= \frac{2 \cdot 125.6 rad}{(8.0 s)^2}=3.93 rad/s^2$