Question

A belt runs on a wheel of radius 30cm. During the time that wheel coasts uniformly to rest from
an initial speed of 2.0revs/s, 25 m of belt length passes over the wheel. Find the deceleration of
the wheel and the number of revolutions it turns while stopping.

Answer 1

Number of turns:

$$n=\frac{l}{2\pi r}$$

The angle covered during deceleration 

$$\theta=2\pi n=\frac{l}{ r}$$

The initial angular velocity

$$\omega=2\pi(2)=4\pi ω$$

Deceleration is

$$\alpha=\frac{\omega^2}{2\theta}=\frac{16\pi^2}{2\frac{25}{0.3}}=0.95rad/s^2$$

The number of revolutions it turns while stopping

$$n=\frac{25}{2\pi (0.3)}=13$$