Answer to Question #197175 in Physics for gibson

Question

A belt runs on the wheel of a hammer mill of radius 20cm. During the time the wheel takes

to coast uniformly to rest from an initial speed of 5rev/s, 30m of belt length over the wheel.

Find for this wheel the,

(a) number revolution it turns while stopping.

(b) deceleration of the wheel.

(c) time it took to come to a stop

Answer 1

(a) We can find the number of revolution the wheel takes while stopping as follows:

$$n=\dfrac{L}{2\pi r}=\dfrac{30\ m}{2\pi\cdot0.2\ m}=24\ rev.$$

(b) Let's first find the angle covered during deceleration:

$$\theta=\dfrac{L}{r}=\dfrac{30\ m}{0.2\ m}= 150\ rad.$$

Then, we can find the deceleration of the wheel from the kinematic equation:

$$\omega^2=\omega_0^2+2\alpha\theta,$$ $$\alpha=\dfrac{\omega^2-\omega_0^2}{2\theta},$$ $$\alpha=\dfrac{0-(5\ \dfrac{rev}{s}\cdot\dfrac{2\pi\ rad}{1\ rev})^2}{2\cdot150\ rad}=-3.3\ \dfrac{rad}{s^2}.$$

The sign minus means that the wheel decelerates.

(c) We can find the time it took to come to a stop from the kinematic equation:

$$\omega=\omega_0+\alpha t,$$ $$t=\dfrac{\omega-\omega_0}{\alpha},$$ $$t=\dfrac{0-5\ \dfrac{rev}{s}\cdot\dfrac{2\pi\ rad}{1\ rev}}{-3.3\ \dfrac{rad}{s^2}}=9.5\ s.$$