Question

A wheel 33 cm in diameter accelerates uniformly from 240 rpm to 360 rpm in 6.5 s. how far will a point on the edge of the wheel have traveled in this time?

Answer 1

First we find the angular acceleration $\alpha$. The angular velocities are $\frac{240}{60} =4 \;rps$ and $\frac{360}{60} =6 \;rps$ The angular velocities and $\alpha$ are related as

$\omega=\omega_0+\alpha t\\ \alpha =\frac{\omega - \omega_0}{t} \\ \alpha =\frac{6-4}{(6.5)} \\ \alpha =0.3077 \;rps^2\\ \alpha =(0.3077 )2\pi =1.933 \;rad/s^2$

Angle turned in 6.5 seconds is

$2 \alpha\theta = \omega^2-\omega_0^2\\ \theta = \frac{ \omega^2-\omega_0^2}{2 \alpha}\\ \theta = \frac{ 6^2-4^2}{2 (1.933)}\\ \theta = 5.173 \; rad$

The distance traveled by a point on the edge of the wheel is $r \theta = \frac{33}{2}(5.173)= 85.35 \;cm$