Question

A wheel with a 0.10-m radius is rotating at 35 rev/s. it then slows uniformly to 15 rev/s over a 3.0-s interval. what is the angular acceleration of a point on the wheel?

Answer 1

The initial angular speed of the wheel is (keeping in mind that $1 rev=2 \pi rad$):
$\omega _i=35 rev/s \cdot ( 2 \pi \frac{rad}{rev})=219.8 rad/s$
The final angular speed instead is:
$\omega _f=15 rev/s \cdot ( 2 \pi \frac{rad}{rev})=94.2 rad/s$
Therefore we can find the value of the angular acceleration:
$\alpha = \frac{\omega _f-\omega _i}{t}= \frac{94.2 rad/s-219.8rad/s}{3.0 s}=-41.87 rad/s^2$
and the negative sign is due to the fact the wheel is decelerating.