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 (rad)/(rev))=219.8 rad/s`
The final angular speed instead is:

`\omega _f=15 rev/s \cdot ( 2 \pi (rad)/(rev))=94.2 rad/s`
Therefore we can find the value of the angular acceleration:

`\alpha = (\omega _f-\omega _i)/(t)= (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.