# Angular Acceleration| Circular Motion| Physics

### Your car tire is rotating at 4.0 rev/s when suddenly you press down hard on the accelerator. After traveling 300 m, the tire’s rotation has increased to 6.5 rev/s . The radius of the tire is 32 cm.

SOLUTION:

PART A

We are given an initial angular velocity of 4.0 rev/s and a final angular acceleration of 6.5 rev/s. Convert this angular velocities to rad/s.

$\:\frac{4.0\:rev}{s}\times \frac{2\pi \:rad}{1\:rev}=8\pi \:\frac{rad}{s}$

$\:\frac{6.5\:rev}{s}\times \frac{2\pi \:rad}{1\:rev}=13\pi \:\frac{rad}{s}$

The car traveled $s=300\:m$. Solve for the angular displacement

$\theta =\frac{s}{r}=\frac{300\:m}{0.32\:m}=937.5\:rad$

Solve for the angular acceleration

$\left(\omega _f\right)^2=\left(\omega _i\right)^2+2\alpha \Delta \theta$

$\left(13\pi \right)^2=\left(8\pi \right)^2+2\alpha \left(937.5\right)$

$\alpha =0.55\:rad/s^2$