Answer to Question #140106 in Physics for Nielle

Question

A car accelerates uniformly from rest and reaches an angular speed of 22.0 m/s in
9.00 s. Assuming the diameter of a tire is 58.0 cm, (a) find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. (b) What is the final angular speed of a tire in revolutions per second?

Answer 1

Given:

$\omega_0=0\: \rm rad/s$

$v=22.0\: \rm m/s$

$t=9.00\: \rm s$

$d=58.0\:\rm cm$

The final angular speed

$$\omega=\frac{v}{R}=\frac{22.0}{0.58/2}=75.9\:\rm rad/s$$

(a)

$$\phi=\frac{\omega+\omega_0}{2}=\rm \frac{75.9+0.00}{2}\times 9.00=341\: rad$$

$$N=\frac{\phi}{2\pi}=\frac{341}{2\pi}=54.4$$

(b)

$$n=2\pi\omega=6.28\times 75.9\:\rm rad/s=477 \:\rm rev/s$$