Answer to Question #169246 in Mechanics | Relativity for April

Question

An automobile accelerates uniformly from rest and reaches a velocity of 22 m/s in 9 s. The tire  diameter is 58 cm. Determine the number of revolutions the tire makes during this motion  and the final angular velocity of the tire in revolutions per second.

Answer 1

a) Let's first find the acceleration of the automobile:

$$a=\dfrac{v-v_0}{t}=\dfrac{22\ \dfrac{m}{s}}{9\ s}=2.44\ \dfrac{m}{s^2}.$$

Then, let's find the distance covered by the automobile during this motion:

$$s=\dfrac{1}{2}at^2=\dfrac{1}{2}\cdot2.44\ \dfrac{m}{s^2}\cdot(9\ s)^2=99\ m.$$

Let's find the distance covered by the tire in one revolution:

$$d=C=\pi d=3.14\cdot0.58\ m=1.82\ m.$$

Finally, we can find the number of revolutions the tire makes during this motion:

$$n=\dfrac{s}{C}=\dfrac{99\ m}{1.82\ \dfrac{m}{rev}}=54.4\ rev.$$

b) We can find the final angular velocity of the tire as follows:

$$v=\omega r,$$ $$\omega=\dfrac{v}{r}=\dfrac{22\ \dfrac{m}{s}}{0.29\ m}=75.9\ \dfrac{rad}{s},$$ $$\omega=75.9\ \dfrac{rad}{s}\cdot\dfrac{1\ rev}{2\pi\ rad}=12.1\ \dfrac{rev}{s}.$$