Answer to Question #154807 in Physics for Nashe

Question

A basketball spinning on someone’s finger undergoes an angular acceleration of - 0.15 rad/s and is traveling at 27 rad/s initially. How many revolutions will it make before coming to a stop?

Answer 1

Let's first find time that the basketball takes to stop:

$$\omega=\omega_0+\alpha t,$$ $$t=-\dfrac{\omega_0}{\alpha}=-\dfrac{27\ \dfrac{rad}{s}}{-0.15\ \dfrac{rad}{s^2}}=180\ s.$$

Then, we can find how many radians the basketball turns through:

$$\theta=\omega_0 t+\dfrac{1}{2}\alpha t^2,$$ $$\theta=27\ \dfrac{rad}{s}\cdot 180\ s+\dfrac{1}{2}\cdot(-0.15\ \dfrac{rad}{s^2})\cdot(180\ s)^2=2430\ rad.$$

Finally, we can find the number of revolutions that the basketball makes before coming to a stop:

$$N_{rev}=2430\ rad\cdot(\dfrac{1\ rev}{2\pi\ rad})=387\ rev.$$

Answer:

$N_{rev}=387\ rev.$