Answer to Question #154807 in Physics for Nashe

Question #154807

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?


1
Expert's answer
2021-01-12T12:13:21-0500

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


ω=ω0+αt,\omega=\omega_0+\alpha t,t=ω0α=27 rads0.15 rads2=180 s.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:


θ=ω0t+12αt2,\theta=\omega_0 t+\dfrac{1}{2}\alpha t^2,θ=27 rads180 s+12(0.15 rads2)(180 s)2=2430 rad.\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:


Nrev=2430 rad(1 rev2π rad)=387 rev.N_{rev}=2430\ rad\cdot(\dfrac{1\ rev}{2\pi\ rad})=387\ rev.

Answer:

Nrev=387 rev.N_{rev}=387\ rev.


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