(adsbygoogle = window.adsbygoogle || []).push({}); 1. A ball of mass m and radius R is both sliding and spinning on a horizontal surface so

that its rotational kinetic energy equals its translational kinetic energy.What is the ratio of the ball’s center-of-mass speed to the speed due to rotation only of a point on the ball’s surface? The moment of inertia of the ball is 0.56mR^{2}.

(For ease, I will refer to omega as w from here on out)

2. Relevant equations

KE = .5(I)(w)^{2}= .5mv^{2}

3. The attempt at a solution

So if I understand it correctly the problem basically wants the ratio of linear velocity to rotational velocity, v to w. So, I set .5Iw^{2}= 1/2mv^{2}

From here, I plugged in the given moment of inertia of the ball.

.5(.56mR^{2})(w)^{2}= .5mv^{2}

Then I cancelled out the .5 and the m,

.56R^{2}w^{2}= v^{2}

Square rooted both sides,

sqrt(.56)Rw = v

but from here I am unsure of how I could possibly eliminate the R, and this causes big problems when finding a ratio. Any suggestions? Did I go wrong somewhere prior to this point? If I could just get that R, it should be easy I would think, but its a variable...

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# Homework Help: Ratio of translational velocity to rotational velocity.

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