- #1

spursfan2110

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**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

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)

## Homework Equations

KE = .5(I)(w)

^{2}= .5mv

^{2}

## 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...