Ratio of translational velocity to rotational velocity.

  • #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 0.56mR2 .

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

Homework Equations



KE = .5(I)(w)2 = .5mv2

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 .5Iw2 = 1/2mv2

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

.5(.56mR2)(w)2 = .5mv2

Then I cancelled out the .5 and the m,

.56R2w2 = v2

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...
 

Answers and Replies

  • #2
kuruman
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Read the problem carefully. You are looking for the ratio of the speed of the center of mass to the speed of a point on the surface of the sphere. The latter quantity is not ω; it is a linear speed (in m/s) so the final answer must be a dimensionless quantity.
 

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