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November 22, 2014

November 22, 2014

Posted by **drwls** on Wednesday, March 21, 2007 at 12:24pm.

V = R w

and I =(2/5) I R^2

Therefore

KE(rotational) = (1/2)(2/5)M R^2(V/R)^2 = (1/5) M V^2

M g H = KE(translational) + KE(rotational)= (7/10)MV^2

V^2 = sqrt[(10/7)gH]

A bowling ball encounters a 0.76m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. If the translational speed of the ball is 3.80m/s at the bottom of the rise, find the translational speed at the top.

Where does the translational speed at the bottom come in?

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