Thursday
April 24, 2014

Homework Help: Physics

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

Equate the increase in potential energy at the higher elevation, M g H, to the decrease in kinetic energy. Make sure you include the kinetic energy of rotation, which is (1/2) I w^2. For a sphere of radius R,
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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