Friday

January 30, 2015

January 30, 2015

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?

- Physics -
**Youthers**, Tuesday, January 27, 2015 at 8:12pmA bicycle wheel has a diameter of 47.6 cm and a mass of 0.809 kg. The bicycle is placed on a stationary stand on rollers and a resistive force of 60.1 N is applied to the rim of the tire. Assume all the mass of the wheel is concentrated on the outside radius.

In order to give the wheel an acceleration of 3.18 rad/s2, what force must be applied by a chain passing over a sprocket with diameter 4.16 cm?

Answer in units of N.

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