A bowling ball is released at teh upper edge of a rampthat is inclined at an ang

Question

A bowling ball is released at teh upper edge of a rampthat is inclined at an angle of 37 degrees with the floor. The rampis 5.0m long. The bowling ball is allowed to roll freely withoutslipping under the influence of gravity down the ramp. The bowlingball has a mass of 4.0kg and a diameter of 40cm. Assume the bowlingball is a uniform sphere. (I= 2/5 MR2) a) Calculate the total kinetic energy (rotational andtranslational) of the bowling ball at the bottom of the ramp. b)Calculate the linear speed of the bowling ball at the bottomof the ramp. c) How would the speed of a pool ball with mass of 400 g anddiameter of 7.0 cm compare with the speed of the bowling ball atthe bottom of the ramp? a) Calculate the total kinetic energy (rotational andtranslational) of the bowling ball at the bottom of the ramp. b)Calculate the linear speed of the bowling ball at the bottomof the ramp. c) How would the speed of a pool ball with mass of 400 g anddiameter of 7.0 cm compare with the speed of the bowling ball atthe bottom of the ramp?

Explanation / Answer

m g h = 1/2 m v2 + 1/2 I2   equals total energy at bottom byconservation (translational + rotational) m g h = 4 * 9.8 * 5 * sin 37 = 118 J 1/2 m v2 + 1/2 I 2 = 1/2 m2 r2 + 1/2 * 2/5 * m 2r2         I = 2 m r2 / 5   for a sphere g h = (1/2 + 1/5) 2r2  = 7 2r2 / 10 v = r = (10 g h / 7) = ((10 * 9.8 * 3) /7) = 6.48 m/s Since the equation for v does not contain m or r  the speed of a pool ball would be the same

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