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Unformatted text preview: h ic e8 e k# i b iC a aha 10 Problem 24 ( a ): Let the marble have mass m and radius R , and let us measure the height above the bottom of the bowl. As the marble goes from the top of the left side down to the bottom of the bowl it starts out with potential energy but no kinetic, and ends up with kinetic energy but no potential. The conservation of energy tells us that U left = mgh = K bot = 1 2 mv 2 bot + 1 2 I 2 bot . As I = 2 5 mR 2 and bot = v bot /R , we have mgh = 7 10 mv 2 bot v 2 bot = 10 7 gh. Now, as the marble climbs up the right side, v drop to 0, but stays at bot . Thus we no longer have = v/R . Only the translational part of the kinetic energy is converted back to potential energy. If h prime is the final height on the right side, then conservation of energy tells us that 0 + 1 2 mv 2 bot + 1 2 I 2 bot = mgh prime + 1 2 I 2 bot bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright K right = 1 5 mv 2 bot , so that mgh prime = 1 2 mv 2 bot = 5 7 mgh h prime = 5 7 h. ( b ): If there is enough friction on the right hand side then the rotational energy is also turned into potential energy, and the final kinetic energy on the right hand side is 0. Thus h prime = h . ( c ): The marble goes higher with friction because on the way up the friction points up. It has to, in order to slow down the rotation of the marble to keep it in step with the translational velocity v ....
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 Summer '08
 Chamberlin
 Mass, Work

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