ps8a20 - Ph 111.01 October 20, 2008 PS 8a (20) Problem 1 =...

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Ph 111.01 October 20, 2008 PS 8a (20) Problem 1 = WalkerP 10-49. A 1.29 kg hoop with a radius of 9.8 cm rolls without slipping and has a linear speed of 1.51 m/s. M R v ω v R (a) Find the translational kinetic energy. J (b) Find the rotational kinetic energy. J (c) Find the total kinetic energy of the hoop. J Solution. We are going to want the angular velocity ω : ( ) () 1.51 m/s 15.41 rad/s 0.098 m v R == = . And the moment of inertia: 2 22 1.29 kg .098 m 0.01239 kg-m IM R = . (a) Calculate the translational kinetic energy: ( )( ) 2 2 11 1.29 kg 1.51 m/s 1.471 m/s KM v = translational (b) Now the rotational kinetic energy: ( )( ) 2 0.01239 kg-m 15.41 rad/s 1.471 J KI = rotational total translational rotational In the case of the hoop, the two kinetic energies are equal! (c) So for the total kinetic energy we just add the results from parts (a) and (b): 1.471 J + 1.471 J = 2.942 J KK K =+ = Problem 2 = Walker P. 10-51. When a pitcher throws a curve ball, the ball is given a fairly rapid spin. A 0.15 kg baseball with a radius of 3.7 cm is thrown with a linear speed of 42 m/s and an angular speed of 36 rad/s. Assume that the ball is a uniform, solid sphere. (a) Calculate how much of its kinetic energy is translational.
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This note was uploaded on 05/13/2010 for the course PHYSICS 111 taught by Professor Staff during the Fall '08 term at S.F. State.

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ps8a20 - Ph 111.01 October 20, 2008 PS 8a (20) Problem 1 =...

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