Nov.5, 2007 - Rotational Dynamics - Solutions

# Nov.5, 2007 - Rotational Dynamics - Solutions - 10.20....

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10.20. IDENTIFY: Only gravity does work, so other 0 W and conservation of energy gives i i f f K U K U . 22 11 f cm cm K Mv I . SET UP: Let f 0 y , so f 0 U and i 0.750 m y . The hoop is released from rest so i 0 K . cm vR . For a hoop with an axis at its center, 2 cm I MR . EXECUTE: (a) Conservation of energy gives if UK . 2 2 2 2 2 2 f () K MR MR MR , so i MR Mgy . 2 i (9.80 m/s )(0.750 m) 33.9 rad/s 0.0800 m gy R . (b) (0.0800 m)(33.9 rad/s) 2.71 m/s EVALUATE: An object released from rest and falling in free-fall for 0.750 m attains a speed of 2 (0.750 m) 3.83 m/s g . The final speed of the hoop is less than this because some of its energy is in kinetic energy of rotation. Or, equivalently, the upward tension causes the magnitude of the net force of the hoop to be less than its weight. 10.22. IDENTIFY: Apply m Fa to the translational motion of the center of mass and zz I to the rotation about the center of mass. SET UP: Let x be down the incline and let the shell be turning in the positive direction. The free-body diagram for the shell is given in Fig.10.22. From Table 9.2, 2 2 cm 3 I mR .

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## This note was uploaded on 04/03/2008 for the course PHYSICS 140 taught by Professor Evrard during the Fall '07 term at University of Michigan.

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Nov.5, 2007 - Rotational Dynamics - Solutions - 10.20....

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