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Nov.5, 2007 - Rotational Dynamics - Solutions

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

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10.20. I DENTIFY : Only gravity does work, so other 0 W and conservation of energy gives i i f f K U K U . 2 2 1 1 f cm cm 2 2 K Mv I . S ET U P : 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 v R . For a hoop with an axis at its center, 2 cm I MR . E XECUTE : (a) Conservation of energy gives i f U K . 2 2 2 2 2 2 1 1 f 2 2 ( ) K MR MR MR , so 2 2 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 v R E VALUATE : 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. I DENTIFY : Apply m F a to the translational motion of the center of mass and z z I to the rotation about the center of mass. S ET U P : 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,
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