ch10-p074 - 74. (a) Constant angular acceleration...

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(b) We use τ = I α , where τ is the torque and I is the rotational inertia. The contribution of the rod to I is M A 2 12 / (Table 10-2(e)), where M is its mass and A is its length. The contribution of each ball is m A /, 2 2 b g where m is the mass of a ball. The total rotational inertia is I Mm =+ = + AA 22 12 2 4 640 120 12 106 2 .. kg m kg m bg b g b g which yields I = 1.53 kg m 2 . The torque, therefore, is τ =⋅ = 153 7 66 117 . kg m rad / s N m. ch c h (c) Since the system comes to rest the mechanical energy that is converted to thermal energy is simply the initial kinetic energy KI i == = × 1 2 1 2 2 39 459 10 0 2 2 4 ω kg m rad / s J. 2 c h c h π (d) We apply Eq. 10-13: θω = + 0 2 2 1 2 23 9 3 2 0 1 2 766 320 tt π b gb g c hb g c hb g rad / s s rad / s s 2 . which yields 3920 rad or (dividing by 2 π ) 624 rev for the value of angular displacement
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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