ch15-p039

# ch15-p039 - 39. (a) We take the angular displacement of the...

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() 2 2 sin 2 1 cos 2 1 1 2 3 2 tT ππ =− = where the trigonometric identity cos 2 θ + sin 2 = 1 is used. Thus, = 22 = 2 0.500 3 2 = 34.2 . Ω− F H I K F H I K F H G I K J π π T t T m sin s rad rad / s a f During another portion of the cycle its angular speed is +34.2 rad/s when its angular displacement is π /2 rad. (c) The angular acceleration is 2 2 cos 2 / . m d dt T T θπ π αθ §· == = ¨¸ ©¹ When = π /4, 2 2 2 = 124 rad/s , 0.500 s 4 α § · ¨ ¸ © ¹ or 2 | | 124 rad/s . = 39. (a) We take the angular displacement of the wheel to be = m cos(2 π t / T ), where m is the amplitude and T is the period. We differentiate with respect to time to find the
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## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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