ch15-p106

# ch15-p106 - 106. (a) The potential energy at the turning...

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22 31 3 0 42 2 dE d d Mv kx Mv a kxv dt dt dt §· § · =+ = + = ¨¸ ¨ ¸ ©¹ © ¹ cm cm cm cm which leads to = 2 3 . a k M x cm F H G I K J Comparing with Eq. 15-8, we see that ω = 23 kM / for this system. Since = 2 π / T , we obtain the desired result: TM k = 2 π /. 106. (a) The potential energy at the turning point is equal (in the absence of friction) to the total kinetic energy (translational plus rotational) as it passes through the equilibrium position: 2 2 2 cm cm cm cm 222 cm cm cm 11 1 1 1 1 2 2 2 2 113 244 m v kx Mv I Mv MR R Mv Mv Mv =+= which leads to Mv kx m cm = / = 0.125 J. The translational kinetic energy is therefore
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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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