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Unformatted text preview: The rotational inertia of the block around the rotational axis is obtained by the parallel axis theorem: I = I COM + mr 2 = (m/2)(a 2 + b 2 ) + mr 2 The period of the oscillation is T(r) = 2 (I/mgr) = 2 [(a 2 + b 2 )/(12gr) + r/g] When the period takes the minimum value, the derivative of the period is zero. For convenience, we consider the function inside the square root, take the derivative of the function and set it equal to zero. For if this function takes a minimum value, the period takes a minimum value too. f(r) = (a 2 + b 2 )/(12gr) + r/g, f(r) = (a 2 + b 2 )/(12gr 2 ) + 1/g = 0 r min = [(a 2 + b 2 )/(12)] = 0.0842 m b) What is the minimum value of the period? Plugging the result of part a) into the expression for the period, we get T(r min ) = 2 [(a 2 + b 2 )/(12gr) + r min /g] = 0.823 s...
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This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.
 Summer '08
 Field
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