qz11sol_3891f10

# qz11sol_3891f10 - The rotational inertia of the block...

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TA: Tomoyuki Nakayama Tuesday, August 31, 2010 PHY 2048: Physic 1, Discussion Section 3891 Quiz 11 (Homework Set #13) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ A rectangular block, with face lengths a = 15.0 m and b = 25.0 cm, is to be suspended on a thin horizontal rod running through a narrow hole in the block. The block is then to be set swinging about the rod like a pendulum, through small angles so that it is in SHM. The figure below shows one possible position of the hole, at distance r from the block's center, along a line connecting the center with a corner. a) For what value of r does the period of the pendulum take the minimum value?
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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.

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