ch15-p049

# ch15-p049 - 49. This is similar to the situation treated in...

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awkward. We pursue the calculus method but choose to work with 12 gT 2 /2 π instead of T (it should be clear that 12 gT 2 /2 π is a minimum whenever T is a minimum). The result is d dx dx dx L x gT L x 12 2 2 2 2 2 0 12 12 π e j di == + =− + which yields / 12 (1.85 m)/ 12 0.53 m xL = as the value of x which should produce the smallest possible value of T . (b) With L = 1.85 m and x = 0.53 m, we obtain T = 2.1 s from the expression derived in part (a). 49. This is similar to the situation treated in Sample Problem 15-5, except that O is no longer at the end of the stick. Referring to the center of mass as
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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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