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Unformatted text preview: )) 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 C (assumed to be the geometric center of the stick), we see that the distance between O and C is h = x. The parallel axis theorem (see Eq. 15-30) leads to I= L2 1 mL2 + mh 2 = m + x 2 ÷. 12 12 Eq. 15-29 gives I T = 2π = 2π mgh L c + x h 2π c + 12 x h = . L2 12 2 2 gx 2 12 gx (a) Minimizing T by graphing (or special calculator functions) is straightforward, but the standard calculus method (setting the derivative equal to zero and solving) is somewhat awkward. We pursue the calculus method but choose to work with 12 gT2/2π instead of T (it should be clear that 12gT2/2π is a minimum whenever T is a minimum). The result is d e j = 0 = d d + 12 xi = − L + 12 12 gT 2 2π dx L2 x 2 dx x2 which yields x = L / 12 = (1.85 m)/ 12 = 0.53 m 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 p...
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This note was uploaded on 05/14/2013 for the course MATH 346 taught by Professor Professormiguelarcones during the Spring '08 term at Binghamton University.

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