awkward. We pursue the calculus method but choose to work with 12gT2/2πinstead of T(it should be clear that 12gT2/2πis a minimum whenever Tis a minimum). The result is ddxdxdxLxgTLx122222201212πejdi==+=−+which yields / 12(1.85 m)/ 120.53 mxL=as the value of xwhich 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 Ois 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.