PHY2048_Fall09_Homework_Set14

PHY2048_Fall09_Homework_Set14 - PHY2048 Fall 2009 Homework...

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Unformatted text preview: PHY2048 Fall 2009 Homework Set 14 Solutions University of Florida Page 1 of 4 Department of Physics PHY 2048 Homework Set 14 Solutions (Chapters 15 & 16) Problem 15-49**: For a physical pendulum we know that Mgh I T rod 2 2 = = , (eq. 1) where in this case h = x and 2 2 12 1 Mx ML I rod + = . It is easier to work with T 2 . If I square (eq. 1) I get + = + = = x x L g Mx ML Mgx I Mgx x T rod 12 4 ) ( 4 4 ) ( 2 2 2 2 12 1 2 2 2 (eq. 2) Taking the derivative with respect to x and setting it equal to zero gives 12 4 1 12 4 ) ( 2 2 2 2 2 2 2 2 = + = + = x L gx x L g dx x dT Hence (a) m m L x 534 . 12 ) 85 . 1 ( 12 min = = = Plugging this into (eq. 2) yields g L g L x L gx T 3 3 4 3 12 2 12 4 2 2 2 min 2 min 2 2 min = = + = and (b) s s m m g L T 074 . 2 ) / 8 . 9 ( 3 ) 85 . 1 ( 3 2 3 3 2 2 min = = = Problem 16-5*: We are given that ) sin( ) , ( + = t kx y t x y m with y m = 6 mm and = 600 rad/s. Let ) sin( ) 6 ( 2 ) , ( 1 1 + = = t kx mm mm t x y and hence 3 1 1 ) sin( = and hence 1 = 19.471 o = 0.3398 rad. Now let 3 1 2 2 ) sin( ) sin( = = + kx and hence 2 = - 19.471 o = -0.3398 rad. Thus, t t t rad = = = = ) ( ) 3398 . ( 2 1...
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PHY2048_Fall09_Homework_Set14 - PHY2048 Fall 2009 Homework...

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