ch10-p065

ch10-p065 - position shows that the change in center of...

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65. Using the parallel axis theorem and items ( e ) and ( h ) in Table 10-2, the rotational inertia is I = 1 12 mL 2 + m ( L /2 ) 2 + 1 2 mR 2 + m ( R + L ) 2 = 10.83 mR 2 , where L = 2 R has been used. If we take the base of the rod to be at the coordinate origin ( x = 0, y = 0) then the center of mass is at y = mL/ 2 + m ( L + R ) m + m = 2 R . Comparing the position shown in the textbook figure to its upside down (inverted)
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Unformatted text preview: position shows that the change in center of mass position (in absolute value) is | y | = 4 R . The corresponding loss in gravitational potential energy is converted into kinetic energy. Thus, K = (2 m ) g (4 R ) = 9.82 rad/s . where Eq. 10-34 has been used....
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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