ch10-p042

# ch10-p042 - OO OO with the one at O in the middle, then the...

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I = 0.083519 ML 2 (0.08352)(0.1000 kg)(1.0000 m) 2 = 8.352 ¡ 10 3 kg m 2 . (b) Comparing to the formula ( e ) in Table 10-2 (which gives roughly I =0.08333 ML 2 ), we find our answer to part (a) is 0.22% lower. 42. (a) Consider three of the disks (starting with the one at point O ): OO . The first one (the one at point O – shown here with the plus sign inside) has rotational inertial (see item (c) in Table 10-2) I = 1 2 mR 2 . The next one (using the parallel-axis theorem) has I = 1 2 mR 2 + mh 2 where h = 2 R . The third one has I = 1 2 mR 2 + m (4 R ) 2 . If we had considered five of the disks
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Unformatted text preview: OO OO with the one at O in the middle, then the total rotational inertia is I = 5( 1 2 mR 2 ) + 2( m (2 R ) 2 + m (4 R ) 2 ). The pattern is now clear and we can write down the total I for the collection of fifteen disks: I = 15( 1 2 mR 2 ) + 2( m (2 R ) 2 + m (4 R ) 2 + m (6 R ) 2 + + m (14 R ) 2 ) = 2255 2 mR 2 . The generalization to N disks (where N is assumed to be an odd number) is I = 1 6 (2 N 2 + 1) NmR 2 . In terms of the total mass ( m = M /15) and the total length ( R = L /30), we obtain...
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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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