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P9 - the disk 2 2 5 1 3 2 1 C C x x W ∆ − = − ∆ −...

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Discussion 9: Name: TAM212, Spring, 2003 A 1 slug homogeneous cylindrical disk is given the counterclockwise velocity of 4 rad/s with the spring unstretched. The spring constant, k, equals 3 lb/ft. If the disk does not slip while rolling, how far will its center point move to left before returning to the right? T W = (1) ( ) = dt W i i v F v v , where the only force doing work is that due to the spring. (Since the wheels are rolling, friction doesn’t do work here) ( ) ( ) 2 1 2 2 2 / 1 δ δ = k W . Note; 0 1 = δ ; C x = δ 2 , where x is positive to the left, and C is the center of mass of
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Unformatted text preview: the disk. ( ) ( ) ( ) ( ) 2 2 5 . 1 3 2 / 1 C C x x W ∆ − = − ∆ − = (2) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 / 1 2 / 1 2 / 1 2 / 1 2 / 1 2 / 1 2 / 1 ω + = ω + = ω + = r v m mr mv I mv T C C C C Because the disk is rolling, r v C / = ω , and thus ( ) 2 2 2 2 2 2 2 / 3 2 / 1 2 / 1 ω = ω + ω = mr r r m T The wheel stops rolling to the left when = ω , so ( ) 2 2 2 / 24 4 1 1 2 / 3 s slugxft x x T − = − = ∆ (3) Putting (1)-(3) together, ( ) ft x x C C 2 2 12 5 . 1 2 = ∆ ⇒ − = ∆ −...
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