HW8 Solutions

I i i i b in this case we obtain 2 2 1 1 2 2 2 2 1 2

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I I I I (b) In this case, we obtain 2 2 1 1 2 2 2 2 1 2 3.3 kg m 450 rev/min 6.6 kg m 900 rev/min 3.3 kg m 6.6 kg m 450 rev min I I I I   or | | 450 rev min . (c) The minus sign indicates that is clockwise, that i s, in the direction of the second disk’s initial angular velocity. 7. We denote the cockroach with subscript 1 and the disk with subscript 2. The cockroach has a mass m 1 = m , while the mass of the disk is m 2 = 4.00 m .

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(a) Initially the angular momentum of the system consisting of the cockroach and the disk is L m v r I m R m R i i i i 1 1 1 2 2 1 0 2 2 0 2 1 2 . After the cockroach has completed its walk, its position (relative to the axis) is r R f 1 2 so the final angular momentum of the system is . 2 1 2 2 2 2 1 R m R m L f f f Then from L f = L i we obtain . 2 1 2 1 4 1 2 2 2 1 0 2 2 2 1 R m R m R m R m f Thus, 2 2 1 2 2 1 0 0 0 0 2 2 1 2 2 1 2 1 ( / ) 2 1 2 1.33 . 4 2 1/ 4 ( / ) 2 1/ 4 2 f m R m R m m m R m R m m With = 0.260 rad/s, we have f =0.347 rad/s. (b) We substitute I = L / into K I 1 2 2 and obtain K L 1 2 . Since we have L i = L f , the kinetic energy ratio becomes 0 0 / 2 1.33. / 2 f f f i i L K K L (c) The cockroach does positive work while walking toward the center of the disk, increasing the total kinetic energy of the system. 8. By angular momentum conservation (Eq. 11-33), the total angular momentum after the explosion must be equal to that before the explosion: p r p r L L L L ( ) L 2 mv p + 1 12 ML 2 = I p + 1 12 ML 2 where one must be careful to avoid confusing the length of the rod ( L = 0.800 m) with the angular momentum symbol. Note that I p = m ( L /2 ) 2 by Eq.10-33, and = v end / r = ( v p 6)/( L /2 ), where the latter relation follows from the penultimate sentence in the problem (and “6” stands for “6.00 m/s” here). Since M = 3 m and = 20 rad/s, we end up with enough information to solve for the particle speed: v p = 11.0 m/s.
9. We make the unconventional choice of clockwise sense as positive, so that the angular velocities in this problem are positive. With r = 0.60 m and I 0 = 0.12 kg ∙ m 2

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