333_Physics ProblemsTechnical Physics

333_Physics ProblemsTechnical Physics - Chapter 11 P11.30...

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Chapter 11 335 P11.30 (a) The total angular momentum of the system of the student, the stool, and the weights about the axis of rotation is given by II I m r total weights student 2 kg m =+=+⋅ 23 0 0 2 ej . Before: r = 100 . m . Thus, I i =+ = 2 3 00 1 00 3 00 9 00 2 .. . . kg m kg m kg m 22 bg af After: r = 0300 m Thus, I f = 2300 300 354 2 . . kg m kg m kg m We now use conservation of angular momentum. ff i i ωω = or f i f i I I = F H G I K J = F H G I K J = 900 0750 191 . . rad s rad s (b) KI ii i == = 1 2 1 2 9 00 0 750 2 53 2 2 ω . kg m rad s J 2 ff f = 1 2 1 2 644 2 2 . kg m rad s J 2 P11.31 (a) Let M = mass of rod and m = mass of each bead. From f f = , we have 1 12 2 1 12 2 2 1 2 2 Mm r r if AA + L N M O Q P L N M O Q P When A = 0500 m , r 1 0100 = m , r 2 0250 = . m , and with other values as stated in the problem, we find f = 920 r a d s . (b) Since there is no external torque on the rod, L = constant and
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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