Chapter 11335P11.30(a)The total angular momentum of the system of the student, the stool, and the weights aboutthe axis of rotation is given byIIImrtotalweightsstudent2kg m=+=+⋅23002ej.Before:r=100. m.Thus,Ii=+⋅=⋅2 3 001 003 009 002....kgmkg mkg m22bgafAfter:r=0300mThus,If⋅=⋅23003003542..kgmkg mkg mWe now use conservation of angular momentum.ff iiωω=orfifiII=FHGIKJ=FHGIKJ=9000750191..rad srad s(b)KIiii==⋅=12129 000 7502 5322ω.kg mrad sJ2fff⋅=121264422.kg mrad sJ2P11.31(a)Let M=mass of rod and m=mass of each bead. From f f=, we have112211222122MmrrifAA+LNMOQPLNMOQPWhen A=0500m,r10100=m,r20250=.m , and with other values as stated in theproblem, we findf=920rads.(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 .