Consider a rod CD which is held horiznotally across a peg transversally .(something like a a st line abt a peg and free to rotate abt any pnt)bt point A(which is not the com) and D is the com . and DA=a
Find the angle rotated by the rod before it slips. and a is the distance between A and D and k is angular accleration
I think this is simple enough at that @ form the horizontal using torque equation mgcos@ a=Ik
k=mgcos@a/I and acc of com mgcos@a^2/I Now using
mgcos@-N=ma we get N as mgcos@(1-ma^2/I) now since it will slip
mgcos(1-ma^2/I)<mgsin@
solving it we get tan@>ul^2/L^2+12a^2 however the asnwer is give as ul^2/l^2+36a^2[
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2 Answers
1this means slip form the peg at a particular momentobviouslygravitybecomes more than gravity