A small ring is located near one end of a rod. Next, the rod is given an angular velocity 'w' along the axis shown.
With what kinetic energy does the ring leave the rod?
Mass of ring='m'. Mass of rod='M'. Length of rod='L'.
The action takes place in a horizontal plane.
root(Ml^2w^2/M+3m) ?
ReplyDeleteif u be the coeff. of fric.,
ReplyDeleteKE=1/2m(w^2l^2-2ugl)
plz confirm...
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ReplyDelete@@khilesh: Inore friction
ReplyDelete@Mayank: Try to keep the answer concise. The current answer gives out a lot of hints.
Sorry, a typing error
ReplyDeleteKE=MmL^2w^2(2M+3m)/2(M+3m)^2 ?
hey i got KE=(1/2)(m)(M^2)(w^2)(l^2)/(M+3m)^2.
ReplyDelete^i meant 'v' and not KE.here is a doubt- if we solve by energy conservation and by conserving angular momentum about the axis,we get 2 different values of 'v'.can anyone explain why?
ReplyDeleteon conserving energy we got diff equation in w and v(RKE + TKE)
ReplyDeleteby conserving angular momentum we got equation in w
1/2Iw^2=1/2Iw'^2+1/2mv^2
ReplyDeletealso,v=w'*L.isnt this the equation you got by conserving energy? what was your diff equation?
catch:
ReplyDeletethe ring has both tangential and radial (radially outwards) velocities.
is the ans (Mm(Lw)^2)/2(M+3m)
ReplyDeleteonly two concepts are applying
ReplyDeleteone is conservation of angular momentum about the O
and conservation of energy(please keep in mind that when the ring just going to leave the rod , it will have both tke and rke)
by the way i got same ans as mayank