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A particle rest on the top of a smooth hemisphere of radius r. It is imparted a horizontal velocity of √(ngr) (root). The angle made by radius vector joining the particle with the verticle, at the instant, the particle looses contact with sphere is Θ.
Find cosΘ.
applying energy conservatin
1/2mu^2 +mgr(1-cosq)=1/2mv^2 eq-1
u^2=ngr (given)
and we have
mgcosq-N=(mv^2)/r eq-2
here N normal reaction ,at this point N is zero
on solving we get
cosq=(n+2)/4
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