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# Q.A particle moves from rest at A on the surface of a smooth cylinder of radius r as shown.At B it leaves the cylinder.The equation relating alpha and beta isa.)3sin(alpha)=2cos(beta)b.)2sin(alpha)=3cos(beta)c.) 3sin(beta)=2cos(alpha)d.)2sin(beta)=3cos(alpha)

10 years ago

let the horizonta lsurface of sphere is at zero potential....radius of sphere os R...

at point A ,

KE is 0 ,PE =mgRcos(alpha)

at point B

KE=mv2 /2 , PE=mgRsin(beta)

applying conservation of energy

total energy initial = total energy final

mv2 /2 = mgR{cos(alpha)- sin(beta)}................1

on resolving weight in two components will give one in direction opposite to normal reaction at point B...

so we can say that centripital force is provided by wsin(beta) only because normal reaction is 0 at this point as the body loses contact....

wsin(beta)=mgsin(beta)=mv2 /R...............2

on solving 1 and 2 we get

3sin(beta)=2cos(alpha)

hence option c is correct....