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A particle of mass m is released from th...

A particle of mass `m` is released from the top of a smooth hemisphere of radius `R` with the horizontal speed `u`. Calculate the angle with verticle where it loses contact with the hemisphere.

A

`sin^(-1)((u^(2))/(3gR)+(2)/(3))`

B

`cos^(-1)((u^(2))/(3gR)+(2)/(3))`

C

`cos^(-1)((u^(2))/(6gR)+(4)/(3))`

D

`sin^(-1)((u^(2))/(6gR)+(2)/(3))`

Text Solution

Verified by Experts

`h=R-Rcostheta`

`v^(2)=u^(2)+2gh=u^(2)+2gR(1-costheta)` ….(i)
`mg costheta-N=(mv^(2))/(R)` ….(ii)
When the particle loses contact
N = 0 ….(iii)
From Eqs. (i), (ii) and (iii), we have
`rArr" "v^(2)=gRcostheta=u^(2)+2gR(1-costheta)`
`rArr" "3gRcostheta=u^(2)+2gR`
`rArr" "costheta=(u^(2))/(3gR)+(2)/(3)`
`rArr" "theta=cos^(-1)((u^(2))/(3gR)+(2)/(3))`
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