A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical.A small block is kept in the bowl at a position where the radius makes an angle `theta` with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is `mu`. Find the range of the angular speed for which the block will not slip.

Answer

A, B, C

Explanation

If et bowl rotates at maximum angular speed, the block tends to slip upwards.
So the frictioN/Al force acts downward.
Here, `r=Rsintheta`
From the free body diagram 1
`R_1-mg costheta-m omega^2_1(Rsintheta)sintheta`
`=0……….i`
[because `r=Rsintheta]` and
`muR_1+mgsintheta-momega^2_1(Rsintheta)costheta`
=0 ii.

Substituting the value of `R_1` from equation i. inequation ii it can be found out that
`omega_1=[(g(sintheta)+mucostheta))/(Rsintheta(costheta-musintheta))]^(1/2)`
again, for minimum speed, the frictioN/Asl force `muR_2` acrts upwasrd,. From free body diagram -2, it can be prove that

`omega_2=[(g(sintheta+mu costheta))/(R sintheta(costhet-musintheta))]^(1/2)`
`:.` the range of speed is betwen `omega_2 and omega_1`