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Question

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 θ with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is μ. Find the range of the angular speed for which the block will not slip.


Solution

If the bowl rotates at maximum angular speed, the block tends to slip upwards.

So the frictional force acts downwards.  

Here,        r=R sin θ

From the free body diagram-1,

R1mg cos θmω21 (R sin θ)sin θ= 0             ...(i)

[because. r=R sin θ] and

μR1+mg sin θmω21 (R sin θ)cos θ= 0             ...(ii)

Substituting the value of R1 from equation (i) in equation (ii), it can be found out that,

ω1=[g (sin θ+μ cos θ)R sin θ (cos θμ sin θ)]1/2

again, for minimum speed, the frictional force μR2 acts upward. From free body diagram-2, it can be proved that,

ω2=[g (sin θ+μ cos θ)R sin θ (cos θμ sin θ)]1/2

the range of speed is between ω2 and  ω1.

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