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 $\theta$ 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 1

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

So the frictional force acts downwards.

Here, $r=Rsin\theta$

From the free body diagram-1,

$R_1-mgcos\theta -m{\omega }_1^2\left(Rsin\theta \right)sin\theta =0...\left(i\right)$

[because. $r=Rsin\theta ]$ and

$\mu R_1+mgsin\theta -m{\omega }_1^2\left(Rsin\theta \right)cos\theta =0...\left(ii\right)$

Substituting the value of $R_1$ from equation (i) in equation (ii), it can be found out that,

${\omega }_1={\left[\frac{g(sin\theta +\mu cos\theta )}{Rsin\theta (cos\theta -\mu sin\theta )}\right]}^{1/2}$

again, for minimum speed, the frictional force $\mu R_2$ acts upward. From free body diagram-2, it can be proved that,

${\omega }_2={\left[\frac{g(sin\theta +\mu cos\theta )}{Rsin\theta (cos\theta -\mu sin\theta )}\right]}^{1/2}$

$\therefore$ the range of speed is between ${\omega }_2$ and ${\omega }_1.$