Question

A hemispherical bowl of raidus r is rotated about its axis of symmetry which is kept vertical. A small block is kept 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 is $\mu$. The maximum speed for which the block will not slip

• A. ${\left[\frac{g(\sin \theta -\mu \cos \theta )}{r\sin \theta (\cos \theta +\mu \sin \theta )}\right]}^{1/2}$
• B. ${\left[\frac{g(\sin \theta +\mu \cos \theta )}{r\sin \theta (\cos \theta +\mu \sin \theta )}\right]}^{1/2}$
• C. ${\left[\frac{g(\sin \theta +\mu \cos \theta )}{r\sin \theta (\cos \theta -\mu \sin \theta )}\right]}^{1/2}$
• D. none of these

Answer 1

Answer: (C)

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

When the bowl rotates, there will be a range of angular speed for which the block will not move.

• When the angular speed of the bowl is minimum in this range, the block will have a tendency to roll down. So, the frictional force will act in the upward direction along the surface of the bowl.
• When the angular speed of the bowl is maximum in this range, the block will have a tendency to roll up. So, the frictional force will act in the downward direction along the surface of the bowl.
  

We only need to consider the second case here.
Lets say the block moves in a horizontal circle with the centre at C as shown in diagram so that the radius is $PC=OP\sin =r\sin \theta$
Its acceleration is therefore ${\omega }^{2}r\sin \theta$. Resolving the forces along PC and applying Newton's second law,
$N\sin \theta +\mu N\cos \theta =m{\omega }^{2}r\sin \theta$ --------(1)
As there is no vertical acceleration,
$N\cos \theta -\mu N\sin \theta =mg$
$\Rightarrow N=\frac{mg}{(cos\theta -sin\theta )}$ ---------(2)
Replacing the value of $N$ from equation (2) in (1)
$\frac{mg(\sin \theta +\mu \cos \theta )}{(\cos \theta -\mu \sin \theta )}=m{\omega }^{2}r\sin \theta$
$\Rightarrow \omega ={\left[\frac{g(\sin \theta +\mu \cos \theta )}{r\sin \theta (\cos \theta -\mu \sin \theta )}\right]}^{1/2}$