A hemispherical bowl of radius R si set rotating about its axis of symmetry which is kept vertical. A small block kept in the bowl rotates with the bowl without slipping on its surface. If the surfaces of the bowl is smooth, and the angle made by the radius through the block with the vertical is `theta`, find the angular speed at which the bowl is rotating.

Explanation

To find the angular speed at which the hemispherical bowl is rotating, we can follow these steps:

### Step 1: Understand the Forces Acting on the Block
In the rotating bowl, the block experiences two main forces: the gravitational force (weight) acting downward and the normal force exerted by the surface of the bowl. Since the surface is smooth, there is no friction.

### Step 2: Analyze the Forces in the Radial Direction
The block is in circular motion, so we need to consider the centripetal force required to keep the block moving in a circle. The component of the normal force acting towards the center of the bowl provides this centripetal force.

- The component of the normal force acting towards the center of the bowl is given by:
\[ N \sin \theta \]
- The required centripetal force for the block moving in a circle of radius \( r \sin \theta \) is:
\[ F_c = m \omega^2 (r \sin \theta) \]

### Step 3: Set Up the Equation for Centripetal Force
Since the block is not slipping, the centripetal force must equal the radial component of the normal force:
\[ N \sin \theta = m \omega^2 (r \sin \theta) \]

### Step 4: Analyze the Forces in the Vertical Direction
In the vertical direction, the weight of the block is balanced by the vertical component of the normal force:
\[ N \cos \theta = mg \]

### Step 5: Divide the Two Equations
Now, we can divide the equation from Step 3 by the equation from Step 4 to eliminate \( N \):
\[ \frac{N \sin \theta}{N \cos \theta} = \frac{m \omega^2 (r \sin \theta)}{mg} \]
This simplifies to:
\[ \tan \theta = \frac{\omega^2 r \sin \theta}{g} \]

### Step 6: Solve for Angular Speed \( \omega \)
Rearranging the equation gives:
\[ \omega^2 = \frac{g \tan \theta}{r \sin \theta} \]
Thus, we can express \( \omega \) as:
\[ \omega = \sqrt{\frac{g \tan \theta}{r \sin \theta}} \]

### Step 7: Simplify the Expression
Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we can substitute:
\[ \omega = \sqrt{\frac{g \sin \theta}{r \sin \theta \cos \theta}} = \sqrt{\frac{g}{r \cos \theta}} \]

### Final Result
The angular speed \( \omega \) at which the bowl is rotating is:
\[ \omega = \sqrt{\frac{g}{r \cos \theta}} \]