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.

Explanation

To solve the problem, we need to analyze the forces acting on the block in the hemispherical bowl and derive the conditions under which the block does not slip. Here's a step-by-step solution:

### Step 1: Understand the Forces Acting on the Block
When the bowl is rotated, the block experiences the following forces:
- Gravitational force \( mg \) acting downward.
- Normal force \( N \) acting perpendicular to the surface of the bowl.
- Frictional force \( f \) acting along the surface of the bowl.

### Step 2: Set Up the Free Body Diagram (FBD)
In the FBD, we can resolve the normal force and frictional force into components:
- The normal force \( N \) acts at an angle \( \theta \) with the vertical.
- The frictional force \( f \) can be expressed as \( f = \mu N \), where \( \mu \) is the coefficient of friction.

### Step 3: Apply Newton's Second Law in the Radial Direction
For the block to rotate in a circle of radius \( r \sin \theta \) (the horizontal distance from the axis of rotation), the net radial force must provide the required centripetal force:
\[ N \sin \theta + f \cos \theta = \frac{mv^2}{r \sin \theta} \]
Substituting \( f = \mu N \):
\[ N \sin \theta + \mu N \cos \theta = \frac{mv^2}{r \sin \theta} \]

### Step 4: Apply Newton's Second Law in the Vertical Direction
In the vertical direction, the forces balance as follows:
\[ N \cos \theta - mg + \mu N \sin \theta = 0 \]
This can be rearranged to find \( N \):
\[ N \cos \theta + \mu N \sin \theta = mg \]
\[ N(\cos \theta + \mu \sin \theta) = mg \]
\[ N = \frac{mg}{\cos \theta + \mu \sin \theta} \]

### Step 5: Substitute \( N \) into the Radial Force Equation
Substituting \( N \) back into the radial force equation:
\[ \left(\frac{mg}{\cos \theta + \mu \sin \theta}\right) \sin \theta + \mu \left(\frac{mg}{\cos \theta + \mu \sin \theta}\right) \cos \theta = \frac{mv^2}{r \sin \theta} \]
This simplifies to:
\[ \frac{mg \sin \theta + \mu mg \cos \theta}{\cos \theta + \mu \sin \theta} = \frac{mv^2}{r \sin \theta} \]

### Step 6: Solve for \( v^2 \)
Cancelling \( m \) from both sides:
\[ \frac{g (\sin \theta + \mu \cos \theta)}{\cos \theta + \mu \sin \theta} = \frac{v^2}{r \sin \theta} \]
Thus,
\[ v^2 = \frac{g (\sin \theta + \mu \cos \theta) r \sin \theta}{\cos \theta + \mu \sin \theta} \]

### Step 7: Relate Linear Velocity to Angular Velocity
The linear velocity \( v \) is related to the angular velocity \( \omega \) by:
\[ v = r \sin \theta \cdot \omega \]
Substituting this into the equation gives:
\[ (r \sin \theta \cdot \omega)^2 = \frac{g (\sin \theta + \mu \cos \theta) r \sin \theta}{\cos \theta + \mu \sin \theta} \]
This leads to:
\[ \omega^2 = \frac{g (\sin \theta + \mu \cos \theta)}{r (\cos \theta + \mu \sin \theta)} \]

### Step 8: Determine the Range of Angular Speed
For the maximum angular speed \( \omega_{\text{max}} \):
\[ \omega_{\text{max}}^2 = \frac{g}{r \sin \theta} \left( \tan \theta + \mu \right) \]
For the minimum angular speed \( \omega_{\text{min}} \):
\[ \omega_{\text{min}}^2 = \frac{g}{r \sin \theta} \left( \tan \theta - \mu \right) \]
Thus, the range of angular speed for which the block will not slip is:
\[ \omega_{\text{min}} \leq \omega \leq \omega_{\text{max}} \]