A hemispherical bowl of radius R is rotating about its own axis (which is vertical) with an angular velocity `omega` . A particle on the frictionless inner surface of the bowl is also rotating with the same `omega` . The particle is a height h from the bottom of the bowl.
(i) Obtain the relation between h and `omega` _________
(ii) Find minimum value of `omega` needed, in order to have a non-zero value of h _____________ .

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry of the Problem
We have a hemispherical bowl with radius \( R \) and a particle located at a height \( h \) from the bottom of the bowl. The particle is rotating with angular velocity \( \omega \).

### Step 2: Identify the Forces Acting on the Particle
The forces acting on the particle are:
- The gravitational force \( mg \) acting downwards.
- The normal force \( N \) exerted by the surface of the bowl acting perpendicular to the surface.

### Step 3: Determine the Geometry of the Particle's Position
The angle \( \theta \) that the radius makes with the vertical can be expressed in terms of \( h \):
\[ \cos \theta = \frac{R - h}{R} \]
\[ \sin \theta = \frac{h}{R} \]

### Step 4: Apply Newton's Second Law in the Radial Direction
In the radial direction, the centripetal force required to keep the particle moving in a circular path is provided by the horizontal component of the normal force:
\[ N \sin \theta = m \frac{v^2}{r} \]
where \( v = \omega (R \cos \theta) \) is the linear velocity of the particle and \( r = R \cos \theta \) is the radius of the circular path.

### Step 5: Substitute for the Velocity
Substituting for \( v \):
\[ N \sin \theta = m \frac{(\omega (R \cos \theta))^2}{R \cos \theta} \]
This simplifies to:
\[ N \sin \theta = m \omega^2 R \cos \theta \]

### Step 6: Apply Newton's Second Law in the Vertical Direction
In the vertical direction, we have:
\[ N \cos \theta - mg = 0 \quad \Rightarrow \quad N \cos \theta = mg \]

### Step 7: Solve for Normal Force \( N \)
From the vertical force balance:
\[ N = \frac{mg}{\cos \theta} \]

### Step 8: Substitute \( N \) into the Radial Equation
Substituting \( N \) into the radial equation:
\[ \frac{mg}{\cos \theta} \sin \theta = m \omega^2 R \cos \theta \]
This simplifies to:
\[ g \tan \theta = \omega^2 R \]

### Step 9: Substitute for \( \tan \theta \)
Using \( \tan \theta = \frac{h}{R - h} \):
\[ g \frac{h}{R - h} = \omega^2 R \]

### Step 10: Rearranging to Find Relation Between \( h \) and \( \omega \)
Rearranging gives:
\[ h g = \omega^2 R (R - h) \]
\[ h g + \omega^2 R h = \omega^2 R^2 \]
\[ h (g + \omega^2 R) = \omega^2 R^2 \]
\[ h = \frac{\omega^2 R^2}{g + \omega^2 R} \]

### Step 11: Find Minimum Value of \( \omega \)
For \( h \) to be non-zero, the denominator must be positive:
\[ g + \omega^2 R > 0 \]
This implies:
\[ \omega^2 R > -g \quad \text{(always true for positive } g \text{)} \]
To find the minimum value of \( \omega \) such that \( h \) is non-zero, we set \( h = 0 \):
\[ \frac{\omega^2 R^2}{g + \omega^2 R} = 0 \]
This occurs when \( g + \omega^2 R = 0 \), leading to:
\[ \omega^2 = \frac{g}{R} \]
Thus, the minimum value of \( \omega \) is:
\[ \omega_{\text{min}} = \sqrt{\frac{g}{R}} \]

### Final Answers
(i) The relation between \( h \) and \( \omega \) is:
\[ h = \frac{\omega^2 R^2}{g + \omega^2 R} \]

(ii) The minimum value of \( \omega \) needed for a non-zero value of \( h \) is:
\[ \omega_{\text{min}} = \sqrt{\frac{g}{R}} \]