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 step by step, we will analyze the forces acting on the particle in the rotating hemispherical bowl and derive the required relations.

### Step 1: Understanding the System
We have a hemispherical bowl of radius \( R \) rotating about its vertical axis with an angular velocity \( \omega \). A particle is located at a height \( h \) from the bottom of the bowl. The forces acting on the particle are:
- The gravitational force \( mg \) acting downwards.
- The normal force \( N \) acting perpendicular to the surface of the bowl.

### Step 2: Geometry of the Problem
The particle is at a height \( h \), which means the distance from the center of the bowl to the particle is \( R - h \). The angle \( \theta \) can be defined as the angle between the vertical axis and the line connecting the center of the bowl to the particle.

Using trigonometry:
- The vertical component of the radius is \( R - h \).
- The horizontal component (radius in the horizontal plane) is \( r = (R - h) \tan(\theta) \).

### Step 3: Applying Newton's Second Law
In the radial direction (horizontal), we have:
\[ N \sin(\theta) = m \omega^2 r \]
In the vertical direction, we have:
\[ N \cos(\theta) = mg \]

### Step 4: Relating Forces
From the vertical component:
\[ N = \frac{mg}{\cos(\theta)} \]

Substituting \( N \) into the radial equation:
\[ \frac{mg}{\cos(\theta)} \sin(\theta) = m \omega^2 r \]

### Step 5: Simplifying the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g \tan(\theta) = \omega^2 r \]

Using the relation \( r = (R - h) \tan(\theta) \):
\[ g \tan(\theta) = \omega^2 (R - h) \tan(\theta) \]

Assuming \( \tan(\theta) \neq 0 \) (which is valid since \( h \) is non-zero), we can divide both sides by \( \tan(\theta) \):
\[ g = \omega^2 (R - h) \]

### Step 6: Finding the Relation between \( h \) and \( \omega \)
Rearranging the equation gives us:
\[ h = R - \frac{g}{\omega^2} \]

This is the relation between \( h \) and \( \omega \).

### Step 7: Finding Minimum Value of \( \omega \)
To find the minimum value of \( \omega \) needed for \( h \) to be non-zero, we set \( h = 0 \):
\[ 0 = R - \frac{g}{\omega^2} \]
This leads to:
\[ \frac{g}{\omega^2} = R \]
Rearranging gives:
\[ \omega^2 = \frac{g}{R} \]
Taking the square root:
\[ \omega = \sqrt{\frac{g}{R}} \]

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

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