A chain of mass m forming a circle of radius R is slipped on a smooth round cone with half- angle ` theta`. Find the tension in the chain if it rotates with a constant angular velocity `omega` about a vertical axis coinciding with the symmetry axis of the cone .

Options

A. `(R omega ^2 + g cot theta ) (m)/(2pi)`
B. `(R omega ^(2) - g cot theta )(m)/(2pi)`
C. `(R omega ^(2) + g cot theta ) (m)/(pi)`
D. `(R omega -g cot theta )(m)/(2 pi)`

Answer

To solve the problem of finding the tension in a chain of mass \( m \) forming a circle of radius \( R \) that is slipped on a smooth cone with half-angle \( \theta \), we can follow these steps:

### Step 1: Analyze the System
Consider a small element of the chain with mass \( dm \). The chain is rotating with a constant angular velocity \( \omega \) about the vertical axis of the cone. The gravitational force acting on this element is \( dm \cdot g \), where \( g \) is the acceleration due to gravity.

**Hint:** Identify the forces acting on the small mass element of the chain.

### Step 2: Identify Forces Acting on the Element
The forces acting on the small element \( dm \) are:
1. The weight \( dm \cdot g \) acting downward.
2. The tension \( T \) in the chain acting along the chain.
3. The normal force \( N \) acting perpendicular to the surface of the cone.

### Step 3: Resolve Forces
Resolve the normal force \( N \) into components:
- The vertical component is \( N \cos \theta \).
- The horizontal component is \( N \sin \theta \).

For the element to be in equilibrium in the vertical direction:
\[ N \cos \theta = dm \cdot g \]

For the horizontal direction, the centripetal force required for circular motion is provided by the horizontal component of the normal force and the tension:
\[ N \sin \theta + T = dm \cdot \omega^2 R \]

### Step 4: Express \( dm \)
The mass element \( dm \) can be expressed in terms of the total mass \( m \) and the angle \( d\alpha \) subtended by the element at the center:
\[ dm = \frac{m}{2\pi} d\alpha \]

### Step 5: Substitute \( dm \) into the Equations
Substituting \( dm \) into the vertical force equation:
\[ N \cos \theta = \frac{m}{2\pi} d\alpha \cdot g \]
From this, we can express \( N \):
\[ N = \frac{m g d\alpha}{2\pi \cos \theta} \]

Substituting \( N \) into the horizontal force equation:
\[ \frac{m g d\alpha}{2\pi \cos \theta} \sin \theta + T = \frac{m}{2\pi} d\alpha \cdot \omega^2 R \]

### Step 6: Isolate Tension \( T \)
Rearranging the equation to solve for \( T \):
\[ T = \frac{m}{2\pi} d\alpha \cdot \omega^2 R - \frac{m g d\alpha}{2\pi \cos \theta} \sin \theta \]
Factoring out \( \frac{m d\alpha}{2\pi} \):
\[ T = \frac{m d\alpha}{2\pi} \left( \omega^2 R - \frac{g \sin \theta}{\cos \theta} \right) \]

### Step 7: Simplify the Expression
Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ T = \frac{m d\alpha}{2\pi} \left( \omega^2 R - g \tan \theta \right) \]

### Final Result
The tension in the chain is given by:
\[ T = \frac{m}{2\pi} \left( \omega^2 R - g \tan \theta \right) \]