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 find the tension in the chain that is rotating around a smooth cone, we can follow these steps:

### Step 1: Understand the System
We have a chain of mass \( m \) forming a circle of radius \( R \) on a smooth cone with a half-angle \( \theta \). The chain rotates with a constant angular velocity \( \omega \) about the vertical axis.

### Step 2: Draw the Free Body Diagram (FBD)
Consider an infinitesimal mass element \( dm \) of the chain. The forces acting on this mass element are:
- The weight \( dm \cdot g \) acting downward.
- The tension \( T \) acting along the chain.
- The normal force \( dN \) acting perpendicular to the surface of the cone.

### Step 3: Analyze Forces in Horizontal and Vertical Directions
1. **Horizontal Forces**: In the horizontal direction, the component of the tension providing the centripetal force is:
\[ 2T \sin(\theta) = dm \cdot \omega^2 \cdot R \]
This can be named as Equation (1).

2. **Vertical Forces**: In the vertical direction, the weight of the mass element is balanced by the vertical component of the normal force:
\[ mg = dN \cdot \cos(\theta) \]
This can be named as Equation (2).

### Step 4: Relate Normal Force to Mass Element
From Equation (2), we can express the normal force \( dN \) in terms of \( mg \):
\[ dN = \frac{mg}{\cos(\theta)} \]

### Step 5: Substitute Normal Force into Horizontal Forces Equation
Substituting \( dN \) into Equation (1):
\[ 2T \sin(\theta) + \frac{mg}{\cos(\theta)} = dm \cdot \omega^2 \cdot R \]

### Step 6: Solve for Tension \( T \)
Rearranging the equation to isolate \( T \):
\[ T = \frac{dm \cdot \omega^2 \cdot R - \frac{mg}{\cos(\theta)}}{2 \sin(\theta)} \]

### Step 7: Integrate Over the Entire Chain
Since \( dm = \frac{m}{2\pi R} d\theta \) (where \( d\theta \) is the infinitesimal angle subtended by the mass element), we can integrate over the full circle to find the total tension in the chain.

### Final Expression for Tension
After integrating and simplifying, we find:
\[ T = \frac{R \omega^2 + g \cot(\theta)}{2\pi} \]

### Conclusion
Thus, the tension in the chain is given by:
\[ T = \frac{R \omega^2 + g \cot(\theta)}{2\pi} \]