A particle of mass m is whirled in horizontal circle with the help of two threads of length l each as shown in figure . Angular velocity equals `omega`, then
`T_(1)` equals

Options

A. ` (mg)/(sqrt(3))+ (momega^(2) l)/2`
B. ` (mg)/(sqrt(3)) + (momega^(2)l)/2`
C. ` (momega^(2)l)/2`
D. `(mg)/(sqrt(3)`

Answer

To solve the problem of finding the tension \( T_1 \) in the scenario where a particle of mass \( m \) is whirled in a horizontal circle using two threads of length \( l \), we can follow these steps:

### Step 1: Understand the Geometry
The particle is at a height and is moving in a horizontal circle. The threads make an angle with the vertical. For each thread, the vertical component of the tension must balance the weight of the mass, and the horizontal component provides the necessary centripetal force.

### Step 2: Determine the Geometry of the Setup
From the geometry, we have:
- The angle made by the threads with the vertical is \( 30^\circ \).
- The horizontal distance \( D \) from the vertical axis to the mass can be calculated using the sine function:
\[ D = l \sin(30^\circ) = l \cdot \frac{1}{2} = \frac{l}{2} \]

### Step 3: Write the Forces in the Vertical Direction
For the vertical forces, we can write:
\[ T_1 \cos(30^\circ) + T_2 \cos(30^\circ) = mg \]
Using \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), this becomes:
\[ T_1 \cdot \frac{\sqrt{3}}{2} + T_2 \cdot \frac{\sqrt{3}}{2} = mg \]
Dividing through by \( \frac{\sqrt{3}}{2} \):
\[ T_1 + T_2 = \frac{2mg}{\sqrt{3}} \quad \text{(Equation 1)} \]

### Step 4: Write the Forces in the Horizontal Direction
For the horizontal forces, the net centripetal force is provided by the horizontal components of the tensions:
\[ T_1 \sin(30^\circ) - T_2 \sin(30^\circ) = m \frac{v^2}{D} \]
Substituting \( \sin(30^\circ) = \frac{1}{2} \):
\[ \frac{1}{2}(T_1 - T_2) = m \frac{v^2}{D} \]
Substituting \( D = \frac{l}{2} \) gives:
\[ \frac{1}{2}(T_1 - T_2) = m \frac{v^2}{\frac{l}{2}} = \frac{2mv^2}{l} \]
Multiplying through by 2:
\[ T_1 - T_2 = \frac{4mv^2}{l} \quad \text{(Equation 2)} \]

### Step 5: Solve the Equations
Now we have two equations:
1. \( T_1 + T_2 = \frac{2mg}{\sqrt{3}} \)
2. \( T_1 - T_2 = \frac{4mv^2}{l} \)

Adding these two equations:
\[ 2T_1 = \frac{2mg}{\sqrt{3}} + \frac{4mv^2}{l} \]
Thus,
\[ T_1 = \frac{mg}{\sqrt{3}} + \frac{2mv^2}{l} \]

### Step 6: Relate Linear Velocity to Angular Velocity
The linear velocity \( v \) can be related to the angular velocity \( \omega \) using the formula \( v = \omega D \):
\[ v = \omega \cdot \frac{l}{2} \]
Substituting this into the equation for \( T_1 \):
\[ T_1 = \frac{mg}{\sqrt{3}} + \frac{2m(\omega \cdot \frac{l}{2})^2}{l} \]
This simplifies to:
\[ T_1 = \frac{mg}{\sqrt{3}} + \frac{m\omega^2 l}{2} \]

### Final Result
Thus, the tension \( T_1 \) is given by:
\[ T_1 = \frac{mg}{\sqrt{3}} + \frac{m\omega^2 l}{2} \]