A particle of mass m attached to an inextensible light string is moving in a vertical circle of radius r. The critical velocity at the highest point is `v_( 0)` to complete the vertical circle. The tension in the string when it becomes horizontal is

Options

A. ` (3mv_(0)^(2))/(r)`
B. `(9 mv_(0)^(2))/(r)`
C. 3mg
D. both (A) and (C) are correct

Answer

To solve the problem of finding the tension in the string when the particle becomes horizontal, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Critical Velocity**:
The critical velocity \( v_0 \) at the highest point is the minimum velocity required for the particle to complete the vertical circle. At this point, the centripetal force is provided solely by the weight of the particle.

2. **Energy Conservation**:
We can use the principle of conservation of mechanical energy. The total mechanical energy at the highest point (point 1) and at the horizontal position (point 2) must be equal.
\[ E_1 = E_2 \]
At the highest point, the energy consists of potential energy and kinetic energy:
\[ E_1 = \frac{1}{2} m v_0^2 + mg(2r) \]
At the horizontal position, the potential energy is lower (height = r), and the energy consists of kinetic energy:
\[ E_2 = \frac{1}{2} m v^2 + mg(r) \]

3. **Setting Up the Equation**:
Setting the energies equal gives us:
\[ \frac{1}{2} m v_0^2 + mg(2r) = \frac{1}{2} m v^2 + mg(r) \]
Simplifying this equation by canceling \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v_0^2 + 2gr = \frac{1}{2} v^2 + gr \]
Rearranging gives:
\[ \frac{1}{2} v^2 = \frac{1}{2} v_0^2 + gr \]
\[ v^2 = v_0^2 + 2gr \]

4. **Finding the Tension at the Horizontal Position**:
When the particle is at the horizontal position, the tension \( T \) in the string provides the necessary centripetal force. The net force towards the center is given by:
\[ T - mg = \frac{mv^2}{r} \]
Rearranging this gives:
\[ T = \frac{mv^2}{r} + mg \]

5. **Substituting for \( v^2 \)**:
From the previous step, we found:
\[ v^2 = v_0^2 + 2gr \]
Substituting this into the tension equation:
\[ T = \frac{m(v_0^2 + 2gr)}{r} + mg \]
Simplifying this:
\[ T = \frac{mv_0^2}{r} + 2mg + mg = \frac{mv_0^2}{r} + 3mg \]

6. **Final Expression for Tension**:
Thus, the tension in the string when it becomes horizontal is:
\[ T = \frac{mv_0^2}{r} + 3mg \]