A light rigid rod of length l is hinged at one end and it is free to rotate in a vertical plane. A particle of mass m is attached to another end of the rod. The particle is released from rest at its highest point. Select the correct statement out of the following

Options

A. Tension in the rod is zero when it makes an angle of `cos^(-1)""((1)/(3))` with the vertical
B. Tension in the rod is zero when it makes an angle of `cos^(-1)""((2)/(3))` with the vertical
C. Speed of partical of mass m is `sqrt((2)/(3)gl)` when the tension in the rod is zero
D. Speed of particle of mass m is `sqrt((4)/(3)gl)` when the tension in the rod is zero

Answer

To solve the problem step by step, we will analyze the motion of the particle attached to the rod when it is released from the highest point.

### Step 1: Understand the Initial Setup
- A light rigid rod of length \( L \) is hinged at one end and free to rotate in a vertical plane.
- A particle of mass \( m \) is attached to the free end of the rod.
- The particle is released from rest at the highest point.

**Hint:** Visualize the setup and identify the forces acting on the mass when it is released.

### Step 2: Determine Initial Potential Energy
- At the highest point, the height \( h \) of the mass \( m \) from the reference point (hinge) is equal to the length of the rod \( L \).
- The initial potential energy \( PE_{initial} \) is given by:
\[ PE_{initial} = mgh = mgL \]

**Hint:** Remember that potential energy is calculated using the formula \( PE = mgh \).

### Step 3: Determine Final Potential Energy
- As the rod rotates and the mass descends to an angle \( \theta \), the height of the mass from the hinge becomes \( L \cos \theta \).
- The final potential energy \( PE_{final} \) is:
\[ PE_{final} = mg(L \cos \theta) \]

**Hint:** The height changes as the mass moves, affecting the potential energy.

### Step 4: Apply Conservation of Energy
- The change in potential energy will convert into kinetic energy. Thus, we can write:
\[ PE_{initial} - PE_{final} = KE \]
- The kinetic energy \( KE \) of the mass at angle \( \theta \) is given by:
\[ KE = \frac{1}{2} mv^2 \]
- Therefore, we have:
\[ mgL - mgL \cos \theta = \frac{1}{2} mv^2 \]

**Hint:** Use the principle of conservation of energy to relate potential and kinetic energy.

### Step 5: Simplify the Equation
- Cancel \( m \) from both sides:
\[ gL(1 - \cos \theta) = \frac{1}{2} v^2 \]
- Rearranging gives:
\[ v^2 = 2gL(1 - \cos \theta) \]

**Hint:** Isolate \( v^2 \) to find the relationship between speed and angle.

### Step 6: Analyze Forces at Angle \( \theta \)
- When the rod is at angle \( \theta \), the forces acting on the mass \( m \) are:
- Weight \( mg \) acting downward.
- Tension \( T \) in the rod acting along the rod.
- The centripetal force required for circular motion is provided by the tension and the component of weight along the rod:
\[ T + mg \cos \theta = \frac{mv^2}{L} \]

**Hint:** Break down the forces acting on the mass into components.

### Step 7: Find Condition for Zero Tension
- Set \( T = 0 \) (the condition when tension is zero):
\[ mg \cos \theta = \frac{mv^2}{L} \]
- Substitute \( v^2 \) from the previous step:
\[ mg \cos \theta = \frac{m(2gL(1 - \cos \theta))}{L} \]
- Simplifying gives:
\[ mg \cos \theta = 2mg(1 - \cos \theta) \]

**Hint:** Setting tension to zero helps to find the angle at which this occurs.

### Step 8: Solve for \( \cos \theta \)
- Cancel \( mg \) from both sides:
\[ \cos \theta = 2(1 - \cos \theta) \]
- Rearranging gives:
\[ 3 \cos \theta = 2 \quad \Rightarrow \quad \cos \theta = \frac{2}{3} \]

**Hint:** Rearranging the equation will help you isolate \( \cos \theta \).

### Step 9: Find the Angle \( \theta \)
- To find \( \theta \), take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{2}{3}\right) \]

**Hint:** Use a calculator or trigonometric tables to find the angle.

### Step 10: Find the Speed of the Particle
- Substitute \( \cos \theta \) back into the equation for \( v^2 \):
\[ v^2 = 2gL\left(1 - \frac{2}{3}\right) = 2gL\left(\frac{1}{3}\right) = \frac{2gL}{3} \]
- Thus, the speed \( v \) is:
\[ v = \sqrt{\frac{2gL}{3}} \]

**Hint:** Make sure to simplify correctly to find the final expression for speed.

### Final Results
- The angle at which tension is zero is \( \theta = \cos^{-1}\left(\frac{2}{3}\right) \).
- The speed of the particle when the tension is zero is \( v = \sqrt{\frac{2gL}{3}} \).