A bob hangs from a rigid support by an inextensible string of length l. It is released from rest when string makes an agngle `60^(@)` with vertical . The speed of the bob at the lowest position is

Options

A. `sqrt( gl)`
B. `sqrt( 3gl)`
C. `sqrt( 2gl)`
D. `sqrt( 5gl)`

Answer

To solve the problem of finding the speed of the bob at the lowest position when it is released from an angle of 60 degrees with the vertical, we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Identify the initial and final positions
- The bob is initially at point A, making an angle of 60 degrees with the vertical.
- The lowest position is point B, where the bob is directly below the point of suspension.

### Step 2: Determine the height difference
- The length of the string is \( L \).
- At point A, the height \( h_A \) can be calculated using trigonometry. The vertical component of the string length when the bob is at 60 degrees is given by:
\[ OC = L \cos(60^\circ) = L \cdot \frac{1}{2} = \frac{L}{2} \]
- Therefore, the height \( h_A \) from the lowest point B to point A is:
\[ h_A = L - OC = L - \frac{L}{2} = \frac{L}{2} \]

### Step 3: Apply the conservation of energy
- The total mechanical energy at point A (initial position) is equal to the total mechanical energy at point B (lowest position).
- At point A, the kinetic energy \( KE_A = 0 \) (since it is released from rest) and the potential energy \( PE_A = mgh_A \).
- At point B, the potential energy \( PE_B = 0 \) (as it is at the lowest point) and the kinetic energy \( KE_B = \frac{1}{2} m v_B^2 \).

### Step 4: Set up the energy conservation equation
\[ KE_A + PE_A = KE_B + PE_B \]
Substituting the values:
\[ 0 + mg\left(\frac{L}{2}\right) = \frac{1}{2} mv_B^2 + 0 \]

### Step 5: Simplify the equation
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g\left(\frac{L}{2}\right) = \frac{1}{2} v_B^2 \]
- Rearranging gives:
\[ v_B^2 = gL \]

### Step 6: Solve for \( v_B \)
Taking the square root:
\[ v_B = \sqrt{gL} \]

### Conclusion
The speed of the bob at the lowest position is:
\[ v_B = \sqrt{gL} \]