A pendulum consists of a small mass m at the end of a string of length L. The pendulum is pulled aside making an angle `theta` with the vertical and released. Taking the suspension point as the axis, at the instant of release

Options

A. torque on the pendulum is mgL `cos theta`
B. torque on the pendulum is mgL `sin theta`
C. angular acceleration of pendulum is `( g cos theta)/( L)`
D. angular acceleration of pendulum is `( g sin theta)/( L)`

Answer

To solve the problem regarding the pendulum, we need to find the torque and angular acceleration at the moment of release when the pendulum is displaced to an angle \( \theta \) from the vertical.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a pendulum consisting of a mass \( m \) attached to a string of length \( L \).
- The pendulum is pulled aside to an angle \( \theta \) and then released.

2. **Identifying Forces**:
- The weight of the mass \( m \) acts downwards and can be resolved into two components:
- \( mg \cos \theta \) (perpendicular to the string)
- \( mg \sin \theta \) (parallel to the direction of motion)

3. **Calculating Torque**:
- The torque \( \tau \) about the pivot point (suspension point) is given by the formula:
\[ \tau = r \times F \]
- Here, \( r \) is the length of the string \( L \) and \( F \) is the force causing the torque, which is \( mg \sin \theta \).
- Since the force is acting perpendicular to the line of action, the torque can be calculated as:
\[ \tau = L \cdot (mg \sin \theta) = mgL \sin \theta \]

4. **Finding Angular Acceleration**:
- The relationship between torque and angular acceleration \( \alpha \) is given by:
\[ \tau = I \alpha \]
- The moment of inertia \( I \) for a point mass at a distance \( L \) from the pivot is:
\[ I = mL^2 \]
- Substituting the values of torque and moment of inertia into the equation gives:
\[ mgL \sin \theta = mL^2 \alpha \]
- Dividing both sides by \( mL^2 \) (and canceling \( m \) and \( L \)):
\[ \alpha = \frac{g \sin \theta}{L} \]

### Final Answers:
- The torque on the pendulum at the moment of release is:
\[ \tau = mgL \sin \theta \]
- The angular acceleration of the pendulum at that moment is:
\[ \alpha = \frac{g \sin \theta}{L} \]