A simple pendulum of length l is pulled aside to make an angle `theta` with the vertical. Find the magnitude of the torque of the weight w of the bob about the point of suspension. When is the torque zero?

Explanation

To solve the problem, we need to find the torque of the weight of the bob about the point of suspension when the pendulum is at an angle \(\theta\) with the vertical. We will also determine when this torque is zero.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Bob:**
- The weight of the bob, \(W\), acts vertically downward at the center of the bob.

2. **Determine the Point of Suspension:**
- Let the point of suspension be denoted as point \(O\).

3. **Calculate the Perpendicular Distance:**
- The torque (\(\tau\)) about point \(O\) is given by the formula:
\[ \tau = \text{Force} \times \text{Perpendicular Distance} \]
- The perpendicular distance from the line of action of the weight \(W\) to the point \(O\) can be expressed as:
\[ R = L \sin \theta \]
- Here, \(L\) is the length of the pendulum.

4. **Express the Torque:**
- Substituting the values into the torque formula, we get:
\[ \tau = W \times (L \sin \theta) = W L \sin \theta \]

5. **Determine When the Torque is Zero:**
- For the torque to be zero, we set the equation for torque to zero:
\[ 0 = W L \sin \theta \]
- Since \(W\) and \(L\) are not zero, we can simplify this to:
\[ \sin \theta = 0 \]
- The values of \(\theta\) for which \(\sin \theta = 0\) are:
\[ \theta = 0^\circ, 180^\circ \]
- However, in the context of a simple pendulum, \(\theta = 180^\circ\) is not a practical position. Therefore, the only relevant angle is:
\[ \theta = 0^\circ \]
- This means the pendulum must be in the vertical position for the torque to be zero.

### Final Answers:
- The magnitude of the torque about the point of suspension \(O\) is:
\[ \tau = W L \sin \theta \]
- The torque is zero when:
\[ \theta = 0^\circ \]