A pendulum of length l = 1m is released from `theta_(0)=60^(@)`. The rate of change of speed of the bob at `theta=30^(@)` is `(g = 10m//s^(2))`

Options

A. `5sqrt(3)m//s^(2)`
B. `5 m//s^(2)`
C. `10 m//s^(2)`
D. `2.5 m//s^(2)`

Explanation

To find the rate of change of speed of the bob at an angle of \( \theta = 30^\circ \), we need to calculate the tangential acceleration of the pendulum bob at that angle. Here’s the step-by-step solution:

### Step 1: Identify the forces acting on the bob
When the pendulum bob is at an angle \( \theta \), the forces acting on it are:
- The gravitational force \( mg \) acting downwards.
- The tension in the string, which acts along the string.

### Step 2: Break down the gravitational force
The gravitational force can be resolved into two components:
1. A component along the direction of the string: \( mg \cos \theta \)
2. A component perpendicular to the string (along the tangent to the path of motion): \( mg \sin \theta \)

### Step 3: Calculate the tangential force
At \( \theta = 30^\circ \):
- The tangential component of the gravitational force is given by:
\[ F_{\text{tangential}} = mg \sin(30^\circ) \]
- Since \( \sin(30^\circ) = \frac{1}{2} \), we can simplify this to:
\[ F_{\text{tangential}} = mg \cdot \frac{1}{2} = \frac{mg}{2} \]

### Step 4: Calculate the tangential acceleration
Using Newton's second law, the tangential acceleration \( a_t \) can be found using:
\[ F = ma \implies a_t = \frac{F_{\text{tangential}}}{m} \]
Substituting the expression for the tangential force:
\[ a_t = \frac{\frac{mg}{2}}{m} = \frac{g}{2} \]

### Step 5: Substitute the value of \( g \)
Given \( g = 10 \, \text{m/s}^2 \):
\[ a_t = \frac{10}{2} = 5 \, \text{m/s}^2 \]

### Conclusion
The rate of change of speed of the bob at \( \theta = 30^\circ \) is \( 5 \, \text{m/s}^2 \).

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