A bob of mass 1 kg is suspended from an inextensible string of length 1 m. When the string makes an angle `60^(@)` with vertical, speed of the bob is 4 m/s
Net acceleration of the bob at this instant is

Options

A. `16 m//s^(2)`
B. `20.4 m//s^(2)`
C. `18.2 m//s^(2)`
D. `10.4 m//s^(2)`

Answer

To solve the problem step by step, we will calculate the net acceleration of the bob suspended from the string when it makes an angle of 60 degrees with the vertical.

### Step 1: Identify the given values
- Mass of the bob, \( m = 1 \, \text{kg} \)
- Length of the string (which acts as the radius), \( R = 1 \, \text{m} \)
- Speed of the bob, \( V = 4 \, \text{m/s} \)
- Angle with the vertical, \( \theta = 60^\circ \)

### Step 2: Calculate the centripetal acceleration
Centripetal acceleration (\( a_c \)) is given by the formula:
\[ a_c = \frac{V^2}{R} \]
Substituting the values:
\[ a_c = \frac{(4 \, \text{m/s})^2}{1 \, \text{m}} = \frac{16}{1} = 16 \, \text{m/s}^2 \]

### Step 3: Determine the tangential acceleration
To find the tangential acceleration (\( a_t \)), we need to calculate the component of the gravitational force acting along the direction of motion. The gravitational force acting on the bob is \( mg \), where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

The tangential component of the gravitational force can be calculated as:
\[ F_t = mg \sin(\theta) \]
Substituting the values:
\[ F_t = 1 \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot \sin(60^\circ) \]
Using \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ F_t = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{N} \]

Now, the tangential acceleration is given by:
\[ a_t = \frac{F_t}{m} = \frac{5\sqrt{3}}{1} = 5\sqrt{3} \, \text{m/s}^2 \]

### Step 4: Calculate the net acceleration
The net acceleration (\( a \)) is the vector sum of the centripetal acceleration and the tangential acceleration. It can be calculated using the Pythagorean theorem:
\[ a = \sqrt{a_c^2 + a_t^2} \]
Substituting the values:
\[ a = \sqrt{(16 \, \text{m/s}^2)^2 + (5\sqrt{3} \, \text{m/s}^2)^2} \]
Calculating each term:
\[ = \sqrt{256 + (5\sqrt{3})^2} \]
\[ = \sqrt{256 + 75} = \sqrt{331} \]
Calculating the square root:
\[ \approx 18.2 \, \text{m/s}^2 \]

### Final Answer
The net acceleration of the bob at the instant when the string makes an angle of \( 60^\circ \) with the vertical is approximately \( 18.2 \, \text{m/s}^2 \).

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