A car is moving with constant speed of 10m/s on a horizontal circular path of radius `10sqrt(3m)` . A bob of mass m suspended through a light string from the roof of the car. What is the angle made by the string with the vertical if the bob is stationary with respect to the car? (g = 10 m/`s^(2)`)

Options

A. `tan^(-1) (sqrt(3))`
B. `tan^(-1) ((1)/(sqrt(2)))`
C. `30^(@)`
D. `45^(@)`

Explanation

To solve the problem, we need to find the angle made by the string with the vertical when the bob is stationary with respect to the car moving in a horizontal circular path. Here’s a step-by-step solution:

### Step 1: Identify the Forces Acting on the Bob
The bob experiences three forces:
1. The gravitational force (weight) acting downwards: \( F_g = mg \)
2. The tension in the string acting along the string: \( T \)
3. The centripetal force required for circular motion acting horizontally: \( F_c = \frac{mv^2}{r} \)

### Step 2: Draw the Free Body Diagram (FBD)
In the FBD of the bob:
- The weight \( mg \) acts downwards.
- The tension \( T \) can be resolved into two components:
- \( T \cos \theta \) acting vertically (balancing the weight)
- \( T \sin \theta \) acting horizontally (providing the centripetal force)

### Step 3: Set Up the Equations
From the FBD, we can establish the following equations:
1. **Vertical Force Balance**: \( T \cos \theta = mg \) (1)
2. **Horizontal Force Balance**: \( T \sin \theta = \frac{mv^2}{r} \) (2)

### Step 4: Divide the Equations
To eliminate \( T \), we can divide equation (2) by equation (1):
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{\frac{mv^2}{r}}{mg} \]
This simplifies to:
\[ \tan \theta = \frac{v^2}{rg} \]

### Step 5: Substitute the Given Values
Given:
- \( v = 10 \, \text{m/s} \)
- \( r = 10\sqrt{3} \, \text{m} \)
- \( g = 10 \, \text{m/s}^2 \)

Substituting these values into the equation:
\[ \tan \theta = \frac{(10)^2}{(10\sqrt{3})(10)} = \frac{100}{100\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Step 6: Find the Angle \( \theta \)
Using the property of tangent:
\[ \tan \theta = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ \]

### Conclusion
The angle made by the string with the vertical is \( \theta = 30^\circ \).

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