A 2 kg stone at the end of a string 1 m long is whirled in a vertical circle at a constant speed. The speed of the stone is `4 m//sec` . The tension in the string will be 52 N , when the stone is

Options

A. At the top of the circle
B. At the bottom of the circle
C. Halfway down
D. None of the above

Answer

To solve the problem of determining the position of the stone when the tension in the string is 52 N, we can follow these steps:

### Step 1: Understand the forces acting on the stone
When the stone is whirled in a vertical circle, two main forces act on it:
1. The gravitational force (weight) acting downwards, which is given by \( W = mg \), where \( m \) is the mass of the stone and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
2. The tension in the string, which acts upwards towards the center of the circle.

### Step 2: Calculate the weight of the stone
Given:
- Mass of the stone, \( m = 2 \, \text{kg} \)
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)

The weight \( W \) of the stone can be calculated as:
\[ W = mg = 2 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 19.62 \, \text{N} \]

### Step 3: Analyze the forces at an angle
When the stone is at an angle \( \theta \) from the vertical, the forces can be resolved into components. The tension \( T \) and the weight \( W \) can be analyzed as follows:
- The vertical component of the tension must balance the weight and provide the necessary centripetal force for circular motion.

### Step 4: Write the equation for forces
At an angle \( \theta \), the vertical component of the tension \( T \) is given by:
\[ T \cos(\theta) = W + \frac{mv^2}{r} \]
Where:
- \( v = 4 \, \text{m/s} \) (speed of the stone)
- \( r = 1 \, \text{m} \) (radius of the circle)
- \( \frac{mv^2}{r} \) is the centripetal force required to keep the stone moving in a circle.

### Step 5: Calculate the centripetal force
Substituting the values into the centripetal force equation:
\[ \frac{mv^2}{r} = \frac{2 \, \text{kg} \times (4 \, \text{m/s})^2}{1 \, \text{m}} = \frac{2 \times 16}{1} = 32 \, \text{N} \]

### Step 6: Substitute values into the force equation
Now substituting \( T = 52 \, \text{N} \) and \( W = 19.62 \, \text{N} \):
\[ 52 \cos(\theta) = 19.62 + 32 \]
\[ 52 \cos(\theta) = 51.62 \]

### Step 7: Solve for \( \cos(\theta) \)
\[ \cos(\theta) = \frac{51.62}{52} \approx 0.992 \]

### Step 8: Calculate \( \theta \)
Now, we can find \( \theta \):
\[ \theta = \cos^{-1}(0.992) \approx 7.1^\circ \]

### Conclusion
The position of the stone when the tension in the string is 52 N is approximately \( 7.1^\circ \) from the vertical.

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