A 2 kg stone at the end of a string 1 m long is whirled in a vertical circle. At some point its speed is 4m/s. The tension of the string is 51.6 newton. At this instant the stone is :

Options

A. at the top of the circle
B. at the bottom of the circle
C. half way down
D. none of these

Answer

To solve the problem step-by-step, we need to analyze the forces acting on the stone when it is whirled in a vertical circle.

### Step 1: Identify the forces acting on the stone
When the stone is at the bottom of the vertical circle, two main forces act on it:
1. The gravitational force (weight) acting downwards: \( F_g = mg \)
2. The tension in the string acting upwards: \( T \)

### Step 2: Write down the equations of motion
At the bottom of the circle, the net force acting on the stone is the difference between the tension and the gravitational force. According to Newton's second law, this net force is also equal to the centripetal force required to keep the stone moving in a circular path:
\[ T - mg = \frac{mv^2}{R} \]
Where:
- \( T \) is the tension in the string,
- \( m \) is the mass of the stone,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( v \) is the speed of the stone,
- \( R \) is the radius of the circle (length of the string).

### Step 3: Substitute the known values
Given:
- Mass \( m = 2 \, \text{kg} \)
- Speed \( v = 4 \, \text{m/s} \)
- Radius \( R = 1 \, \text{m} \)
- Tension \( T = 51.6 \, \text{N} \)

Now, substituting these values into the equation:
\[ 51.6 - (2 \cdot 9.8) = \frac{2 \cdot (4^2)}{1} \]

### Step 4: Calculate the gravitational force
Calculate \( mg \):
\[ mg = 2 \cdot 9.8 = 19.6 \, \text{N} \]

### Step 5: Calculate the centripetal force
Calculate \( \frac{mv^2}{R} \):
\[ \frac{mv^2}{R} = \frac{2 \cdot (4^2)}{1} = \frac{2 \cdot 16}{1} = 32 \, \text{N} \]

### Step 6: Solve for tension
Now substituting back into the equation:
\[ 51.6 - 19.6 = 32 \]
\[ 32 = 32 \]

### Conclusion
The calculations confirm that the tension in the string matches the given value when the stone is at the bottom of the vertical circle. Therefore, at this instant, the stone is at the **bottom of the circle**.

### Final Answer
The stone is at the **bottom of the circle**.

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