A 2kg stone is swung in a vertical circle by attaching it at the end of a string of length 2m. If the string can with stand a tension 140.6N, the maximum speed with which the stone can be rotated is

Options

A. `22ms^(-1)`
B. `44ms^(-1)`
C. `33ms^(-1)`
D. `11ms^(-1)`

Answer

To solve the problem, we need to find the maximum speed at which a 2 kg stone can be swung in a vertical circle with a string length of 2 m, given that the maximum tension the string can withstand is 140.6 N.

### Step-by-Step Solution:

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

2. **Write the Equation for Forces at the Bottom**:
At the bottom of the swing, the tension must provide the centripetal force required to keep the stone moving in a circle, in addition to balancing the weight of the stone. The equation can be written as:
\[ T = F_c + F_g \]
where \( F_c = \frac{mv^2}{r} \) is the centripetal force, and \( F_g = mg \).

3. **Substitute the Known Values**:
Given:
- Mass \( m = 2 \, \text{kg} \)
- Length of the string (radius) \( r = 2 \, \text{m} \)
- Maximum tension \( T = 140.6 \, \text{N} \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)

The equation becomes:
\[ 140.6 = \frac{2v^2}{2} + 2 \cdot 9.8 \]

4. **Simplify the Equation**:
Simplifying the equation:
\[ 140.6 = v^2 + 19.6 \]

5. **Solve for \( v^2 \)**:
Rearranging gives:
\[ v^2 = 140.6 - 19.6 \]
\[ v^2 = 121 \]

6. **Calculate \( v \)**:
Taking the square root:
\[ v = \sqrt{121} = 11 \, \text{m/s} \]

### Final Answer:
The maximum speed with which the stone can be rotated is \( 11 \, \text{m/s} \).

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