A stone of mas 1 kg is tied to one end of a string of length 0.5 m. It is whirled in a vertical circle. If the maximum tension in the string is 58.8 N, the velocity at the top is

Options

A. `1.82 ms^(-1)`
B. `2.2 ms^(-1)`
C. `3.26 ms^(-1)`
D. `2.87 ms^(-1)`

Answer

To solve the problem, we need to find the velocity of the stone at the top of the vertical circle. We will use the concepts of tension, centripetal force, and conservation of energy.

### Step-by-Step Solution:

1. **Identify the Forces at the Bottom of the Circle**:
At the bottom of the vertical circle, the forces acting on the stone are the tension in the string (T) and the weight of the stone (mg). The centripetal force required to keep the stone moving in a circle is provided by the net force acting on it. The equation for the maximum tension at the bottom can be given by:
\[ T_{\text{max}} = m \frac{v^2}{r} + mg \]
where:
- \( T_{\text{max}} = 58.8 \, \text{N} \) (maximum tension)
- \( m = 1 \, \text{kg} \) (mass of the stone)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( r = 0.5 \, \text{m} \) (length of the string)

2. **Substituting Values**:
Substitute the known values into the equation:
\[ 58.8 = 1 \cdot \frac{v^2}{0.5} + 1 \cdot 9.8 \]
Simplifying this gives:
\[ 58.8 = \frac{v^2}{0.5} + 9.8 \]

3. **Rearranging the Equation**:
Rearranging the equation to isolate \( v^2 \):
\[ 58.8 - 9.8 = \frac{v^2}{0.5} \]
\[ 49 = \frac{v^2}{0.5} \]

4. **Solving for \( v^2 \)**:
Multiply both sides by 0.5:
\[ v^2 = 49 \times 0.5 = 24.5 \]

5. **Finding Velocity at the Bottom**:
Taking the square root to find \( v \):
\[ v = \sqrt{24.5} \approx 4.95 \, \text{m/s} \]

6. **Using Conservation of Energy**:
Now, we apply the conservation of energy to find the velocity at the top of the circle. The total mechanical energy at the bottom equals the total mechanical energy at the top:
\[ \text{Potential Energy at Bottom} + \text{Kinetic Energy at Bottom} = \text{Potential Energy at Top} + \text{Kinetic Energy at Top} \]
\[ 0 + \frac{1}{2} m v^2 = mgh + \frac{1}{2} m v_{\text{top}}^2 \]
where \( h = 2r = 1 \, \text{m} \).

7. **Substituting Values**:
Substitute the known values:
\[ \frac{1}{2} \cdot 1 \cdot 24.5 = 1 \cdot 9.8 \cdot 1 + \frac{1}{2} \cdot 1 \cdot v_{\text{top}}^2 \]
\[ 12.25 = 9.8 + \frac{1}{2} v_{\text{top}}^2 \]

8. **Rearranging to Solve for \( v_{\text{top}}^2 \)**:
Rearranging gives:
\[ 12.25 - 9.8 = \frac{1}{2} v_{\text{top}}^2 \]
\[ 2.45 = \frac{1}{2} v_{\text{top}}^2 \]
\[ v_{\text{top}}^2 = 2.45 \times 2 = 4.9 \]

9. **Finding Velocity at the Top**:
Taking the square root to find \( v_{\text{top}} \):
\[ v_{\text{top}} = \sqrt{4.9} \approx 2.21 \, \text{m/s} \]

### Final Answer:
The velocity of the stone at the top of the vertical circle is approximately **2.21 m/s**.