If a stone of mass m is rotated in a vertical circular path of radius 1m, the critical velocity is

Options

A. 6.32 m/s
B. 3.13 m/s
C. 9.48 m/s
D. 12.64m/s

Answer

To find the critical velocity for a stone of mass \( m \) rotating in a vertical circular path of radius \( r = 1 \, \text{m} \), we can follow these steps:

### Step 1: Understand the Concept of Critical Velocity
Critical velocity is the minimum velocity required at the highest point of the circular path for the stone to complete the circular motion without falling. At this point, the gravitational force must provide the necessary centripetal force.

### Step 2: Identify the Forces at the Highest Point
At the highest point (let's call it point C), the forces acting on the stone are:
- The weight of the stone (\( mg \)) acting downwards.
- The tension in the string (\( T_C \)) also acting downwards.

For the stone to maintain circular motion at this point, the net force must equal the centripetal force required to keep the stone moving in a circle.

### Step 3: Write the Equation for Centripetal Force
The centripetal force required at the highest point is given by:
\[ F_{\text{centripetal}} = \frac{mv_C^2}{r} \]
where \( v_C \) is the velocity at the highest point and \( r \) is the radius of the circle.

At the highest point, the net force can be expressed as:
\[ mg + T_C = \frac{mv_C^2}{r} \]

### Step 4: Set Tension to Zero for Critical Velocity
For critical velocity, the tension \( T_C \) can be considered as zero (i.e., the stone is just about to lose contact with the string). Thus, the equation simplifies to:
\[ mg = \frac{mv_C^2}{r} \]

### Step 5: Cancel Mass \( m \) from Both Sides
Since the mass \( m \) appears on both sides of the equation, we can cancel it out:
\[ g = \frac{v_C^2}{r} \]

### Step 6: Rearrange to Find Critical Velocity
Rearranging the equation gives us:
\[ v_C^2 = g \cdot r \]
Taking the square root of both sides, we find:
\[ v_C = \sqrt{g \cdot r} \]

### Step 7: Substitute Values
Given:
- \( g \approx 10 \, \text{m/s}^2 \) (for simplicity)
- \( r = 1 \, \text{m} \)

Substituting these values:
\[ v_C = \sqrt{10 \cdot 1} = \sqrt{10} \approx 3.16 \, \text{m/s} \]

### Step 8: Final Answer
The critical velocity \( v_C \) is approximately \( 3.16 \, \text{m/s} \). Among the options provided, the closest value is \( 3.13 \, \text{m/s} \).

### Conclusion
Thus, the critical velocity for the stone is approximately \( 3.13 \, \text{m/s} \).

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