A stone tied to a string is rotated a vertical circle. The minimum speed of the stone during a complete vertical circular motion.

Options

A. is independent of the mass of the stone
B. is independent of the length of the string
C. decreases with increasing mass of the stone
D. decreases with increasing length of the string

Answer

To find the minimum speed of a stone tied to a string that is rotated in a vertical circle, we need to analyze the forces acting on the stone at the uppermost point of the circle.

### Step-by-Step Solution:

1. **Identify the Forces at the Uppermost Point:**
At the uppermost point of the vertical circle, the forces acting on the stone are:
- The gravitational force (weight) acting downward, which is \( mg \).
- The tension in the string, \( T_b \), also acting downward.

2. **Centripetal Force Requirement:**
For the stone to move in a circular path, there must be a net centripetal force acting towards the center of the circle. At the uppermost point, the centripetal force is provided by the sum of the tension in the string and the weight of the stone:
\[ T_b + mg = \frac{mv_b^2}{r} \]
where \( v_b \) is the speed of the stone at the uppermost point and \( r \) is the radius of the circle (length of the string).

3. **Condition for Minimum Speed:**
The minimum speed occurs when the tension in the string is zero (i.e., \( T_b = 0 \)). This is the condition for just completing the circular motion without the string going slack. Thus, we can rewrite the equation:
\[ 0 + mg = \frac{mv_b^2}{r} \]

4. **Canceling Mass:**
Since mass \( m \) appears on both sides of the equation, we can cancel it out (assuming \( m \neq 0 \)):
\[ g = \frac{v_b^2}{r} \]

5. **Solving for Minimum Speed:**
Rearranging the equation to solve for \( v_b \):
\[ v_b^2 = gr \]
Taking the square root of both sides gives us:
\[ v_b = \sqrt{gr} \]

### Final Answer:
The minimum speed of the stone at the uppermost point during a complete vertical circular motion is:
\[ v_b = \sqrt{gr} \]