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 can follow these steps:

### Step 1: Understanding the Forces at the Highest Point
At the highest point of the circular motion, the forces acting on the stone are its weight (mg) and the tension in the string (T). For the stone to maintain circular motion at this point, the centripetal force required must be provided by the gravitational force and the tension in the string.

### Step 2: Setting Up the Equation
At the highest point, the centripetal force (mv²/L) must equal the gravitational force (mg) plus the tension (T):
\[ T + mg = \frac{mv^2}{L} \]
For the minimum speed condition, the tension T can be considered as zero (T = 0). Therefore, we have:
\[ mg = \frac{mv^2}{L} \]

### Step 3: Simplifying the Equation
We can cancel the mass (m) from both sides of the equation (assuming m ≠ 0):
\[ g = \frac{v^2}{L} \]
Rearranging gives us:
\[ v^2 = gL \]
Taking the square root, we find:
\[ v = \sqrt{gL} \]

### Step 4: Conservation of Energy
To find the minimum speed at the lowest point, we can use the conservation of mechanical energy. The total mechanical energy at the highest point (potential + kinetic) must equal the total mechanical energy at the lowest point.

1. At the highest point (height = 2L):
- Potential Energy (PE) = mg(2L)
- Kinetic Energy (KE) = \(\frac{1}{2}mv^2\)

2. At the lowest point (height = 0):
- Potential Energy (PE) = 0
- Kinetic Energy (KE) = \(\frac{1}{2}mu^2\)

Setting the energies equal:
\[ mg(2L) + \frac{1}{2}mv^2 = \frac{1}{2}mu^2 \]

### Step 5: Substituting for v
From the previous step, we know \( v = \sqrt{gL} \). Substituting this into the energy equation:
\[ mg(2L) + \frac{1}{2}m(gL) = \frac{1}{2}mu^2 \]

### Step 6: Simplifying the Energy Equation
Cancelling m from all terms:
\[ g(2L) + \frac{1}{2}gL = \frac{1}{2}u^2 \]
Combining terms:
\[ \frac{5gL}{2} = \frac{1}{2}u^2 \]
Multiplying through by 2:
\[ 5gL = u^2 \]
Taking the square root gives us the minimum speed:
\[ u = \sqrt{5gL} \]

### Final Answer
The minimum speed of the stone during a complete vertical circular motion is:
\[ u = \sqrt{5gL} \]

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