A ball of mass m is rotated in a vertical circle with constant speed.The difference in tension at the top and botton would be

Options

A. 6mg
B. 5mg
C. 2mg
D. mg

Answer

To find the difference in tension at the top and bottom of a ball rotating in a vertical circle with constant speed, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces at the Top and Bottom:**
- At the **top** of the vertical circle, the forces acting on the ball are:
- Weight of the ball (mg) acting downwards.
- Tension in the string (T_top) acting downwards.
- The centripetal force required for circular motion is provided by the sum of tension and weight.
- At the **bottom** of the vertical circle, the forces acting on the ball are:
- Weight of the ball (mg) acting downwards.
- Tension in the string (T_bottom) acting upwards.
- The centripetal force required for circular motion is provided by the difference between tension and weight.

2. **Write the Equations for Forces:**
- At the **top** of the circle:
\[ T_{top} + mg = \frac{mv^2}{r} \]
Rearranging gives:
\[ T_{top} = \frac{mv^2}{r} - mg \]
- At the **bottom** of the circle:
\[ T_{bottom} - mg = \frac{mv^2}{r} \]
Rearranging gives:
\[ T_{bottom} = \frac{mv^2}{r} + mg \]

3. **Calculate the Difference in Tension:**
- Now, we need to find the difference between the tension at the bottom and the tension at the top:
\[ T_{bottom} - T_{top} = \left(\frac{mv^2}{r} + mg\right) - \left(\frac{mv^2}{r} - mg\right) \]
- Simplifying this expression:
\[ T_{bottom} - T_{top} = \frac{mv^2}{r} + mg - \frac{mv^2}{r} + mg = 2mg \]

4. **Conclusion:**
- The difference in tension at the bottom and top of the vertical circle is:
\[ T_{bottom} - T_{top} = 2mg \]

### Final Answer:
The difference in tension at the top and bottom of the vertical circle is \(2mg\).

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