A ball of mass `M` is rotating in a conical pendulum by a string of length `L`. If radius of circular path is `L/sqrt(2)`, find the velocity of mass

Options

A. `sqrt(gL)`
B. `sqrt((gL)/2)`
C. `sqrt(2gL)`
D. `sqrt((gL)/sqrt(2))`

Answer

To find the velocity of a ball of mass \( M \) rotating in a conical pendulum with a string of length \( L \) and a radius of circular path \( \frac{L}{\sqrt{2}} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components of Forces**:
- The tension \( T \) in the string can be resolved into two components:
- Vertical component: \( T \cos \theta \)
- Horizontal component: \( T \sin \theta \)
- Here, \( \theta \) is the angle the string makes with the vertical.

2. **Determine the Angle**:
- Given that the radius of the circular path is \( \frac{L}{\sqrt{2}} \), we can find the angle \( \theta \).
- In a conical pendulum, the relationship between the radius \( r \), the length of the string \( L \), and the angle \( \theta \) is given by:
\[ r = L \sin \theta \]
- Thus, we have:
\[ \frac{L}{\sqrt{2}} = L \sin \theta \implies \sin \theta = \frac{1}{\sqrt{2}} \implies \theta = 45^\circ \]

3. **Apply Newton's Second Law**:
- In the vertical direction, the forces must balance:
\[ T \cos \theta = Mg \]
- In the horizontal direction, the net force provides the centripetal force:
\[ T \sin \theta = \frac{Mv^2}{r} \]

4. **Substitute the Known Values**:
- Substitute \( r = \frac{L}{\sqrt{2}} \) and \( \theta = 45^\circ \) into the equations:
- For the vertical direction:
\[ T \cos 45^\circ = Mg \implies T \frac{1}{\sqrt{2}} = Mg \implies T = Mg \sqrt{2} \]
- For the horizontal direction:
\[ T \sin 45^\circ = \frac{Mv^2}{\frac{L}{\sqrt{2}}} \implies T \frac{1}{\sqrt{2}} = \frac{Mv^2 \sqrt{2}}{L} \]

5. **Equate the Tension Expressions**:
- Substitute \( T = Mg \sqrt{2} \) into the horizontal equation:
\[ Mg \sqrt{2} \cdot \frac{1}{\sqrt{2}} = \frac{Mv^2 \sqrt{2}}{L} \]
- This simplifies to:
\[ Mg = \frac{Mv^2 \sqrt{2}}{L} \]

6. **Cancel Mass \( M \)**:
- Since \( M \) is non-zero, we can cancel it from both sides:
\[ g = \frac{v^2 \sqrt{2}}{L} \]

7. **Solve for Velocity \( v \)**:
- Rearranging gives:
\[ v^2 = \frac{gL}{\sqrt{2}} \implies v = \sqrt{\frac{gL}{\sqrt{2}}} \]

### Final Result:
Thus, the velocity of the mass \( v \) is:
\[ v = \sqrt{\frac{gL}{\sqrt{2}}} \]