A uniform rod of length `l`, mass `m`, cross-sectional area `A` and Young's modulus `Y` is rotated in horizontal plane about a fixed vertical axis passing through one end, with a constant angular velocity `omega`. Find the total extension in the rod due to the tension produced in the rod.

Options

A. `(2m omega^(2)l^(2))/(3AY)`
B. `(m omega^(2)l^(2))/(AY)`
C. `(m omega^(2)l^(2))/(3 AY)`
D. `(5 m omega^(2) l^(2))/(48 AY)`

Answer

To solve the problem of finding the total extension in a uniform rod of length `l`, mass `m`, cross-sectional area `A`, and Young's modulus `Y` when it is rotated about a fixed vertical axis with a constant angular velocity `ω`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Mass Element**:
- Consider a small element of the rod at a distance `x` from the axis of rotation with a length `dx`.
- The mass of this small element can be expressed as:
\[ dm = \frac{m}{l} \cdot dx = \mu \cdot dx \]
where \(\mu = \frac{m}{l}\) is the mass per unit length.

2. **Calculate the Centripetal Force**:
- The centripetal force acting on this mass element due to its rotation is given by:
\[ dT = dm \cdot \omega^2 \cdot x = \mu \cdot \omega^2 \cdot x \cdot dx \]

3. **Relate Tension to the Extension**:
- The tension at a distance `x` from the axis of rotation is the sum of the contributions from all elements of the rod from `x` to `l`. Thus, we can write:
\[ T(x) = \int_x^l dT = \int_x^l \mu \cdot \omega^2 \cdot x \cdot dx \]
- Evaluating this integral gives:
\[ T(x) = \mu \cdot \omega^2 \int_x^l x \cdot dx = \mu \cdot \omega^2 \left[ \frac{x^2}{2} \right]_x^l = \mu \cdot \omega^2 \left( \frac{l^2}{2} - \frac{x^2}{2} \right) \]

4. **Calculate the Extension of Each Element**:
- The extension \(dL\) of the small element \(dx\) can be expressed using Young's modulus:
\[ Y = \frac{T}{A} \cdot \frac{dL}{dx} \implies dL = \frac{T \cdot dx}{Y \cdot A} \]
- Substituting for \(T\):
\[ dL = \frac{\mu \cdot \omega^2 \left( \frac{l^2}{2} - \frac{x^2}{2} \right) \cdot dx}{Y \cdot A} \]

5. **Integrate to Find Total Extension**:
- The total extension \(L\) of the rod is obtained by integrating \(dL\) from \(0\) to \(l\):
\[ L = \int_0^l dL = \int_0^l \frac{\mu \cdot \omega^2 \left( \frac{l^2}{2} - \frac{x^2}{2} \right)}{Y \cdot A} \cdot dx \]
- Evaluating this integral gives:
\[ L = \frac{\mu \cdot \omega^2}{2Y \cdot A} \left[ l^2x - \frac{x^3}{3} \right]_0^l = \frac{\mu \cdot \omega^2}{2Y \cdot A} \left( l^3 - \frac{l^3}{3} \right) = \frac{\mu \cdot \omega^2 l^3}{3Y \cdot A} \]

6. **Substitute for \(\mu\)**:
- Recall that \(\mu = \frac{m}{l}\):
\[ L = \frac{m \cdot \omega^2 l^2}{3Y \cdot A} \]

### Final Result:
The total extension in the rod due to the tension produced is:
\[ L = \frac{m \cdot \omega^2 \cdot l^2}{3Y \cdot A} \]