A uniform rod of length `l` is being rotated in a horizontal plane with a constant angular speed about an axis passing through one of its ends. If the tension generated in the rod due to rotation is T(x) at a distance x from the axis. Then which of the following graphs depicts it most closely?

Options

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Answer

To solve the problem of determining the tension in a rotating uniform rod, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a uniform rod of length \( l \) rotating about one of its ends with a constant angular speed \( \omega \).
- We need to find the tension \( T(x) \) at a distance \( x \) from the axis of rotation.

2. **Consider a Small Element**:
- Take a small element of the rod at a distance \( x \) with a length \( dx \).
- The mass of this small element can be expressed as:
\[ dm = \frac{m}{l} \cdot dx \]
where \( m \) is the total mass of the rod.

3. **Identify Forces Acting on the Element**:
- The tension at distance \( x \) is \( T(x) \).
- The tension at the next segment \( x + dx \) is \( T(x) + dT \).
- The net force acting towards the center (due to rotation) on this small element is:
\[ T(x) - (T(x) + dT) = -dT \]

4. **Centripetal Force Requirement**:
- The centripetal force required for the small element to maintain its circular motion is given by:
\[ F_c = dm \cdot \omega^2 \cdot x = \left(\frac{m}{l} \cdot dx\right) \cdot \omega^2 \cdot x \]

5. **Set Up the Equation**:
- Equating the centripetal force to the change in tension gives:
\[ -dT = \frac{m}{l} \cdot dx \cdot \omega^2 \cdot x \]
- Rearranging this, we have:
\[ dT = -\frac{m \omega^2}{l} \cdot x \cdot dx \]

6. **Integrate to Find Tension**:
- Integrate both sides to find \( T(x) \):
\[ \int dT = -\frac{m \omega^2}{l} \int x \, dx \]
- The limits for \( x \) will be from \( 0 \) to \( L \) (the length of the rod), and for \( T \) from \( T(0) \) to \( T(L) \):
\[ T(x) - T(0) = -\frac{m \omega^2}{l} \left[\frac{x^2}{2}\right]_{0}^{x} \]
- This gives:
\[ T(x) - T(0) = -\frac{m \omega^2}{l} \cdot \frac{x^2}{2} \]

7. **Determine Tension at the Ends**:
- At \( x = L \), the tension \( T(L) = 0 \) (since there is no mass at the end of the rod).
- Therefore, we can find \( T(0) \):
\[ 0 - T(0) = -\frac{m \omega^2}{l} \cdot \frac{L^2}{2} \]
- Thus, \( T(0) = \frac{m \omega^2 L^2}{2l} \).

8. **Final Expression for Tension**:
- Substituting back, we find:
\[ T(x) = T(0) - \frac{m \omega^2}{2l} x^2 = \frac{m \omega^2 L^2}{2l} - \frac{m \omega^2}{2l} x^2 \]
- Simplifying gives:
\[ T(x) = \frac{m \omega^2}{2l} (L^2 - x^2) \]

9. **Graphical Representation**:
- The expression \( T(x) = \frac{m \omega^2}{2l} (L^2 - x^2) \) indicates that the tension decreases quadratically from \( T(0) \) to \( 0 \) at \( x = L \).
- Therefore, the graph of \( T(x) \) versus \( x \) will be a downward-opening parabola, starting from a maximum value at \( x = 0 \) and reaching zero at \( x = L \).