A uniform rod rotating in gravity free region with certain constant angular velocity. The variation of tensile stress with distance x from axis of rotation is best represented by which of the following graphs.

Options

A.
B.
C.
D.

Answer

To solve the problem of determining the variation of tensile stress with distance \( x \) from the axis of rotation in a uniform rod rotating in a gravity-free region, we can follow these steps:

### Step 1: Understand the System
We have a uniform rod rotating about an axis with a constant angular velocity \( \omega \). We need to find how the tensile stress varies with distance \( x \) from the axis of rotation.

### Step 2: Define the Differential Element
Consider a small element of the rod at a distance \( x \) from the axis of rotation. Let the length of this element be \( dx \) and its mass be \( dm \).

### Step 3: Identify Forces Acting on the Element
The forces acting on this differential element include:
- Tension \( T \) at one end of the element.
- Tension \( T + dT \) at the other end of the element.
- Centrifugal force acting outward, given by \( F_c = dm \cdot \omega^2 \cdot x \).

### Step 4: Set Up the Force Balance Equation
For the element to be in equilibrium, the net force must be zero. Thus, we can write:
\[ T + dT - T - dm \cdot \omega^2 \cdot x = 0 \]
This simplifies to:
\[ dT = dm \cdot \omega^2 \cdot x \]

### Step 5: Express \( dm \) in Terms of \( dx \)
Since the rod is uniform, the mass per unit length \( \lambda \) is constant. Therefore, we can express \( dm \) as:
\[ dm = \lambda \cdot dx \]
where \( \lambda = \frac{m}{L} \) (mass per unit length).

### Step 6: Substitute \( dm \) into the Equation
Substituting \( dm \) into the force balance equation gives:
\[ dT = \lambda \cdot dx \cdot \omega^2 \cdot x \]

### Step 7: Integrate to Find Tension
Integrating both sides, we have:
\[ \int dT = \int \lambda \cdot \omega^2 \cdot x \, dx \]
This results in:
\[ T = \frac{\lambda \cdot \omega^2}{2} x^2 + C \]
where \( C \) is the constant of integration.

### Step 8: Determine the Constant of Integration
At the free end of the rod (where \( x = L \)), the tension \( T \) is zero. Thus:
\[ 0 = \frac{\lambda \cdot \omega^2}{2} L^2 + C \implies C = -\frac{\lambda \cdot \omega^2}{2} L^2 \]
Substituting back, we get:
\[ T = \frac{\lambda \cdot \omega^2}{2} (x^2 - L^2) \]

### Step 9: Relate Tension to Stress
The tensile stress \( \sigma \) at a distance \( x \) is given by:
\[ \sigma = \frac{T}{A} \]
where \( A \) is the cross-sectional area of the rod. Thus:
\[ \sigma = \frac{\lambda \cdot \omega^2}{2A} (L^2 - x^2) \]

### Step 10: Analyze the Resulting Equation
The equation for stress can be rewritten as:
\[ \sigma = k (L^2 - x^2) \]
where \( k = \frac{\lambda \cdot \omega^2}{2A} \) is a constant. This is a downward-opening parabola.

### Conclusion
The variation of tensile stress with distance \( x \) from the axis of rotation is best represented by a downward-opening parabola.