A rod of length L is pivoted at one end and is rotated with as uniform angular velocity in a horizontal plane. Let `T_1 and T_2` be the tensions at the points L/4 and 3L/4 away from the pivoted ends.

Options

A. `T_1gtT_2`
B. `T_2gtT_1`
C. `T_1=T_2`
D. `The relation between `T_1 and T_2` depends on whether the rod rotates clockwise or anticlockwise.

Answer

To solve the problem of comparing the tensions \( T_1 \) and \( T_2 \) at points \( \frac{L}{4} \) and \( \frac{3L}{4} \) respectively on a rotating rod, we can follow these steps:

### Step 1: Understand the System
The rod of length \( L \) is pivoted at one end and rotates with a uniform angular velocity \( \omega \). Each point on the rod experiences a centripetal acceleration due to this rotation.

### Step 2: Define the Tensions
Let \( T_1 \) be the tension at the point \( \frac{L}{4} \) from the pivot, and \( T_2 \) be the tension at the point \( \frac{3L}{4} \) from the pivot.

### Step 3: Consider a Small Element
Take a small element \( dx \) of the rod at a distance \( x \) from the pivot. The mass of this small element can be expressed as \( dm = \frac{M}{L} dx \), where \( M \) is the total mass of the rod.

### Step 4: Apply Newton's Second Law
For the small element \( dx \) at distance \( x \), the net force acting on it in the radial direction (towards the pivot) is given by:
\[ T_1 - T_2 = dm \cdot \omega^2 \cdot x \]
Substituting \( dm \):
\[ T_1 - T_2 = \left(\frac{M}{L} dx\right) \cdot \omega^2 \cdot x \]

### Step 5: Analyze the Relationship
From the equation \( T_1 - T_2 = \frac{M \omega^2}{L} x \, dx \), we can see that as \( x \) increases, \( T_1 \) must be greater than \( T_2 \) because the term \( \frac{M \omega^2}{L} x \, dx \) is positive. This indicates that tension decreases as we move away from the pivot.

### Step 6: Compare Tensions at Specific Points
Now, we need to compare \( T_1 \) (at \( \frac{L}{4} \)) and \( T_2 \) (at \( \frac{3L}{4} \)):
- Since \( \frac{L}{4} < \frac{3L}{4} \), we conclude that \( T_1 > T_2 \).

### Conclusion
Thus, the tension at \( \frac{L}{4} \) (i.e., \( T_1 \)) is greater than the tension at \( \frac{3L}{4} \) (i.e., \( T_2 \)):
\[ T_1 > T_2 \]

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