#### Question

A rod of length L is pivoted at one end and is rotated with a 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.

T

_{1}> T_{2}T

_{2}> T_{1}T

_{1}= T_{2}The relation between T

_{1}and T_{2}depends on whether the rod rotates clockwise or anticlockwise.

#### Solution

T_{1 }> T_{2 }

Let the angular velocity of the rod be \[\omega\] .

Distance of the centre of mass of portion of the rod on the right side of L/4 from the pivoted end :

\[r_1 = \frac{L}{4} + \frac{1}{2}\left( \frac{3L}{4} \right) = \frac{5L}{8}\]

Mass of the rod on the right side of L/4 from the pivoted end : \[\text{m}_1 = \frac{3}{4}\text{M}\]

At point L/4, we have :

\[T_1 = \text{ m}_1 \omega^2 \text{ r}_1 \]

\[ = \frac{3}{4}\text{ M } \omega^2 \frac{5}{8}\text{ L} = \frac{15}{32}\text{ M }\omega^2 \text{ L}\]

Distance of the centre of mass of rod on the right side of 3L/4 from the pivoted end :

\[\text{r}_1 = \frac{1}{2}\left( \frac{L}{4} \right) + \frac{3L}{4} = \frac{7L}{8}\]

Mass of the rod on the right side of L/4 from the pivoted end : \[\text{m}_1 = \frac{1}{4}\text{M}\]

At point 3L/4, we have :

\[\text{T}_2 = \text{m}_2 \omega^2 \text{r}_2 \]

\[ = \frac{1}{4}\text{M} \omega^2 \frac{7}{8}\text{L} = \frac{7}{32}\text{M} \omega^2 \text{L}\]

∴ T_{1} > T_{2 }