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Question
A rod of length L is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. Let T1 and T2 be the tensions at the points L/4 and 3L/4 away from the pivoted ends.
Options
T1 > T2
T2 > T1
T1 = T2
The relation between T1 and T2 depends on whether the rod rotates clockwise or anticlockwise.
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Solution
T1 > T2
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 :
\[\text{T}_2 = \text{m}_2 \omega^2 \text{r}_2 \]
\[ = \frac{1}{4}\text{M} \omega^2 \frac{7}{8}\text{L} = \frac{7}{32}\text{M} \omega^2 \text{L}\]
∴ T1 > T2
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