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.

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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 = L/4 + $\frac{1}{2}$\left( 3L/4 \right) = 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( L/4 \right) + 3L/4 = 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} $\therefore$ T1 > T2 [Solution diagram: 66637_solution_1.png]

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