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. Then:

• A. $T_1>T_2$
• B. $T_2>T_1$
• C. $T_1=T_2$
• D. the relation between $T_1$ and $T_2$ depends on whether the rod rotates clockwise or anticlockwise

Answer 1

The correct option is A $T_1>T_2$

At $L/4$ from the pivoted end:
Tension, $T_1=$ centrifugal force on the rod behind this point (of $3L/4$ length)

$\Rightarrow$ $T_1=m^{\prime }{\omega }^{2}r=\frac{M}{L}\frac{3}{4}L{\omega }^2(\frac{L}{4}+\frac{3L}{8})$ (because $r$ is the distance of COM)

$\Rightarrow$ $T_1=\frac{15}{32}M{\omega }^{2}L$
At $3L/4$ from the pivot end:
Tension, $T_2=$ centrifugal force on the rod behind this point (of $L/4$ length)

$\Rightarrow$ $T_2=m^{\prime }{\omega }^{2}r=\frac{M}{L}\frac{1}{4}L{\omega }^2(\frac{3L}{4}+\frac{L}{8})$ (because $r$ is distance of COM)

$\Rightarrow$ $T_2=\frac{7}{32}M{\omega }^{2}L$

$\therefore$ $T_1>T_2$