Question

A rod of length L is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. The value of x for point P, where tension is $\frac{1}{4}$ of the tension at O.

Answer 1

Tension at $O$:

$T_o=\frac{M}{L}{\omega }^2\times L=M{\omega }^2$

where, $M$ is mass per unit length, $L$ is length of the rod and $\omega$ its angular velocity.

Tension at $P$:

$T_P=\frac{M}{L}{\omega }^{2}x=\frac{1}{4}M{\omega }^2=\frac{1}{4}{T}_o$

$⟹x=\frac{1}{4}$