First slide

Moment of intertia

Question

Point masses $m_{1} and m_{2}$ are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity $ω_{0}$ is minimum is given by

Moderate

A

$x = \frac{m_{2} L}{m_{1} + m_{2}}$
B

$x = \frac{m_{1} L}{m_{1} + m_{2}}$
C

$x = \frac{m_{1}}{m_{2}} L$
D

$x = \frac{m_{2}}{m_{1}} L$

Solution

The position of point P on rod through which the axis should pass so that the work required to set the rod rotating with minimum angular velocity $ω_{0}$ is their centre of mass
so $m_{1} x = m_{2} (L - x) \Rightarrow x = \frac{m_{2} L}{m_{1} + m_{2}}$

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