First slide

Pure rotational motion and its kinetic energy

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

Difficult

A

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

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

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

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

Solution

Moment of inertia of the system about the axis of rotation (through point P ) is
$I = m_{1} x^{2} + m_{2} (L - x)^{2}$
$By work energy theorem,$
$Work done to set the rod rotating with angular velocity ω_{0}$
$= Increase in rotational kinetic energy$
$W = \frac{1}{2} I ω_{0}^{2} = \frac{1}{2} [m_{1} x^{2} + m_{2} (L - x)^{2}] ω_{0}^{2}$
$For W to be minimum, \frac{d W}{d x} = 0$
$i.e. \frac{1}{2} [2 m_{1} x + 2 m_{2} (L - x) (- 1)] ω_{0}^{2} = 0$
$or m_{1} x - m_{2} (L - x) = 0 (∵ ω_{0} \neq 0)$
$or (m_{1} + m_{2}) x = m_{2} L or x = \frac{m_{2} L}{m_{1} + m_{2}}$

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