Point masses and 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 re

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Solution

Q.
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 [2015]

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

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

3 $x = \frac{m_{1} L}{m_{1} + m_{2}}$

4 $x = \frac{m_{1}}{m_{2}} L$

Ans.
(2)

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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