Rotational Motion Question 85

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 $ {\omega_0} $ is minimum, is given by

Options:

A) $ x=\frac{m _1L}{m _1+m _2} $

B) $ x=\frac{m _1}{m _2}L $

C) $ x=\frac{m _2}{m _1}L $

D) $ x=\frac{m _2L}{m _1+m _2} $

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Answer:

Correct Answer: D

Solution:

  • As two point masses $ m _1 $ and $ m _2 $ are placed at opposite ends of a rigid rod of length L and negligible mass .

    Total moment of inertia of the rod

    $ I=m _1x^{2}+m _2{{(L-x)}^{2}} $

    $ I=m _1x^{2}+m _2L^{2}+m _2x^{2}-2m _2Lx $

    As, $ I $ is minimum i.e.

    $ \frac{dI}{dx}=2m _1x+0+2xm _2-2m _2L=0 $

    $ \Rightarrow $ $ x(2m _1+2m _2)=2m _2L $

    $ \Rightarrow $ $ x=\frac{m _2L}{m _1+m _2} $

    When $ I $ is minimum, then work done on rotating a rod $ 1/2I{{\omega }^{2}} $ with angular velocity $ {\omega_0} $ will be minimum.

    Shortcut Way 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 $ {\omega_0} $ is their centre of mass, we have

    $ m _1x=m _2(L-x)\Rightarrow x=\frac{m _2L}{m _1+m _2} $



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