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:
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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} $
BETA
