Rotational Motion Question 85

Question: Point masses m1 and m2 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 ω0 is minimum, is given by

Options:

A) x=m1Lm1+m2

B) x=m1m2L

C) x=m2m1L

D) x=m2Lm1+m2

Show Answer

Answer:

Correct Answer: D

Solution:

  • As two point masses m1 and m2 are placed at opposite ends of a rigid rod of length L and negligible mass .

    Total moment of inertia of the rod

    I=m1x2+m2(Lx)2

    I=m1x2+m2L2+m2x22m2Lx

    As, I is minimum i.e.

    dIdx=2m1x+0+2xm22m2L=0

    x(2m1+2m2)=2m2L

    x=m2Lm1+m2

    When I is minimum, then work done on rotating a rod 1/2Iω2 with angular velocity ω0 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 ω0 is their centre of mass, we have

    m1x=m2(Lx)x=m2Lm1+m2



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