Questions

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

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a

x=m2m1L

b

x=m2Lm1+m2

c

x=m1Lm1+m2

d

x=m1m2L

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detailed solution

Correct option is B

Moment of inertia of the system about the axis of rotation (through point P ) isI=m1x2+m2(L-x)2 By work energy theorem, Work done to set the rod rotating with angular velocity ω0= Increase in rotational kinetic energy W=12Iω02=12m1x2+m2(L-x)2ω02 For W to be minimum, dWdx=0 i.e. 122m1x+2m2(L-x)(-1)ω02=0 or m1x-m2(L-x)=0 ∵ω0≠0 or m1+m2x=m2L or x=m2Lm1+m2

Similar Questions

Three point masses, each of m, are placed at the corners of an equilateral triangle of side *l*. Then the moment of inertia of this system about an axis along one side of the triangle is

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