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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:

x=m1m2L

x=m1L/(m1+m2)

x=m2m1L

x=m2L/(m1+m2)

Solution:

MI of m1 about the axis: I1=m1x^2
MI of m2 about the axis: I2=m2(L−x)^2
KE is rotational.Total KE is E=1/2I1ω0^2+1/2I2ω0^2=1/2ω0^2(m1x^2+m2(L−x)^2)
Work done is change in KE.
To minimize E, differentiate wrt x and equate to zero.
m1x−m2(L−x)=0
⇒x=m2L/(m1+m2)