K.E.=21Iω2
I is min. about the centre of mass
So, (m1)(x)=(m2)(L−x)
x=m1+m2m2L
Point masses m1 and m2 are placed at the opposite ends of 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 L 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:

Held on 30 Apr 2015 · Verified 9 Jul 2026.
x=m2m1L
x=m1m2L
x=m1+m2m2L
x=m!+m2m1L
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