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 0 is minimum, is given by [figure]
- x = m2m1L
- x = m2Lm1 + m2
- x = m1Lm1 + m2
- x = m1m2L
Answer
(2) x = m2Lm1 + m2
Solution
Work required W = 12I 02, so W is minimum when I is minimum.
I = m1x2 + m2(L - x)2
For minimum I: dIdx = 2m1x - 2m2(L - x) = 0
m1x = m2(L - x)
∴ x = m2Lm1 + m2 (the axis passes through the centre of mass)
Answer: (2)