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Point masses m(1) and m(2) are placed at...

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 perpendicualr 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 :

A

`x= (m_(1))/(m_(2))L`

B

`x= (m_(2))/(m_(1))L`

C

`x= (m_(2)L)/(m_(1) + m_(2))`

D

`x= (m_(1)L)/(m_(1) + m_(2))`

Text Solution

Verified by Experts

The correct Answer is:
C

`K.E.= (1)/(2) I omega_(0)^(2)`
I is min. about the centre of mass
So, `(m_(1)) (x)= (m_(2)) (L-x) rArr x= (m_(2) L)/(m_(1) + m_(2))`
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