# 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 is minimum, is given by : Option 1) Option 2) Option 3) Option 4)

As we learned in concept

Moment of inertia for system of particle -

$\dpi{100} I= m_{1}r_{1}^{2}+m_{2}r_{2}^{2}+.........m_{n}r_{n}^{2}$

$\dpi{100} = \sum_{i=1}^{n}\: m_{i}r_{i}^{2}$

- wherein

Applied when masses are placed discretely.

Total M.I of rod

$I = m_{1}x^{2}+m_{2}(L-x)^{2}$

$I = m_{1}x^{2}+m_{2}L^{2}+m_{2}x^{2}-m_{2}Lx$

As I is minimum i.e.

$\frac{di}{dx} = 2m_{1}x+0+2xm_{2}-2m_{2}L=0$

=>$x(2m_{1}+2m_{2})= 2m_{2}L$

$x = \frac{m_{2}L}{m_{1}+m_{2}}$

Option 1)

this is the incorrect option

Option 2)

this is the incorrect option

Option 3)

this is the correct option

Option 4)

this is the incorrect option

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