Point masses m _ 1 and m _ 2 are placed at the opposite ends of...

Point masses ${m}_{1}$ and ${m}_{2}$ are placed at the opposite ends of rigid rod of length $\text{L}$ , and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point $\text{L}$ 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:
Point masses m _ 1 and m _ 2 are placed at the opposite ends of rigid rod of length L , and negligible mass. The rod is

Options

A $x = \dfrac{{m}_{1}}{{m}_{2}}L$
B $x = \dfrac{{m}_{2}}{{m}_{1}}L$
C $x = \dfrac{{m}_{2}L}{{m}_{1} + {m}_{2}}$
D $x = \dfrac{{m}_{1}L}{{m}_{!} + {m}_{2}}$

Correct answer

C. $x = \dfrac{{m}_{2}L}{{m}_{1} + {m}_{2}}$

Step-by-step solution

K . E . = \dfrac{1}{2}I{\omega}^{2}

I is min. about the centre of mass

So, \left({m}_{1}\right)

\left(x\right) = \left({m}_{2}\right)\left(L - x\right)

x = \dfrac{{m}_{2}L}{{m}_{1} + {m}_{2}}

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