Point masses m_1 and m_2 are placed | vedclass

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Question

Moment of inertia of the system about the axis of rotation (through point $P$) is

$I = {m_1}{x^2} + {m_2}{\left( {L - x} \right)^2}$

By work ene

Vedclass · Accepted Answer

$x= $$\frac{{{m_2}L}}{{{m_1} + {m_2}}}$

Vedclass · Answer

Moment of inertia of the system about the axis of rotation (through point $P$) is

$I = {m_1}{x^2} + {m_2}{\left( {L - x} ight)^2}$

By work energy theorem,

Work done to set the rod rotating with angular

velocity ${{\omega _0}}$ = Increase in rotational kinetic energy 

$W = \frac{1}{2}I\omega _0^2 = \frac{1}{2}\left[ {{m_1}{x^2} + {m_2}{{\left( {L - x} ight)}^2}} ight]\omega _0^2$

For $W$ to be mimimum,$\frac{{dW}}{{dx}} = 0$

$i.e.,\,\frac{1}{2}\left[ {2{m_1}x + 2{m_2}\left( {L - x} ight)\left( { - 1} ight)} ight]\omega _0^2 = 0$
or ${m_1}x - {m_2}\left( {L - x} ight) = 0 ( {{\omega _0} 
e 0}$
or ${{m_1} + {m_2}} x = {m_2}L\,or\,x = \frac{{{m_2}L}} {{{m_1} + {m_2}}}$

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