- Home
- Standard 11
- Physics
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 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 $\omega_0$ is minimum, is given by
$x= $$\frac{{{m_2}L}}{{{m_1} + {m_2}}}$
$x=$ $\frac{{{m_1}L}}{{{m_1} + {m_2}}}$
$x= $$\frac{{{m_1}L}}{{{m_2}}}$
$x=$$\frac{{{m_2}L}}{{{m_1}}}$
Solution
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 energy theorem,
Work done to set the rod rotating with angular
velocity ${{\omega _0}}$ = Increase in rotational kinetic energy
$W = \frac{1}{2}I\omega _0^2 = \frac{1}{2}\left[ {{m_1}{x^2} + {m_2}{{\left( {L – x} \right)}^2}} \right]\omega _0^2$
For $W$ to be mimimum,$\frac{{dW}}{{dx}} = 0$
$i.e.,\,\frac{1}{2}\left[ {2{m_1}x + 2{m_2}\left( {L – x} \right)\left( { – 1} \right)} \right]\omega _0^2 = 0$
or ${m_1}x – {m_2}\left( {L – x} \right) = 0 ( {{\omega _0} \ne 0}$
or ${{m_1} + {m_2}} x = {m_2}L\,or\,x = \frac{{{m_2}L}} {{{m_1} + {m_2}}}$
