Question 647: Centre of Mass and Rotational Motion

Question

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
![](https://neetcare-image.cdn.examgoal.net/7k9eik78u7crr/652c36a4-6324-42db-9c9c-1a65dcb2c39a/bb919770-5b9f-11ea-a4e3-ff1fddf1e89e/file-7k9eik78u7crs-341w.jpg)

Accepted Answer

$K = {1 \over 2}\left( {{m_1}{x^2}} \right)\omega _0^2 + {1 \over 2}{m_2}{\left( {L - x} \right)^2}\omega _0^2$
$dK/dx = 0$
${m_1}x = {m_2}\left( {L - x} \right)$
${m_1}x = {m_2}L - {m_2}x$
$ \Rightarrow x = {{{m_2}L} \over {{m_1} + {m_2}}}$

Answer

$x = {{{m_2}} \over {{m_1}}}L$

Answer

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

Answer

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

Answer

$x = {{{m_1}} \over {{m_2}}}L$

Centre of Mass and Rotational Motion

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