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

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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