Question 229: Centre of Mass and Rotational Motion

Question

Consider a two particle system with particles having masses ${m_1}$ and ${m_2}$. If the first particle is pushed towards the center of mass through a distance $d,$ by what distance should the second particle is moved, so as to keep the center of mass at the same position?

Accepted Answer

Initially,
$0 = {{{m_1}\left( { - {x_1}} ight) + {m_2}{x_2}} \over {{m_1} + {m_2}}} \Rightarrow {m_1}{x_1} = {m_2}{x_2}$
Finally,
mass m
1
moved towards the center a distance d so the distance of mass m
1
from the origin is x
1
- d, and now let mass m
2
need to move d' to keep the center at the origin.
$	herefore$ $0 = {{{m_1}\left( {d - {x_1}} ight) + {m_2}\left( {{x_2} - d'} ight)} \over {{m_1} + {m_2}}}$
$ \Rightarrow 0 = {m_1}d - {m_1}{x_1} + {m_2}{x_2} - {m_2}d'$
$ \Rightarrow d' = {{{m_1}} \over {{m_2}}}d$

Answer

${{{m_2}} \over {{m_1}}}\,\,d$

Answer

${{{m_1}} \over {{m_1} + {m_2}}}d$

Answer

${{{m_1}} \over {{m_2}}}d$

Answer

$d$

Centre of Mass and Rotational Motion

Consider a two particle system with particles having masses {m_1} and {m_2}. If the first particle is pushed towards the center of mass through a distance d, by what distance should the second particle is moved, so as to keep the center of mass at the same position?

Choose the correct answer: