Question 320: Centre of Mass and Rotational Motion

Question

A thin uniform bar of length $$L$$ and mass $$8$$ m lies on a smooth horizontal table. Two point masses m and 2 m are moving in the same horizontal plane from opposite sides of the bar with speeds 2$$\upsilon $$ and $$\upsilon $$ respectively.  The masses stick to the bar after collision at a distance $${L \over 3}$$ and $${L \over 6}$$ respectively from the center of the bar.  If the br starts rotating about its center of mass as a result of collision, the angular speed of the bar will be :

Accepted Answer

Centre of mass after the collision will be at O. Therefore,
$$2m{L \over 6} = m{L \over 3}$$
Now, using conservation of angular momentum, we get
$$2m{L \over 6} 	imes v + m{L \over 3} 	imes 2v = I\omega $$
Therefore,
$$2m{L \over 6} 	imes v + m{L \over 3} 	imes 2v = \left[ {8m{{{L^2}} \over {12}} + 2m{{\left( {{L \over 6}} ight)}^2} + m{{\left( {{L \over 3}} ight)}^2}} ight]\omega $$
$$ \Rightarrow \left( {{L \over 6} + {L \over 3}} ight)2mv = \left[ {8m{{{L^2}} \over {12}} + {{2m{L^2}} \over {36}} + {{m{L^2}} \over 9}} ight]\omega $$
$$ \Rightarrow {{3L} \over 6} 	imes 2mV = {5 \over 6}m{L^2}\omega $$
$$ \Rightarrow Lmv = {5 \over 6}m{L^2}\omega  \Rightarrow \omega  = {{6Lmv} \over {5m{L^2}}} \Rightarrow \omega  = {6 \over 5}{v \over L}$$

Answer

$${v \over {5L}}$$

Answer

$${6v \over {5L}}$$

Answer

$${3v \over {5L}}$$

Answer

$${v \over {6L}}$$

Centre of Mass and Rotational Motion

Choose the correct answer: