Question

Angular velocity of the rod after collision is

• A. $11/17$ $rad/s$
• B. $30/41rad/s$
• C. $19/17$ $rad/s$
• D. $30/17$ $rad/s$

Answer 1

The correct option is B $30/41rad/s$

COM is at $2.2m$ from bottom (full system).

Angular momentum about COM is conserved in internal changes.

But first lets get into COM frame.
$V_{COM}$ $=(2\times 10+1\times 10)/5=6m/s$

Velocities in COM frame:

$V\left(2kg\right)=V\left(1kg\right)=4m/s,V\left(rod\right)=-6m/s$
$L_i$ $=2\times 0.8\times 4+2\times 0.2\times 6-1\times 1.2\times 4=4$
$L_f$ $=I\omega$
$I_{COM}=\frac{M{l}^2}{12}+M{(\frac{l}{2}-x_{com})}^2+m_1(x_1-x_{com}^2)+m_2(x_2-x_{com}{)}^2$

$I_{COM}=\frac{2\left(4^2\right)}{12}+2(0.2{)}^2+2(0.8{)}^2+1(1.2{)}^2=\frac{16.4}{3}$

$4=\frac{16.4\omega }{3}$

$\omega =\frac{30}{41}rad/s$