Question

Suppose the particle of the previous problem has a mass $m$ and a speed $\nu$ before the collision and it sticks to the rod after the collision. The rod has a mass $M$.
(a) Find the velocity of the centre of mass $C$ of the system constituting "the rod plus the particle".?
(b) Find the velocity of the particle with respect to $C$ before the collision.
(c) Find the velocity of the rod with respect to $C$ before the collision.
(d) Find the angular momentum of the particle and of the rod about the centre of mass $C$ before the collision.
(e) Find the moment of inertia of the system about the vertical axis through the centre of mass $C$ after the collision.
(f) Find the velocity of the centre of mass $C$ and the angular velocity of the system about the centre of mass after the collision.

Answer 1

a) If we take the two bodies as a system therefore total external force $=0$Applying L.C.L.M:$mV=(M+m)v^{\prime }$$\Rightarrow v^{\prime }=\frac{mv}{M+m}$
b) Let the velocity of the particle w.r.t the centre of mass $=V^{\prime }$$\Rightarrow v^{\prime }=\frac{m\times 0+Mv}{M+m}\Rightarrow v^{\prime }=\frac{Mv}{M+m}$
c) If the body moves towards the rod with a velocity of $v$, i.e. the rod is moving with a velocity - $v$ towards the particle.Therefore the velocity of the rod w.r.t the centre of mass $=V^{-}$$\Rightarrow V^{-}=\frac{M\times O=m\times v}{M+m}=\frac{-mv}{M+m}$
d) The distance of the centre of mass from the particle$=\frac{M\times I/2+m\times O}{M+m}=\frac{M\times I/2}{(M+m)}$Therefore angular momentum of the particle before the collision$=I\omega =M{r}^{2}cm\omega$${=m\left\{\left(mI/2\right)/(M+m)\right\}}^2\times V/\left(I/2\right)$$=(m{M}^{2}vI/2(M+m))$Distance fo the centre of mass from the centre of mass of the rod $=$$=R_{cm}^1=\frac{M\times 0+m\times \left(I/2\right)}{(M+m)}=\frac{\left(mI/2\right)}{(M+m)}$Therefore angular momentum of the rod about the centre of mass$=M{V}_{cm}{R}_1^{cm}$$=\left|\frac{-M{m}^{2}Iv}{2{(M+m)}^2}\right|=\left|\frac{M{m}^{2}Iv}{2{(M+m)}^2}\right|$ (If we consider magniture only)
e) Moment of inertia of the system $=M.Iduetorod+M.Iduetoparticle$$=\frac{M{I}^2}{12}+\frac{M{\left(mI/2\right)}^2}{{(M+m)}^2}+\frac{m{\left(MI/s\right)}^2}{{(M+m)}^2}$$=\frac{M{I}^2(M+4m)}{12(M+m)}$
f) Velocity of the centre of mass $V_m=\frac{M\times 0+mV}{(M+m)}=\frac{mV}{(M+m)}$(Velocity of centre of mass of the system before the collision = Velocity of cenre of mass of the system after collision)(Because external force $=0$)Angular velocity of the system about the centre of mass,$P_{cm}=I_{cm}\omega$$\Rightarrow M\overrightarrow{V_M}\times \overrightarrow{r_M}+m\overrightarrow{v_M}\times \overrightarrow{r_M}=I_{cm}\omega$$\Rightarrow M\times \frac{mv}{(M+m)}\times \frac{mI}{2(M+m)}+m\times \frac{Mv}{(M+m)}\times \frac{MI}{2(M+m)}=\frac{M{I}^{2}2(M+4m)}{12(M+m)}\times \omega$$\Rightarrow \frac{M{m}^{2}vI+m{M}^{2}vI}{2{(M+m)}^2}=\frac{M{I}^2(M+4m)}{12(M+m)}\times \omega$$\Rightarrow \frac{Mm/(M+m)}{{(M+m)}^2}=\frac{M{I}^2(M+m)}{12(M+m)}\times \omega$$\Rightarrow \frac{6mv}{(M+4m)}=\omega$