Suppose the particle of the previous problem has a mass m and a speed v before the collision and it sticks to the rod after the collision. The rod has a mass M:
- Find the velocity of the centre of mass C of the system constituting ''The rod plus the particle''.
- Find the velocity of the particle with respect to C before the collision.
- Find the velocity of the rod with respect to C before the collision.
- Find the angular momentum of the particle and of the rod about the centre of mass C before the collision.
- Find the moment of inertia of the system about the vertical axis through the centre of mass C after the collision.
- Find the velocity of the centre of mass C and the angular velocity of the system about the centre of mass after the collision.
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- If we take the two bodies as a system therefore total external force = 0
$\text{mV}=(\text{M}+\text{m})\text{v}'$
$\Rightarrow\text{v}'=\frac{\text{m}\text{v}}{\text{M}+\text{m}}$
- Let the velocity of the particle w.r.t. the centre of mass = V'
$\Rightarrow\text{v}'=\frac{\text{Mv}}{\text{M}+\text{m}}$
- 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.
$\Rightarrow\text{V}^{-}=\frac{\text{M}\times\text{O}=\text{m}\times\text{v}}{\text{M}+\text{m}}=\frac{-\text{mv}}{\text{M}+\text{m}}$
- The distance of the centre of mass from the particle
Therefore angular momentum of the particle before the collision
$=\text{l}\omega=\text{Mr}^2\text{cm}\omega$
$=\text{m}\Bigg\{\frac{\text{m}\big(\frac{1}{2}\big)}{(\text{M}+\text{m})}\Bigg\}^2\times\frac{\text{V}}{\frac{1}{2}}$
$=\frac{(\text{mM}^2\text{vl})}{2(\text{M}+\text{m})}$
Distance of the centre of mass from the centre of mass of the rod $=\text{R}^1_{\text{cm}}=\frac{\text{M}\times0+\text{m}\times\big(\frac{1}{2}\big)}{(\text{M}+\text{m})}=\frac{\big(\frac{\text{ml}}{2}\big)}{(\text{M}+\text{m})}$
Therefore angular momentum of the rod about the centre of mass
$=\text{MV}_{\text{cm}}\text{R}_{\text{cm}}^1$
$=\text{M}\times\Big\{\frac{(-\text{mv})}{(\text{M}+\text{m})}\Big\}\Bigg\{\frac{\big(\frac{\text{ml}}2{}\big)}{(\text{M}+\text{m})}\Bigg\}$
$=\Bigg|\frac{-\text{Mm}^2\text{lv}}{2(\text{M}+\text{m})^2}\Bigg|=\frac{\text{Mm}^2\text{lv}}{2(\text{M}+\text{m})^2}$ (If we consider the magnitude only)
- Moment of inertia of the system = M.I. due to rod + M.I. due to particle
$=\frac{\text{Ml}^2(\text{M}+\text{4m})}{12(\text{M}+\text{m})}$
- Velocity of the centre of mass $\text{V}_{\text{m}}=\frac{\text{M}\times0+\text{mV}}{(\text{M}+\text{m})}=\frac{\text{mV}}{(\text{M}+\text{m})}$
(Because External force = 0)
Angular velocity of the system about the centre of mass,
$\text{P}_{\text{cm}}=\text{l}_{\text{cm}}\omega$
$\Rightarrow\text{M}\vec{\text{V}}_{\text{M}}\times\vec{\text{r}}_{\text{m}}+\text{m}\vec{\text{v}}_{\text{m}}\times\vec{\text{r}}_{\text{m}}=\text{l}_{\text{cm}}\omega$
$\Rightarrow\text{M}\times\frac{\text{mv}}{(\text{M}+\text{m})}\times\frac{\text{ml}}{2(\text{M}+\text{m})}+\text{m}\times\frac{\text{Mv}}{(\text{M}+\text{m})}\times\frac{\text{Ml}}{2(\text{M}+\text{m})}$
$=\frac{\text{Ml}^2(\text{M}+\text{4m})}{12(\text{M}+\text{m})}\times\omega$
$\Rightarrow\frac{\text{M}\text{m}^2\text{vl}+\text{m}\text{M}^2\text{vl}}{2(\text{M}+\text{m})^2}=\frac{\text{Ml}^2(\text{M}+\text{m})}{12(\text{M}+\text{m})}\times\omega$
$\Rightarrow\frac{\frac{\text{Mm}}{(\text{M}+\text{m})}}{2(\text{M}+\text{m})^2}=\frac{\text{Ml}^2(\text{M}+\text{m})}{12(\text{M}+\text{m})}\times\omega$
$\Rightarrow\frac{6\text{mv}}{(\text{M}+\text{4m})\text{l}}=\omega$
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