Question

Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass :
(a) Show $p=p_i^{\prime }+m_{i}V$
where $p_i$ is the momentum of the $i$th particle (of mass $m_i$) and $p_i^{\prime }=m_i{v}_i^{\prime }$. Note $v_i^{\prime }$ is the velocity of the $i$th particle relative to the centre of mass.
Also, prove using the definition of the center of mass $\sum p_i^{\prime }=0$
(b) Show $K=K^{\prime }+\frac{1}{2}M{V}^2$
where $K$ is the total kinetic energy of the system of particles, $K^{\prime }$ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and $M{V}^2/2$ is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system). The result has been used in Sec. 7.14.
(c) Show $L=L^{\prime }+R\times MV$
where $L^{\prime }=r_i^{\prime }\times p_i^{\prime }$ is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember $r_i^{\prime }=r_i-R$; rest of the notation is the standard notation used in the chapter. Note $L^{\prime }$ and $MR\times V$ can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.
(d) Show $\frac{d{L}^{\prime }}{dt}=\sum r_i^{\prime }\times \frac{d{p}^{\prime }}{dt}$
Further, show that
$\frac{d{L}^{\prime }}{dt}={\tau }_{ext}^{\prime }$
where ${\tau }_{ext}^{\prime }$ is the sum of all external torques acting on the system about the centre of mass.
(Hint : Use the definition of centre of mass and Newtons Third Law. Assume the internal forces between any two particles act along the line joining the particles.)

Answer 1

(a) Let us consider a system of n particles of masses $m_1$ $m_2$ .......... $m_n$. Let their velocities wrt ground be $\overrightarrow{V_1}g\overrightarrow{V_2}g,......\overrightarrow{V_n}g$

The moment of the system of the particles is $\overrightarrow{p}=m_1\overrightarrow{V_1}g+m_2\overrightarrow{V_2}g+.......+m_n\overrightarrow{V_n}g$.

$\Rightarrow \overrightarrow{p}=m_1(\overrightarrow{V_{1c}}+\overrightarrow{V_g})+m_2(\overrightarrow{V_{2c}}+\overrightarrow{V_g})+..........$ where $\overrightarrow{V_{1c}},\overrightarrow{V_{2c}}$ are velocities if individual masses with respect to the centre of mass and $V_g$ is the velocity of the centre of mass of the system wrt ground.

After adding up all of the term we get,

$\overrightarrow{p}=p_i^{\prime }+m_i\overrightarrow{V}$

(b) The kinetic energy of n- particle of the systme

$K=\frac{1}{2}{m}_1{V}_{1g}^2+\frac{1}{2}{m}_2{V}_{2g}^2+.....$

On simplification we get

$K=K^{\prime }+\frac{1}{2}M{V}^2$ where $V=V-g$

(c) The angular momentum of the sphere about O is $L=I_0\omega$

But according to the perpendicular axes theorem

$I_0=I_C+M{R}^2$

$\therefore L=(I_C+M{R}^2)\omega$

$L=I_C\omega +M{R}^2\omega$

$L=I_C\omega +M\left(R\omega \right)R$

$L=I_C\omega +MVR$

$L={\overrightarrow{L}}^{\prime }+\overrightarrow{R}\times M\overrightarrow{V}$

(d) The angular momentum of the system wrt centre of mass

${\overrightarrow{L}}^{\prime }=\sum r_i^{\prime }\times p_i^{\prime }$

Differentiating wrt time, we get

$\frac{d{L}^{\prime }}{dt}=\sum {\overrightarrow{r_i}}^{\prime }\times \frac{d{p}^{\prime }}{dt}$

We know that $\frac{d{L}^{\prime }}{dt}=\overrightarrow{{\tau }_{ext}}$

where $\overrightarrow{{\tau }_{ext}}$ is the sum of the external torques acting on the system about the centre of the mass.