Login/Sign Up
Home » Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:(i) Show where is the angular momentum of the system about the centre of mass with velocities considered with respect to the centre of mass. Note , rest of the notation is the standard notation used in the lesson. Note L’ and MR can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.(ii) Prove that : Further prove that: Where t’ext is the sum of all external torques acting on the system about the centre of mass. (Clue : A pply Newton’s Third Law and the definition of centre of mass. Consider that internal forces between any two particles act along the line connecting the particles.)
Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:
(i) Show L=L^{\prime}+R \times M V where L^{\prime}=\Sigma r^{\prime}{ }_{i} \times p_{i}^{\prime} is the angular momentum of the system about the centre of mass with velocities considered with respect to the centre of mass. Note r_{i}=r_{i}-R, rest of the notation is the standard notation used in the lesson. Note L’ and MR \times V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.
(ii) Prove that : \mathrm{dL}^{\prime} / \mathrm{dt}=\sum \mathrm{r}_{\mathrm{i}}^{\prime} \mathrm{x} \mathrm{dp}^{\prime} / \mathrm{dt} Further prove that: \mathrm{dL}^{\prime} / \mathrm{dt}=\mathrm{T}^{\prime} \mathrm{ext} Where t’ext is the sum of all external torques acting on the system about the centre of mass. (Clue : A pply Newton’s Third Law and the definition of centre of mass. Consider that internal forces between any two particles act along the line connecting the particles.)
CBSE | Class 11 | NCERT | Physics | System of Particles and Rotational Motion
Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:
(i) Show L=L^{\prime}+R \times M V where L^{\prime}=\Sigma r^{\prime}{ }_{i} \times p_{i}^{\prime} is the angular momentum of the system about the centre of mass with velocities considered with respect to the centre of mass. Note r_{i}=r_{i}-R, rest of the notation is the standard notation used in the lesson. Note L’ and MR \times V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.
(ii) Prove that : \mathrm{dL}^{\prime} / \mathrm{dt}=\sum \mathrm{r}_{\mathrm{i}}^{\prime} \mathrm{x} \mathrm{dp}^{\prime} / \mathrm{dt} Further prove that: \mathrm{dL}^{\prime} / \mathrm{dt}=\mathrm{T}^{\prime} \mathrm{ext} Where t’ext is the sum of all external torques acting on the system about the centre of mass. (Clue : A pply Newton’s Third Law and the definition of centre of mass. Consider that internal forces between any two particles act along the line connecting the particles.)

Here \vec{r}_{i}=\vec{r}_{i}+\vec{R}+R \ldots

(1)

also, \vec{V}_{i}=\vec{V}_{i}+\vec{V} \ldots \ldots .

(2)

Where \vec{r}_{i}^{\overrightarrow{3}} and \vec{v}_{i}^{\overrightarrow{3}} denote the radius vector and velocity of the \mathrm{i}^{\text {th }} particle refers to the new origin as the centre of mass \mathrm{O}‘, and \vec{V} is the velocity of the centre of mass with respect to O.

 

(i) Total angular momentum of the system of particles is calculated below:

\vec{L}=\vec{r}_{i} \times \vec{p}

=\left(\vec{r}_{i}+\vec{R}\right) \times \sum_{i} m_{i}\left(\vec{V}_{i}+\vec{V}\right)

=\sum_{i}\left(\vec{R} \times m_{i} \vec{V}\right)+\sum_{i}\left(\vec{r}_{i} \times m_{i} \vec{V}_{i}\right)+\left(\sum_{i} m_{i} r_{i} \vec{i}\right) \times \vec{V}+\overrightarrow{R \times} \sum_{i} m_{i} \vec{V}_{i}

=\sum_{i}\left(\vec{R} \times m_{i} \vec{V}\right)+\sum_{i}\left(r_{i}^{\vec{H}} \times m_{i} \vec{V}_{i}\right)+\left(\sum_{i} m_{i} r_{i} \vec{i}\right) \times \vec{V}+\vec{R} \times \frac{d}{d t}\left(\sum_{i} m_{i} \vec{r}_{i}\right)

\text { However, we know } \sum_{i} m_{i} \vec{r}_{i}=0

\text { Since, } \sum_{i} m_{i} \vec{r}_{i}^{7}=\sum_{i} m_{i}\left(\vec{r}_{i}-\vec{R}\right)=M \vec{R}-M \vec{R}=0

According to the definition of centre of mass, we can write,

\sum_{i}\left(\vec{R} \times m_{i} \vec{V}\right)=\vec{R} \times M \vec{V}

Such that, \vec{L}=\vec{R} \times M \vec{V}+\sum_{i} \vec{r}_{i} \times \vec{P}_{i}

Given, \vec{L}=\sum \vec{r}_{i} \times \vec{p}_{i}

Thus, we have ; \vec{L}=\vec{R} \times M \vec{V}+\vec{L}

(ii) From previous solution, we have

\vec{L}^{\prime}=\sum \overrightarrow{r_{i}} \times \vec{P}_{i} \frac{d \vec{D}^{\prime}}{d t}=\sum \overrightarrow{r_{i}} \times \frac{d \vec{P}_{i}}{d t}+\sum \frac{d r^{7}}{d t} \times \vec{P}_{i}

=\sum \overrightarrow{r_{i}}{\vec{T}} \times \frac{d \vec{P}_{i}}{d t}

=\sum \vec{r}_{i} \times \overrightarrow{F_{i}^{e x t}}=T_{e e x t}^{\vec{t}}

Total torque can be calculated as,

=\tau_{\epsilon e x t}^{\overrightarrow{e x t}}=\sum \overrightarrow{r_{i}}{\vec{r}} \times \overrightarrow{F_{i}^{\overrightarrow{e x t}}}

=\sum\left(\vec{r}_{i}+\vec{R}\right) \times F_{i}^{\overrightarrow{e x t}}

=\tau_{e x t}^{\vec{\prime}}+T_{0}^{\overrightarrow{(e x t)}}

Where, \tau_{e x t}^{\overrightarrow{-x} t} is the net torque about the centre of mass as origin and \tau_{0}^{\vec{e} x t} is about the origin \mathrm{O}.

\tau_{e x t}=\sum \vec{r}_{i} \times F_{i}^{\overrightarrow{e x t}}

=\sum r^{\vec{r}}_{i} \times \frac{d \vec{P}_{i}}{d t}

=\frac{d}{d t} \sum\left(r_{i}{ }^{\vec{H}} \times \vec{P}_{i}\right)=\frac{d \vec{D}}{d t}

As a result, we have, \frac{d \vec{L}^{\prime}}{d t}=\tau_{\text {ext }}^{\vec{t}}

i

More Material