Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:

(i) Show  p = p’i + miV

Where pi is the momentum of the ith particle (of mass mi ) and p’i = mivi‘.  Note v’i is the velocity of the ith particle with respect to the centre of mass.

Also, verify using the definition of the centre of mass that Σp’i = 0

(ii) Prove that K = K′ + $$\dfrac{1}{2}$$MV2

Where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of the system when the particle velocities are taken relative to the centre of mass and MV2 /2 is the kinetic energy of the translation of the system as a whole.

(iii) Show L = L’+ R × MV

where L’ = ∑r’i × p’i 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 = ri – R, rest of the notation is the standard notation used in the lesson. Note L’ and  MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

(iv) Prove that :

$$\dfrac{dL’}{dt}$$= ∑ r’i x $$\dfrac{dp’}{dt}$$

Further prove that :

$$\dfrac{dL’}{dt}$$ = τ’ext

Where τ’ext is the sum of all external torques acting on the system about the centre of mass. ( Clue : Apply  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.)

Asked by Sudhanshu | 1 year ago |  146

##### Solution :-

Here $$\vec{r_{i}} = \vec{r’_{i}} + \vec{R} + R$$ . . . . . . (1)

and, $$\vec{V_{i}} = \vec{V’_{i}} + \vec{V}$$ . . . . . . . . .(2)

Where, $$\vec{r’_{i}}rand \vec{v’_{i}}$$ represent the radius vector and velocity of the ith particle referred to centre of mass O’ as the

new origin and $$\vec{V}$$ is the velocity of centre of mass with respect to O.

(i) Momentum of ith particle

$$\vec{p’} = m_{i}\vec{V’_{i}}$$

$$m_{i}(\vec{V’_{i}} + \vec{V})$$                     [ From equation (1) ]

Or, $$\vec{P} = m_{i}\vec{V} + \vec{P_{i}}$$

(ii) Kinetic energy of system of particles

K = $$\dfrac{1}{2}\sum m_{i}V^{2}$$

=$$\dfrac{1}{2}\sum m_{i}\vec{V_{i}}.\vec{V_{i}}$$

=$$\dfrac{1}{2}\sum m_{i}(\vec{V’_{i}} +\vec{V})(\vec{V’_{i}}+ \vec{V})$$

$$\dfrac{1}{2}\sum m_{i}(\vec{V’^{2}_{i}} +\vec{V^{2}} + 2\vec{V’_{i}}\vec{V})$$

$$\dfrac{1}{2}\sum m_{i}V’^{2}_{i} +\dfrac{1}{2}\sum m_{i}V^{2} +\sum m_{i}\vec{V’_{i}}\vec{V}$$

$$\dfrac{1}{2}$$ MV2 + K’

Where M = $$\sum m_{i}$$ = total mass of the system.

K’ = $$\frac{1}{2}\sum m_{i}V’^{2}_{i}$$

= kinetic energy of motion about the centre of mass.

Or, $$\dfrac{1}{2}$$ Mv2 = kinetic energy of motion of centre of mass.( Proved )

Since, $$\sum_{i}m_{i}\vec{V’_{i}}\vec{V}= \sum m_{i}\dfrac{d\vec{r_{i}}}{dt}\vec{V}$$ =0

(iii) Total angular momentum of the system of particles.

$$\vec{L} = \vec{r_{i}}\times \vec{p}$$

=$$(\vec{r_{i}} +\vec{R})\times \sum_i m_{i}(\vec{V’_{i}}+\vec{V})$$

$$\sum _i(\vec{R} \times m_{i}\vec{V}) + \sum _i(\vec{r’_{i}} \times m_{i}\vec{V’_{i}})+$$

$$(\sum _i m_{i}\vec{r’_{i}})\times \vec{V}+\vec{R\times }d ​ (∑ i ​ m i ​ r’ i ​ ​ )$$

However, we  know $$\sum_i m_{i}\vec{r’_{i}}=0$$

Since, $$\sum_i m_{i}\vec{r’_{i}}= \sum_i m_{i}(\vec{r’_{i}} – \vec{R})=M\vec{R} -M\vec{R}=0$$

According to the definition of centre of mass,

$$\sum_i (\vec{R} \times m_{i}\vec{V}) = \vec{R}\times M\vec{V}$$

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 ;$$\vec{L} = \vec{R} \times M\vec{V} + \vec{L’}$$

Answered by Sudhanshu | 1 year ago

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