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Q7.33(c):
Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass : (c) Show $L = L' + R \times MV$ where $L' = \sum r'_i \times p'_i$ is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember $r'_i = r_i - R$; rest of the notation is the standard notation used in the chapter. Note $L'$ and $M \times R V$ can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.

Solution :

Step 1: Defining Initial Vectors

Given the system of particles, the position vector of the $i$-th particle relative to the origin $O$ is denoted by $\vec{r}_i$. The position vector of the Centre of Mass (CM) is denoted by $\vec{R}$. We define the position vector of the $i$-th particle relative to the CM as:

$\vec{r}_i' = \vec{r}_i - \vec{R} \implies \vec{r}_i = \vec{R} + \vec{r}_i'$ [Vector Addition Principle]

Step 2: Expressing Velocity Vectors

Differentiating the position vector with respect to time $t$ gives the velocity of the $i$-th particle $\vec{v}_i$:

$\vec{v}_i = \frac{d\vec{r}_i}{dt} = \frac{d\vec{R}}{dt} + \frac{d\vec{r}_i'}{dt}$

Let $\vec{V} = \frac{d\vec{R}}{dt}$ be the velocity of the CM and $\vec{v}_i' = \frac{d\vec{r}_i'}{dt}$ be the velocity of the $i$-th particle relative to the CM. Thus:

$\vec{v}_i = \vec{V} + \vec{v}_i'$

Step 3: Expanding the Total Angular Momentum ($L$)

The total angular momentum $L$ of the system about origin $O$ is defined as the sum of the cross products of position and linear momentum of all particles:

$L = \sum (\vec{r}_i \times \vec{p}_i) = \sum m_i (\vec{r}_i \times \vec{v}_i)$

Substituting $\vec{r}_i = \vec{R} + \vec{r}_i'$ and $\vec{v}_i = \vec{V} + \vec{v}_i'$:

$L = \sum m_i [(\vec{R} + \vec{r}_i') \times (\vec{V} + \vec{v}_i')]$

Step 4: Distributive Expansion and Simplification

Expanding the cross product term-by-term:

$L = \sum m_i (\vec{R} \times \vec{V}) + \sum m_i (\vec{R} \times \vec{v}_i') + \sum m_i (\vec{r}_i' \times \vec{V}) + \sum m_i (\vec{r}_i' \times \vec{v}_i')$

Since $\vec{R}$ and $\vec{V}$ are constant with respect to the summation index $i$, we factor them out:

$L = (\vec{R} \times \vec{V}) \sum m_i + \vec{R} \times \sum (m_i \vec{v}_i') + (\sum m_i \vec{r}_i') \times \vec{V} + \sum (\vec{r}_i' \times m_i \vec{v}_i')$

Step 5: Applying Centre of Mass Properties

We utilize the following fundamental properties of the CM:

  • $\sum m_i = M$ (Total mass of the system)
  • $\sum m_i \vec{r}_i' = 0$ [By definition of CM, the first moment of mass about the CM is zero]
  • $\frac{d}{dt} (\sum m_i \vec{r}_i') = \sum m_i \vec{v}_i' = 0$ [Differentiating the above relation]

Step 6: Final Substitution

Substituting these identities back into the expanded equation from Step 4:

$L = (\vec{R} \times \vec{V}) M + \vec{R} \times (0) + (0) \times \vec{V} + \sum (\vec{r}_i' \times \vec{p}_i')$

Recognizing that $\sum (\vec{r}_i' \times \vec{p}_i') = L'$, the equation simplifies to:

$L = M(\vec{R} \times \vec{V}) + L'$

Rearranging the terms gives the required relation:

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

CM R r'_i m_i

Final Solution: $L = L' + \vec{R} \times M\vec{V}$, where $L'$ represents the angular momentum relative to the centre of mass and $\vec{R} \times M\vec{V}$ represents the angular momentum of the centre of mass about the origin.


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