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Let's say that a three dimensional object with continuous mass distribution is undergoing rotational motion about an axis that lies on the centre of mass. The translational velocity of the centre of mass is $\vec{0}$.

I understand that the angular momentum is not zero because the direction of the $\vec{r} \times d\vec{p}$ vector is same for all points of the object so they add up to form the total angular momentum.

However I failed to derive quantitively that the linear momentum of the object is equal to $\vec{0}$. I tried to use symmetry or geometry in calculating the integral $$\vec{p} = \int dm \ \vec{v}$$ but for a random continuous mass distribution, with non-constant density $\rho(\vec{r})$, it wasn't easy.

Is there any good mathematical justification that clearly shows the above quantity is zero? (For example, I have seen the reasoning that it is a time derivative of the coordinates of COM relative to the COM so it should be zero but that heavily relies on physical intuition.)

abouttostart
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2 Answers2

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Each particle $m_i$ located at $\boldsymbol{r}_i$ relative to the center of mass has linear velocity $\boldsymbol{v}_i = \boldsymbol{\omega}\times \boldsymbol{r}_i$. Just as you add up all the masses $m=\sum_i m_i$ to get the total mass, you add up all the momenta to get the total translational momentum

$$ \boldsymbol{p} = \sum_i m_i \boldsymbol{v}_i = \sum_i m_i ( \boldsymbol{\omega} \times \boldsymbol{r}_i ) = \boldsymbol{\omega}\times \sum_i m_i \boldsymbol{r}_i$$

But by the definition of center of mass $\sum_i m_i \boldsymbol{r}_i =0$, so $\boldsymbol{p} = 0$

See this answer with a lot more details on deriving momentum (linear and angular) for a co-rotating group of particles.

John Alexiou
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  • your sum is not zero, unless you are at the center of mass! –  Aug 23 '20 at 13:36
  • to decompose $ v_i $ correctly see here https://en.wikipedia.org/wiki/Angular_velocity#Rigid_body_considerations –  Aug 23 '20 at 14:31
  • @Wolphramjonny - correct, but the question states "undergoing rotational motion about an axis that lies on the centre of mass". So $\boldsymbol{v}i = \boldsymbol{\omega} \times \boldsymbol{r}_i$ instead of the more general case $$\boldsymbol{v}_i = \boldsymbol{v}{\rm COM} + \boldsymbol{\omega} \times \boldsymbol{r}_i$$ – John Alexiou Aug 23 '20 at 16:28
  • yes, the axis goes through the center of mass, but the location of the center of mass in your coordinate system could be located anywhere –  Aug 23 '20 at 18:11
  • @Wolphramjonny , he defined $\vec{r}$ as the relative vector from the COM, therefore $ \sum m_i \vec{r}_i$ is indeed $\vec{0}$. – CasualScience Aug 23 '20 at 19:16
  • I fail to see where the OP assumed that, and it is not necessary to assume, anyways. –  Aug 23 '20 at 19:24
  • @Wolphramjonny read my first sentence. $r_i$ is defined relative to the center of mass. I did this for simplification. I understand how the general case works (look into linked answer), but the question was specifically how momenta cancel out for rotational motion. – John Alexiou Aug 23 '20 at 21:24
  • @Wolphramjonny, John states it... you don't have $\vec{v} = \vec{\omega} \times \vec{r}$ unless $\vec{r}$ is a displacement vector from the axis of rotation. – CasualScience Aug 23 '20 at 21:24
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The velocity of a point $dm$ is $r\omega$, where $r$ is the radial distance of the point to the center of mass.

Orient a coordinate system at the center of mass, your integral takes the form

$$ \vec{p} = \int \limits_0^M \vec{\omega}\times\vec{r} ~ dm = \vec{\omega} \times \int \int \int r \rho(\vec{r}) ~dV$$

But notice the integral on the right is precisely the definition of the center of mass, in this coordinate frame, that is at the origin, with coordinates $\vec{0}$.

Therefore,

$$p = \omega \cdot 0 = 0.$$

CasualScience
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