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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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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,

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

Something Different
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