Physics Asked on July 17, 2021
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 $vec{r} times dvec{p}$ 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.)
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.
Correct answer by John Alexiou on July 17, 2021
The velocity of a point $dm$ is $romega$, 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
$$ vec{p} = int limits_0^M vec{omega}timesvec{r} ~ dm = vec{omega} times int int int r rho(vec{r}) ~dV$$
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,
$$p = omega cdot 0 = 0.$$
Answered by Bobak Hashemi on July 17, 2021
Notice that $vec{p} = int dm vec{v}=Mvec{v}_{CM}=0$ by assumption.
Answered by Wolphram jonny on July 17, 2021
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