So I know that $$F=ma$$ ($F$ is force, $m$ is mass, and $a$ is acceleration). I also know that $$p=m\int a dt=mv$$ (Where $p$ is momentum and $v$ is velocity). So I was wondering if the following object has some importance: $$m\int\int a(dt)^2=mx$$
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3$\begingroup$ I am voting to reopen because I learned from the aswer something not known before, the term and use of "moment of mass" and the answer should be open to searches $\endgroup$– anna vCommented Dec 10, 2022 at 6:48
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2 Answers
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The mass $m$ times position $\vec{x}$ of a particle is called the "moment of mass" of that particle.
If there are several particles, the moment of mass is the sum of moments of mass of each particle individually.
The moment of mass of a system, divided by the total mass of the system, is an important quantity called the center of mass.
The quantity $m \int \vec{v} \mathrm{d}t$ is not equal to $m\vec{x},$ the moment of mass. Instead, it's the change in the moment of mass over the time of the integral.
answered Dec 9, 2022 at 18:28
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Mass moment or Moment of a mass, is yet another vector quantity that we define for a purpose. Especially when we look at obtaining a centre of mass of a continuous or a discrete distribution, we first need to fix one particular point, here let's say we fix that point to be our Centre of mass, and considering we want to calculate the Centre of mass, we now for a discrete system measure mass moment of each mass and then summate over and divide by total no of such particles in our discrete system.
But in the above Expression we integrate v with respect to dt which will yield us the displacement but the product of this is never a 'Mass Moment'
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1$\begingroup$ Yes, moment is not dynamic. Here distance is given in terms of time. It is weighted trajectory of particle if ploted with time. $\endgroup$ Commented Dec 10, 2022 at 14:05