CollinsArg
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What does the Center of Mass equation mean?
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https://en.wikipedia.org/wiki/Weighted_arithmetic_meanCollinsArg said:
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It's a weighted average of the positions of the point masses 1 and 2. This means for instance that if the mass 1 is made large enough, the COM is close to the position of 1 no matter where 2 is.
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Here's a way to see this weighted average.
For simplicity, work in 1 dimension.
Suppose I had five identical blocks, each with mass 1 kg,
that I place at various positions on a number line. Where is the [straight-]average position of those five blocks?
[itex]r_{avg}=\displaystyle\frac{r_1+r_2+r_3+r_4+r_5}{1+1+1+1+1}[/itex]
Suppose there were two clumps... so that [itex]r_1=r_4[/itex] (so there are 2 at [itex]r_1[/itex]) and [itex]r_2=r_3=r_5[/itex] (so there are 3 at [itex]r_2[/itex]).
Then
[itex] \begin{eqnarray*}<br /> r_{avg}<br /> &=&\frac{(r_1+r_4)+ (r_2+r_3+r_5)}{(1+1)+(1+1+1)}\\<br /> &=&\frac{(2)r_1+ (3)r_2}{(2)+(3)}\\<br /> \end{eqnarray*}[/itex]
Instead of grouping by count, we could group by mass. (In this case of identical blocks, it doesn't matter numerically... but this suggests that you can take any quantitative property and conceptually break it down into units of that property). Think of multiplying the above by [itex]1=\frac{{\rm\ kg}}{{\rm\ kg}}[/itex].
[itex] \begin{eqnarray*}<br /> r_{avg}<br /> &=&\frac{(2{\rm\ kg})r_1+ (3{\rm\ kg})r_2}{(2{\rm\ kg})+(3{\rm\ kg})}\\<br /> \end{eqnarray*}[/itex]
The physical reason why the center-of-mass is defined this way is because
the total momentum of a system of particles [itex]\vec P=m_1\vec v_1+m_2\vec v_2[/itex]
can be modeled as a particle of mass [itex]M=m_1+m_2[/itex]
with velocity [itex]\vec V=\frac{\vec P}{M}=\frac{m_1\vec v_1+m_2\vec v_2}{m_1+m_2}=\frac{d}{dt}\left( \frac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2} \right)=\frac{d}{dt}\vec R[/itex]
equal to the velocity of the center-of-mass.
(Off topic, but related.
By the way, the "average-velocity" is a time-weighted average of velocity:
[itex]\vec v_{avg}=\frac{\vec v_1 \Delta t_1 +\vec v_2 \Delta t_2}{\Delta t_1+\Delta t_2}=\frac{\Delta \vec x_1+\Delta \vec x_2}{\Delta t_1+\Delta t_2}=\frac{\Delta \vec x_{total}}{\Delta t_{total}}[/itex].
It annoys me that the textbook definition skips this interpretation by defining [itex]\vec v_{avg}\equiv \frac{\Delta \vec x_{total}}{\Delta t_{total}}[/itex].
Then it proceeds to define the [instantaneous] velocity as a limiting case of the average-velocity.
Maybe they skip the interpretation because it's strange to define a quantity in terms of its average.)
For simplicity, work in 1 dimension.
Suppose I had five identical blocks, each with mass 1 kg,
that I place at various positions on a number line. Where is the [straight-]average position of those five blocks?
[itex]r_{avg}=\displaystyle\frac{r_1+r_2+r_3+r_4+r_5}{1+1+1+1+1}[/itex]
Suppose there were two clumps... so that [itex]r_1=r_4[/itex] (so there are 2 at [itex]r_1[/itex]) and [itex]r_2=r_3=r_5[/itex] (so there are 3 at [itex]r_2[/itex]).
Then
[itex] \begin{eqnarray*}<br /> r_{avg}<br /> &=&\frac{(r_1+r_4)+ (r_2+r_3+r_5)}{(1+1)+(1+1+1)}\\<br /> &=&\frac{(2)r_1+ (3)r_2}{(2)+(3)}\\<br /> \end{eqnarray*}[/itex]
Instead of grouping by count, we could group by mass. (In this case of identical blocks, it doesn't matter numerically... but this suggests that you can take any quantitative property and conceptually break it down into units of that property). Think of multiplying the above by [itex]1=\frac{{\rm\ kg}}{{\rm\ kg}}[/itex].
[itex] \begin{eqnarray*}<br /> r_{avg}<br /> &=&\frac{(2{\rm\ kg})r_1+ (3{\rm\ kg})r_2}{(2{\rm\ kg})+(3{\rm\ kg})}\\<br /> \end{eqnarray*}[/itex]
The physical reason why the center-of-mass is defined this way is because
the total momentum of a system of particles [itex]\vec P=m_1\vec v_1+m_2\vec v_2[/itex]
can be modeled as a particle of mass [itex]M=m_1+m_2[/itex]
with velocity [itex]\vec V=\frac{\vec P}{M}=\frac{m_1\vec v_1+m_2\vec v_2}{m_1+m_2}=\frac{d}{dt}\left( \frac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2} \right)=\frac{d}{dt}\vec R[/itex]
equal to the velocity of the center-of-mass.
(Off topic, but related.
By the way, the "average-velocity" is a time-weighted average of velocity:
[itex]\vec v_{avg}=\frac{\vec v_1 \Delta t_1 +\vec v_2 \Delta t_2}{\Delta t_1+\Delta t_2}=\frac{\Delta \vec x_1+\Delta \vec x_2}{\Delta t_1+\Delta t_2}=\frac{\Delta \vec x_{total}}{\Delta t_{total}}[/itex].
It annoys me that the textbook definition skips this interpretation by defining [itex]\vec v_{avg}\equiv \frac{\Delta \vec x_{total}}{\Delta t_{total}}[/itex].
Then it proceeds to define the [instantaneous] velocity as a limiting case of the average-velocity.
Maybe they skip the interpretation because it's strange to define a quantity in terms of its average.)