hi

The center of mass is special point in body and is moving as if whole mass was in it.
Velocity of center of mass V(C)= ( m1*v1 + m2*v2... )/(m1 + m2...)

I understand how we derived formula for velocity of center of mass, but I don't know how we derived formula for position of center of mass.

c_m = (m1 p1 + m2 p2)/(m1 + m2)
where p1 and p2 are the vector positions of m1 and m2.

So can anybody tell me how we got the above formula so I will know why it works?

thank you in advance

Newton's Second Law is F = dp/dt. Summing Newton's Second Law for a system of masses and separating the forces into internal and external forces yields

F1_ext + F2_ext + ... + sum(internal forces) = dp1/dt + dp2/dt + ...

The sum of the internal forces vanishes if Newton's Third Law applies. Denoting the sum of the external forces as F_tot,

F_tot = F1_ext + F2_ext + ...= dp1/dt + dp2/dt + ...

We want a coordinate system in which Newton's Second Law applies and that allows us to treat the system of masses as an aggregate. Denoting this central momentum as p_c,

dp_c/dt = F_tot = dp1/dt + dp2/dt + ...

Thus the center of momentum defining form is

dp_c/dt = dp1/dt + dp2/dt + ...

Now assume constant masses for the various components. Integrating once with respect to time yields the velocity equation, and integrating that with respect to time yields the position CoM formula.

Alternatively, think of the position CoM formula as definitional.

I'm confused. How can you derive the way you did velocity of center of mass from changes in momentum? As far as I know one derives velocity of center of mass from current momentum of the system:

P...momentum of a system

P=m(whole_system) * V(c)

V(c) = P/m(whole_system) = ( m1*v1 + m2*v2 ... )/( m1+m2 ... )

Alternatively, think of the position CoM formula as definitional.Hi Boris, I agree with D H here. I think you should probably consider the formula as the definition of the center of mass. It is just the average position of the masses (technically it is a weighted average with the masses being the weighting factors for the average).

The interesting thing about the center of mass isn't so much its derivation but the nice properties it has. Like the center of mass is unaffected by internal forces (e.g. between rocket and exhaust). And any external force which does not pass through the center of mass will change the angular momentum of the system.

-Dale

Hi Boris, I agree with D H here. I think you should probably consider the formula as the definition of the center of mass. It is just the average position of the masses (technically it is a weighted average with the masses being the weighting factors for the average).
-Dale

Hmm, I'm afraid you guys lost me. :(

What do you mean by "I should consider the formula as the definition of the center of mass" ?

What formula do you mean I should consider as definition ... formula for velocity of center or mass or for its position?

(technically it is a weighted average with the masses being the weighting factors for the average).

That's another thing that confuses me. For me an average is for example (4+6+8)/3=6 .
I realise if one object has very large mass compared to other objects, then velocity of center will be closer to that object's velocity than others, but i still don't regard formula for velocity of center of mass as something that gives us an average speed value(it does gives us speed closer to object with largest mass but I don't view that as average)

Hi boris,

If you define the position of the center of mass of an N particle system as x = (m_1 x_1 + m_2 x_2 + ... + m_N x_N)/(m_1 + m_2 + ... + m_N) and then differentiate this formula with respect to time, you recover your expression for the center of mass velocity. Alternatively, if you would like to start from the momentum as you have done, you can define the center of mass velocity as v = (m_1 v_1 + ... m_N v_N)/(m_1 + ... + m_N). You can now obtain the center of mass position by integrating the center of mass velocity. In both cases all you need to know is that the mass is independent of time and that differentiation/integration is additive. Hopefully you can now see why the two definitions are equivalent to each other and how one expression cen be obtained from from the other.

What formula do you mean I should consider as definition ... formula for velocity of center or mass or for its position?I would consider the position of the center of mass as the definition. The velocity of the center of mass is then the rate of change of the position, just like the standard definition of velocity. But as PM mentioned above, it doesn't really matter which you consider as the definition, you get the other one easily by differentiation or integration.

i still don't regard formula for velocity of center of mass as something that gives us an average speed value(it does gives us speed closer to object with largest mass but I don't view that as average)Look at this page for more information about weighted averages ( http://en.wikipedia.org/wiki/Weighted_average ). A normal average like the kind you wrote is simply a special case of a weighted average: the case where all the weights are equal to 1. If you look at the formula for the weighted average and then look at the formulae for position and velocity of the center of mass you will see that they are weighted averages of the positions and velocities with the weights being given by the mass.

-Dale

If you define the position of the center of mass of an N particle system as x = (m_1 x_1 + m_2 x_2 + ... + m_N x_N)/(m_1 + m_2 + ... + m_N) and then differentiate this formula with respect to time, you recover your expression for the center of mass velocity

You mean x(time=1) - x(time=0) = dx -> dx/dtime = velocity_center.
where x(time=1) is position of center of mass at time=1 and x(time=0) position of center of mass at time=0 ?



center of mass velocity as v = (m_1 v_1 + ... m_N v_N)/(m_1 + ... + m_N). You can now obtain the center of mass position by integrating the center of mass velocity.

Integrating the center of mass velocity into what formula? I'm truly sorry but I am not following you


I would consider the position of the center of mass as the definition


When you say I should consider it as definition you mean I shouldn't be bothered how we got that definition and instead I should just use it?

sorry guys...I know I'm not the sharpest out there :(

No boris, I meant the position of the center of mass is given by the mass of particle one times the position of particle one plus the mass of particle two times the position of particle two ... all divided by the total mass of the system. If you take this as your definition of the position of the center of mass then you can obtain the formula you wrote for the velocity of the center of mass by performing the calculus operation of differentiating with respect to time. On the other hand, you can define the center of mass velocity first as the mass of particle one times the velocity of particle one plus the mass of particle two times the velocity of particle two .... all divided by the total mass. You can then find the previous formula I gave for the position of the center of mass by performing the calculus operation of integrating with respect to time. Remember, the velocity is the time derivative of the position so the position is the time integral of the velocity.

You can then find the previous formula I gave for the position of the center of mass by performing the calculus operation of integrating with respect to time. Remember, the velocity is the time derivative of the position so the position is the time integral of the velocity.

Besides me being mentally challenged, I think there is also language barier here since I have no clue what you mean by position is time integral of velocity.

In any case, for me to derive position of center of mass from velocity:

V(c)= Center_position / time
Center_position = V(c) * time = (m1*v1 + m2*v2...)*time / (m1+m2...)

but now what?

When you say I should consider it as definition you mean I shouldn't be bothered how we got that definition and instead I should just use it?Yes, that is exactly what I am suggesting. I simply consider the center of mass as the (weighted) average position. Then you can look at the interesting properties that the center of mass has.

The other approach you can take is the one that DH covered in the first response. Remember, DH said, "We want a coordinate system in which Newton's Second Law applies and that allows us to treat the system of masses as an aggregate" and from that derived the formula. Basically, you state the property that you want it to have and derive the necessary formula.

So it is either start with the formula and then find the properties or start with the desired properties and find the formula. I prefer the first way in this case because it makes sense to me why you would call the (weighted) average position the "center of mass" independent of any properties it might have.

Integrating the center of mass velocity into what formula? Sorry if I misunderstand your comment, but are you not taking a calculus class also? PM is talking about integration, not plugging something into a formula. Integration and differentiation are basic operations of calculus just like addition and subtraction are basic operations of arithmetic. It is essentially the math developed to describe classical physics. If you differentiate position you get velocity, if you integrate velocity you get position. That is what PM was talking about, but unless you also have calculus it won't mean anything. Most physics formulas can only be properly derived using calculus.

-Dale

Besides me being mentally challenged, I think there is also language barier here since I have no clue what you mean by position is time integral of velocity.

The problem is not a language barrier, it is a math barrier. Physics has been calculus-based since Newton. An understanding of the calculus is a pre-requisite for understanding physics at anything beyond the "Physics for Poets" level. You need at least an introductory calculus background to understand why the definition of the center of mass is so useful.

I assume the following example is differentiation at work. If that's not it, then
blah,no I don't know what differentiation is!

s1=3m
t1=1s

s2=6m
t2=2s

speed = (s2-s1)/(t2-t1) = ds/dt = 3m/s


Anyhow,I quit highschool at age 15 and now 5 years later I decided i shall finish it.So I'm learning this stuff at home and when the time comes I will go and take the tests. So cut me some slack since it's not easy learning this on your own :)


In any case, can't you guys show me with formulas the stuff you talked about in your posts ?

Hi boris, I would hope no one here ever mocks you for trying to learn or for not yet knowing. That may not be the case unfortunately, but I certainly won't do it. I do, however, need to know what level you're at in order to be able to help you effectively. Calculus is very important tool that you will need to understand the basics of in order to really grasp most of physics.

The example you gave is not quite the derivative, but it's close. What you have calculated is the average velocity between the two times t1 and t2. The derivative of position is actually defined as the limit as t1 approaches t2 of the quantity (s2-s1)/(t2-t1). Now I suspect that these concepts may be somewhat foreign to you, but the good news is that many calculus concepts are very intuitive. In this case, the exact velocity is actually defined as the limit of the average velocity as the time over which you are averaging tends to zero (the limit procedure I was talking about above). In the case of a particle moving with constant velocity, the average velocity is always the same, right? The position is therefore s = vt + s0 where s0 is your starting point and v is your velocity. Now follow the procedure above to calculate the velocity: s(t1) = v*t1 + s0, s(t2) = v*t2 + s0, and so the difference is s(t2)-s(t1) = v*(t2-t1) now divide this by (t2-t1) to obtain (s2-s1)/(t2-t1) = v for the average velocity. Now this a constant so the limit as t1 approaches t2 of (s2-s1)/(t2-t1) is just v and thus the exact velocity is v as promised.

Let's return to the formula I gave above for the position of the center of mass. You know the velocity of the center of mass is v_c = (m_1 v_1 + ... + m_N v_N )/(m_1 + ... + m_N). Now suppose that all the particles are moving with constant velocity. Clearly the center of mass velocity is also constant. Let's multiply the whole equation by t remembering that v_1 t = x_1 (if I assume all the particles start at position 0) and v_2 t = x_2 etc. This means that v_c t = (m_1 v_1 t + ... + m_N v_N t)/(m_1 + ... + m_N) where all I did was multiply each term by t. Now v_c t = x_c, the position of the center of mass so that we can rewrite the above formula as x_c = (m_1 x_1 + ... + m_N x_N )/(m_1 + ... + m_N), and this is the expression I gave to you above for the position of the center of mass. I only proved this to you in the case of constant velocities but it actually holds in general. In calculus, the velocity is the derivative of position, and the procedure for going from velocity back to position is called integration, a kind of "reverse differentiation" if you will.

Here are some online links to help you with your calculus:
http://archives.math.utk.edu/visual.calculus/
http://www.math.hmc.edu/calculus/tutorials/

Here is a good physics reference site:
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

So I'm learning this stuff at home and when the time comes I will go and take the tests. So cut me some slack since it's not easy learning this on your own :)Well, I am certainly willing to help however I can. I wasn't asking to be antagonistic or anything. It is just that math and physics are so intertwined that I need to know your math background in order to properly explain things. You can learn all of the concepts that you need to know for High School physics without using calculus, but you will not be able to derive many of the formulas until you take calculus.

That being said, did you understand the weighted average? If so, then just remember that the center of mass is the weighted average of the positions and the velocity of the center of mass is the weighted average of the velocities. In each case the weights are the masses. With that definition of the center of mass you don't need calculus.

-Dale

Hi Boris,

As you indicated yourself in your opening post, the center of mass concept simply dumps a number of masses into one hypothetical point mass i.e. the location of the center of mass X is for two masses m1 and m2 at the locations x1 and x2 given by the equation

(m1+m2)*X = m1*x1 + m2*x2

and therefore

X= (m1*x1 + m2*x2)/(m1+m2).

If you differentiate this with regard to time you get the velocity V of the center of mass:

V = (m1*v1 + m2*v2)/(m1+m2) .


For some more information on this see for instance http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html .

Thomas

thanx for your help. Not that it matters or anything, but I hope you don't mind if I delve into your posts tommorow since I won't be at home today

As for my math background ... I already took the tests for 1. and 2. year of highschool ( and did it:) ) and will now learn physics before continuing with math for 3. and 4. year of highschool


Don't know if it helps, but so far I learned

-functions(linear,quadratic,exponential,inverse, sine and cosine)
-logarithms
-vectors
-a bit of limits
-sequences
-polynomials
-bit of ellipses and hyperbola
-geometry

But no calculus as of yet

cheers

so far I learned

-functions(linear,quadratic,exponential,inverse, sine and cosine)
-logarithms
-vectors
-a bit of limits
-sequences
-polynomials
-bit of ellipses and hyperbola
-geometry

But no calculus as of yetThe vectors are essential to physics, as is the related trigonometry. Be sure that you are familiar and comfortable with vector addition, multiplication of a vector and a scalar, breaking a vector up into components, and recovering the magnitude and direction of a vector from the components. Those will be essential concepts that you will use in free-body diagrams and throughout physics. Later on you will also need the vector dot product (for work) and cross product (for torque).

You will use polynomials, ellipses, and other functions quite a bit so it is good that you have been exposed to them. However, the derivation of the polynomial formulas that you will use (e.g. 1/2 a t^2 + v0 t + x0) usually requires calculus. Just be patient for that part, eventually it will be easy but at this point it is not possible.

Limits are the first and most fundamental part of calculus. Be sure that you understand them when you go to take calculus, but they will not directly help you much at this point.

By the way, are you comfortable now with the center of mass idea? Also, I highly recommend the hyperphysics site that tsmid mentioned above. I have the main page ( http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html ) bookmarked and I still refer to it fairly often.

-Dale

I assume the following example is differentiation at work. If that's not it, then
blah,no I don't know what differentiation is!

s1=3m
t1=1s

s2=6m
t2=2s

speed = (s2-s1)/(t2-t1) = ds/dt = 3m/s


Anyhow,I quit highschool at age 15 and now 5 years later I decided i shall finish it.So I'm learning this stuff at home and when the time comes I will go and take the tests. So cut me some slack since it's not easy learning this on your own :)


In any case, can't you guys show me with formulas the stuff you talked about in your posts ?

First, regarding the slack, will do, and good luck with your exams.

Second, regarding differentiation, you are getting close. Differentiation is defined as the the ratio of the change in the dependent variable ("s" in this case) with respect to change in the independent variable ("t") in the limit that the independent variable change goes to zero. To make it more concrete, in your example, you need to make t2 become closer and closer to t1.

Third, regarding showing equations, we do do that, but doing so can be difficult in this forum. This forum just doesn't provide the tools needed to go full throttle with the math.

thank you for your help and for the links

cheers