Angular Momentum about Center of Mass

[Pifly310]

Hello, I have a question about the angular momentum about the center of mass. I am using Goldstein's Classical Mechanics. In the formula, it is stated that angular momentum about the center of mass (L) = R x Mv + SUM-->i of (r'i x p'i)

With R being the center of mass vector, M being the total mass, v being the velocity of the center of mass and r'i and p'i being the vector and momentum of the i particles with respect to the center of mass.

My question is about the second term, if written instead as SUM-->i of (r'i x mi*v'i) = SUM-->i of (r'i*mi x v'i) using this form, SUM-->i(r'i*mi) is a null vector and the term would go to zero leaving only the first term in the equation for L.


Thanks.

 

[sigmaro]

it wouldn't be a null vector even though mi is small it exists
but i think there is another mistake there, if this is angular momentum wrt center of mass why is there a R x Mv
Lcm=SUM-->i of (r'i x p'i)
and total momentum wrt origin is
Lo=R x Mv + SUM-->i of (r'i x p'i)

 

[Pifly310]

The SUM-->i of (r'i*mi) is equal to zero because r'i are the coordinates of the particles with respect to the center of mass, and by definition of center of mass, this value should sum to zero and nullify the last term. It can be shown that SUM --> i of (r'i*mi) = SUM-->i of ((ri - R)*mi) with R again being the position of the center of mass and ri being the position vectors of the i particles. The SUM-->i of ((ri-R)*mi) = SUM-->i of (ri*mi-R*mi). Yet the SUM-->i of (ri*mi) is equal to RM where M is the total mass, and SUM-->i of (R*mi) = RM as well, so this summation equals zero.

 

[Sourabh N]

But v_i cannot be taken out of the summation (which is over the label i)!

I'd elaborate if you want.

 

[Pifly310]

Yes, that would be helpful. The term before that that goes to zero is the SUM-->i of (r'i x mi*V) and the author takes out mi to join it with r'i in the summation. Is this because V is a constant vector that is crossed with each of the r'i terms whereas v'i is not? What would happened to SUM-->i of (r'i x V) without the mi terms?

 

[Sourabh N]

Yes, that would be helpful. The term before that that goes to zero is the SUM-->i of (r'i x mi*V) and the author takes out mi to join it with r'i in the summation. Is this because V is a constant vector that is crossed with each of the r'i terms whereas v'i is not? What would happened to SUM-->i of (r'i x V) without the mi terms?

Yes! Because V is a constant vector, it can be taken out of summation, but since v_i have the index i attached, they can't be taken out of the summation.

If it's not clear, think of it this way. Had you had an integral sign instead of summation, V is a constant function, whereas v_i becomes v(x) where x is some dummy variable which you are integrating over. V can be taken out of the integral, v(x) can't.

 

[Pifly310]

OK thanks. That makes sense to me if the the terms are multiplied, but since they are crossed, in each summation term for SUM--> of (r'i x mi*V), this could be written as SUM--> of (|r'i||V|sin(theta_i)mi) where theta_i is the angle between r'i and V, and then each term would have a unique theta_i and after expanding out the terms, you could factor out the V but not necessarily the theta_i's and couldn't form the SUM--> i of (r'i*mi)? I could see using the argument that each of the r'i's has an opposite whose cross with V would cancel it out in the summation because they are with respect to the center of mass, but I don't see how you could use the argument of factoring out the m'i terms.

 

[Sourabh N]

Okay. Center of mass remains fixed means $$\Sigma$$$$\vec{r}$$$$_{i}$$m$$_{i}$$ = 0. This implies $$\Sigma$$r$$_{i,x}$$m$$_{i}$$ = 0 and same for y and z. Note that summation is over i and by i,x I mean x coordinate of i'th particle.

Assume V is along z (or x, or y, doesn't matter), so that V_z = V, V_x = V_y =0. Then the cross product will give you two terms, -$$\Sigma$$r$$_{i,x}$$m$$_{i}$$*V along y and $$\Sigma$$r$$_{i,y}$$m$$_{i}$$*V along x. Each of them go to Zero, hence the cross product is zero.

You can choose to choose an arbitrary(but constant) direction for V. It's only more tedious, the result is same.

 

[Pifly310]

I understand that argument, i am still unsure of how one could factor out the m'i terms though, since with the cross product an angle is introduced that is not the same for each term, i.e. the SUM--> of (r'i x mi*V) = SUM--> of (|r'i||V|sin(theta_i)mi) and since each term has unique theta_i, after writing out the terms, you could factor out the V but not the theta_i's and therefore you would not be able to form V*SUM--> i of (r'i*mi). I'm not sure where i am going wrong here. Thanks for clarifying the last summation, I understand it does in fact go to zero, now I'm trying to justify the method the text uses (actually Kleppner and Kolenkow now) in factoring out the SUM--> of (r'i*mi).

 

[Sourabh N]

I understand that argument, i am still unsure of how one could factor out the m'i terms though, since with the cross product an angle is introduced that is not the same for each term, i.e. the SUM--> of (r'i x mi*V) = SUM--> of (|r'i||V|sin(theta_i)mi) and since each term has unique theta_i, after writing out the terms, you could factor out the V but not the theta_i's and therefore you would not be able to form V*SUM--> i of (r'i*mi). I'm not sure where i am going wrong here. Thanks for clarifying the last summation, I understand it does in fact go to zero, now I'm trying to justify the method the text uses (actually Kleppner and Kolenkow now) in factoring out the SUM--> of (r'i*mi).

You're getting confused between the vectors and scalars, I think. When you write SUM--> of (r'i*mi) r'i is NOT a scalar, it's the vector r_i. I presume what the text does is :
$$\Sigma_{i}$$($$\vec{r}$$$$_{i}$$m$$_{i}$$$$\times$$V) = $$\Sigma_{i}$$($$\vec{r}$$$$_{i}$$m$$_{i}$$)$$\times$$V , which is allowed since V is constant. And $$\Sigma_{i}$$$$\vec{r}$$$$_{i}$$m$$_{i}$$ = 0. $$\Sigma_{i}$$r$$_{i}$$m$$_{i}$$ is NOT equal to zero, since r$$_{i}$$ is the norm (hence a positive quantity) and so is m$$_{i}$$

 

[Pifly310]

Ah i think I figured it out. When written as SUM--> of (|r'i||V|sin(theta_i)mi), you could write out the terms factor out V, and since V is constant, the r'i*sin(theta_i) terms would just be the components of the r'i vectors perpendicular to V. Since the sum of the r'i vectors in all directions goes to zero, this component goes to zero as well. The text writes SUM-->i of (r'i x mi*V) = SUM-->i of ((r'i*mi) x V). By this logic, mi is always a scalar, and scalar *Vector_a x Vector_b = Vector_a x scalar*Vector_b, so couldn't SUM-->i of ((r'i x mi*v'i) be written as SUM-->i of ((r'i*mi) x v'i) even if v'i changes because of the indices. I just don't see why not even if v'i is not a constant vector.

 

[Sourabh N]

Yes, you can write it like that. But I don't see it serving any purpose here. Clarify your point?

 

[Pifly310]

Hmm, but when the text writes SUM-->i of (r'i x mi*V) = SUM-->i of ((r'i*mi) x V), it then says since SUM-->i of (r'i*mi) = 0, this term equals zero. If the SUM-->i of ((r'i x mi*v'i) can be written as SUM-->i of ((r'i*mi) x v'i), wouldn't this go to zero by the same logic? Did the text leave step out?

 

[Sourabh N]

No. You see,

Hmm, but when the text writes SUM-->i of (r'i x mi*V) = SUM-->i of ((r'i*mi) x V), it then says since SUM-->i of (r'i*mi) = 0, this term equals zero.

V can be taken out of summation, since it is same for all values of i.

If the SUM-->i of ((r'i x mi*v'i) can be written as SUM-->i of ((r'i*mi) x v'i), wouldn't this go to zero by the same logic? Did the text leave step out?

v'i cannot be taken out of the summation, because it depends on the value of i.

Okay consider this. In this example, I take everything to be scalar, for clarity. Suppose i goes from 1 to 3. So r_1*m_1 + r_2*m_2 + r_3*m_3 = 0.
Let V = 2. And v_1 = 1, v_2 = 2, v_3 = 3.

Now you see, Sum over i (r_i*m_i*V) = Sum over i (r_i*m_i) *V = (r_1*m_1 + r_2*m_2 + r_3*m_3)*V = 0. BUT, Sum over i (r_i*m_i*v_i) = r_1*m_1*v_1 + r_2*m_2*v_2 + r_3*m_3*v_3 is clearly not zero.

 

[Pifly310]

Right, this is very clear to me. But this is using the cross product, which introduces unique sin(theta) terms for the vectors and while V can be factored out, those sin(theta) terms which are indexed by i cannot. So I don't see how one could simply factor out the V from a cross product, whereas it is clear you could do that with multiplication. Following this, I'm not sure how the text uses SUM-->i of ((r'i*m'i) x V) at all, since while the SUM--> of (r'i*m'i) goes to zero, when the individual terms are crossed with V, the theta terms are introduced and when expanding the sum, the SUM-->i of (r'i*mi) cannot be factored out because of these sin(theta)_i terms that are different for different i's.

 

[Sourabh N]

Refer to Post #8 for explanation of how this works out when you have cross product.
It's kind of tricky to see the summation going to zero when you write it in terms of theta_i, the angle between r_i and V and atleast I can't think of any easy way to write it here. Maybe someone else can help. Do you understand the way I did in #8?

 

[Pifly310]

Yes, I understand #8 and generally the derivation of the angular momentum about the center of mass. Now its the ability to factor out the SUM-->i of (r'i*mi) before carrying through with the cross product.

 

[Pifly310]

I've attached a copy of the page where the derivation is done if this helps. I believe they also forget to dot the R vector in the second equality for L at the top of the page. T

 

[Sourabh N]

Yep. Looks like a typo.

 

[Pifly310]

Ok. SUM-->i of (r'i x miV) = |r'1|*m1*|V|*sin(theta1) + |r'2|*m2*|V|*sin(theta2) +...
|V| can be factored out and = |V|*(|r'1|*m1*(theta1) + |r'2|*m2*(theta2) ...)
which equals SUM--> of (r'1*mi) x V = |V|*|r'1|*m1*sin(theta1) + |V|*|r'2|*m2*sin(theta2)...

Since SUM-->i of (r'i x miV) can be written as SUM-->i of (r'i*mi) x V, SUM -->i of (r'i*mi) = 0 therefore the SUM-->i of (r'i x miV) = 0 as well.

That makes a lot of sense and I see why the sum can be factored out now as opposed to when v'i or anything indexed with i is in the place of V or another constant vector.

Thanks Sourabh, you've been very helpful.