Recent content by 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...

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

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.

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...

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...

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...

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...

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...

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...

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)...

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...