Physics question: What is the equation for angular momentum?

Status
Not open for further replies.

BigToque

Lifer
Oct 10, 1999
11,700
0
76
My physics text says that angular momentum has the equation L=Iw, and in a number of places in the text says that "I" (moment of inertia) is = ∑mr^2.

If you substitute that into the equation you get L = m(r^2)w

If I go to the article for angular momentum on wikipedia (http://en.wikipedia.org/wiki/Angular_momentum), it says angular momentum is L=mrv.

So I've got L = m(r^2)w and L=mrw... So which is it and why?

I've got a practice exam question and the answer requires the equation be L=mrw, my professor says it's L=mrw, and wikipedia says L=mrw, so clearly angular momentum is L=mrw

But I don't see L=mrw in my textbook ANYWHERE. All I see is L=Iw, and that I=∑mr^2.

What am I missing?
 

Born2bwire

Diamond Member
Oct 28, 2005
9,840
6
71
Yeah, the main problem here is that angular velocity (\omega) is not the same as linear velocity (v). They are related as \omega = v/r (assuming that the trajectory is a circle). These are, of course, only applicable to specific problems like where the trajectory is a circle. Angular velocity and angular momentum exist in more general cases but the relationships are different. In general, we use the cross product to denote the exact expression as esun state above. If the vectors r and v are at right angles (like they are in a circle trajectory) then the cross product simplifies to the product of the magnitudes of the vectors (rv).
 

BigToque

Lifer
Oct 10, 1999
11,700
0
76
You cannot ignore the cross in the expression L = r x mv.

http://en.wikipedia.org/wiki/Cross_product

And you also cannot just drop the sum when you plug I into the equation for L.

v is also not the same thing as w.

I have to be honest, I've never even heard of a cross product. I also assumed dropping the sum would be fine if you were working with only a single object.

Here's the question I have that I don't understand how to get the answer to:

A 1500-kg satellite orbits a planet in a circular orbit of radius 6.2*10^6 m. What is the angular momentum of the satellite in it's orbit around the planet if the satellite completes one orbit every 1.5*10^4 s?

I've got:
m=1500kg
r=6.2*10^6 m
w=2597 m/s (2πr/T)

When mentioning angular momentum, the textbook makes reference to P=mv and saying L=Iw is pretty much the same.

So if L=Iw, and I've got w, how am I supposed to get m and r into the place where I've got I?

The text says I=∑mr^2, but if you make the equation for angular momentum L=(∑mr^2)w, you don't get the right answer.

So what funny business needs to happen to get rid of the ∑ and the ^2 out of the equation?
 

esun

Platinum Member
Nov 12, 2001
2,214
0
0
What equation are you using for omega? Did you write 2*pi*r/T? If so that is incorrect.

Consider: what are the units of angular velocity?
 
Status
Not open for further replies.