Total Angular Momentum

1. Nov 9, 2009

lmnt

The problem statement
Icm=$$\frac{2}{5}$$mr2
$$\omega$$=v/r
m=mass of object

There's some things I don't really get about total angular momentum in a rigid body. Suppose a perfectly spherical object is rolling in uniform circular motion (ie. in a loop). Find the total angular momentum.

2. Relevant equations
L=mvr for particle, and L=Icm$$\omega$$ for rigid body

3. Attempt at solution

Since it is rolling, it has angular momentum due to its own rotation around its axis, like a normal object that's just spinning right? So then L = Icm$$\omega$$ and I would use $$\omega$$=v/r with r=radius of sphere

But then, it also has angular momentum around the centre of the loop-the-loop that it rolls around. Here's where i get lost. Would i consider the object as a particle and use L=mvr, or do i use L = Icm$$\omega$$ again, and which radius do i use when i sub in for $$\omega$$? the radius of the loop or the radius of the mass itself?

I get that the total angular momentum for a situation like this would require adding together the seperate momentums but i don't get how exactly they are seperated.

Last edited: Nov 9, 2009
2. Nov 9, 2009

indr0008

actually when u are asked to calculate the angular momentum of such system, what u need to do is to apply the mv X r (with v and r being vectors) to all the particles in the rigid body, but that will be complicated
but thankfully, the calculation can be simplified (in most simple problems) by separating the angular mementum in the way you have described, they will give the same and the correct answer just like if u do it rigorously for each and every particles in the rigid body

hope that helps :)

3. Nov 10, 2009

lmnt

Thanks, that makes it clear, i thought i was simplifying it too much, so its good to know that I don't have to do the less simplified way with all the rigorous work.

4. Nov 10, 2009

ideasrule

For the sphere's angular momentum about the center of the loop, you use mvr, where r is the distance from the loop's center to the sphere's center. For the sphere's angular momentum about its own center of mass, you'd use 2/5MR^2. For the total angular momentum you'd add them together.