How to find angular momentum of a body about an axis other than the axis of rotation?

The problem at hand is finding the angular momentum of a cylinder (rotating about its center of mass, with the axis of rotation along its length) about an axis on the edge of the cylinder (this axis is also parallel to the axis of rotation).

Some general explanation will help too.

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 You first find its moment of inertia about the new axis by using the Parallel Axis Theorum. http://en.wikipedia.org/wiki/Parallel_axis_theorem Then Angular Momentum = Iω
 Are you sure about that? I mean, the cylinder isn't rotating about that axis. I used conservation of angular momentum in a particular question using that method and the answers came out to be wrong. If, for example, you consider a particle with mass M moving along the radius then the angular momentum along the new axis using the method you suggested is: ( MR^2 + MR^2 ) w However if you simply use Mv x r, it changes with time. EDIT: OK, maybe not the best example, but I think my point's valid.

Recognitions:
Homework Help

How to find angular momentum of a body about an axis other than the axis of rotation?

You have two frames of reference with parallel axes, one with origin at the CM of the rotating cylinder, the origin of the other one somewhere on the x axis at distance D (which is equal to R now, but do it for the general case.) Write the position vector and velocity of a mass element of the cylinder in the new system and use the definition of the angular momentum to get the contribution of the mass element. Integrate for the whole cylinder.

ehild

 Hmmm... I'll do it. No shortcuts? I wanted to know this so that certain problems would be faster to solve.
 Yes, sorry guys, I misread the question. (Didn't think this would be in the Introductory Physics Forum!)
 Recognitions: Homework Help Well, it is fun to find out things. One shortcut is that you need not bother about integrating with respect to z. ehild

Blog Entries: 27
Recognitions:
Gold Member
Homework Help
 Quote by cantgetaname Some general explanation will help too.
hi cantgetaname!

the total angular momentum about an axis is the angular momentum about that axis of the whole mass concentrated at the c.o.m., plus the angular momentum about a parallel axis through the c.o.m.

(ie total angular momentum = orbital angular momentum + spin)

(same as ∑ (d + ri) x (ω x miri) = d x (ω x ∑ miri) + ∑ ri x (ω x miri) = d x (ω x mtotalrc.o.m.) + ∑ ri x (ω x miri))

if, as in this case, the body is rotating about its c.o.m., then vc.o.m. = 0,

so L = mdvc.o.m. + Ic.o.m.ω, = 0 + Ic.o.m.ω = Ic.o.m.ω

but if the body is rotating about the new axis, then vc.o.m. = dω,

so L = mdvc.o.m. + Ic.o.m.ω, = (md2 + Ic.o.m.)ω, = Inew axisω

 Recognitions: Homework Help If I understood well the CM of the cylinder does not rotate about the new axis. Does it? ehild

Blog Entries: 27
Recognitions:
Gold Member
Homework Help
 Quote by ehild If I understood well the CM of the cylinder does not rotate about the new axis. Does it? ehild
oops! i didn't read the question properly and i didn't even read what i'd written properly

i've edited my previous post to correct it

thanks, ehild!

 Quote by tiny-tim if, as in this case, the body is rotating about its c.o.m., then vc.o.m. = 0, so L = mdvc.o.m. + Ic.o.m.ω, = 0 + Ic.o.m.ω = Ic.o.m.ω but if the body is rotating about the new axis, then vc.o.m. = dω, so L = mdvc.o.m. + Ic.o.m.ω, = (md2 + Ic.o.m.)ω, = Inew axisω