# Angular Momentum of Rotating Cylinder

• cantgetaname
In summary, the conversation discusses finding the angular momentum of a cylinder rotating about its center of mass along its length, and how to calculate it for an axis on the edge of the cylinder that is parallel to the axis of rotation. The solution involves using the Parallel Axis Theorem to find the moment of inertia about the new axis, and then using the formula for angular momentum. The conversation also clarifies that if the cylinder is rotating about its center of mass, the angular momentum will be the same for any point along the axis, but it will vary if the axis of rotation is shifted.
cantgetaname
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.

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.

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!)

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Well, it is fun to find out things. One shortcut is that you need not bother about integrating with respect to z.

ehild

cantgetaname said:
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

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If I understood well the CM of the cylinder does not rotate about the new axis. Does it?

ehild

oops!

ehild said:
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!

tiny-tim said:
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
So if I'm not wrong, along the new axis the angular momentum will be the same as the one in the middle? (in this case)

that's right …

if the (instantaneous) axis of rotation passes through the centre of mass, then the angular momentum tensor is the same about any point, and the angular momentum is the same about any two parallel axes

(a bit like a couple being the same about any point)

## 1. What is angular momentum and how is it related to a rotating cylinder?

Angular momentum is a measure of the rotational motion of an object. In the case of a rotating cylinder, the angular momentum is the product of the moment of inertia and the angular velocity. This relationship shows that as the cylinder spins faster, its angular momentum increases.

## 2. How is the angular momentum of a rotating cylinder calculated?

The angular momentum of a rotating cylinder can be calculated by multiplying its moment of inertia (a measure of its resistance to rotational motion) by its angular velocity (the rate at which it rotates). The formula is L = Iω, where L is angular momentum, I is moment of inertia, and ω is angular velocity.

## 3. How does the angular momentum of a rotating cylinder change if its moment of inertia is increased while its angular velocity remains constant?

If the moment of inertia of a rotating cylinder is increased while its angular velocity remains constant, its angular momentum will also increase. This is because the moment of inertia is directly proportional to the angular momentum, and as the moment of inertia increases, so does the angular momentum.

## 4. What happens to the angular momentum of a rotating cylinder if its angular velocity is increased while its moment of inertia remains constant?

If the angular velocity of a rotating cylinder is increased while its moment of inertia remains constant, its angular momentum will also increase. This is because the angular velocity is directly proportional to the angular momentum, and as the angular velocity increases, so does the angular momentum.

## 5. How does the angular momentum of a rotating cylinder affect its stability?

The angular momentum of a rotating cylinder affects its stability by providing a stabilizing force known as gyroscopic stability. As the cylinder rotates, its angular momentum creates a force that resists any changes in its rotational motion. This helps to keep the cylinder stable and prevent it from tipping over.

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