# Problem finding angular momentum

1. Jun 15, 2014

### kelvin490

Angular momentum L is defined by summation of mrxv about a certain point. For a rigid body that rotate about an axis that is not the axis of symmetry, in general the direction of L is keep on changing. Also we get different L if we take about different point.

Iω is constant for constant ω and I is summation of ∫r2dm where r is the perpendicular distance between the element and the rotation axis. This value is constant even the rigid body is rotating about an axis that is not the axis of symmetry, in contrary with the above argument.

What is the restriction for using Iω as the value of angular momentum?

2. Jun 15, 2014

### king vitamin

Take a look at these lecture notes: http://farside.ph.utexas.edu/teaching/301/lectures/node119.html

The main result is that, if the object's axis of rotation is $\mathbf{k}$, then $\mathbf{k}\cdot \mathbf{L} = I_k \omega$, where $I_k$ is computed with respect to the axis of rotation (which I believe is what you meant in your second paragraph). So the projection of angular momentum upon the rotation axis is constant and equal to Iω, but the other components are not, and they change with time (the angular momentum will precess around the rotation axis).

So the answer to your question is basically that you can only use Iω for angular momentum if the angular momentum is along a symmetry axis, so there is no torque.

3. Jun 15, 2014

### mattt

If you have a XYZ frame (it does not matter if it is inertial or non-inertial, this will be a pure kinematics question), and you have a system of point particles, and each one of this particles describes a circle around Z-axis with exactly the same angular velocity (which can vary with time, but must be the same angular velocity for all particles of the system), then the total angular momentum vector of this system of point particles with respect to the origin point O of this XYZ-frame is:

$$\vec{L}(t) = \sum_i \vec{L}_i(t) = -\omega (t) \left(\sum_i m_i z_i(t) x_i(t)\right)\vec{u}_x - \omega (t) \left(\sum_i m_i z_i(t) y_i(t)\right)\vec{u}_y + \omega (t) I \vec{u}_z$$

(where "I" is the moment of inertia of the system of particles with respect to the Z-axis).

If the system of particles is symmetric with respect to the Z-axis, or if the system of particles is symmetric with respect to the XY-plane, then:

$$\vec{L}(t) = \sum_i \vec{L}_i(t) = \omega (t) I \vec{u}_z$$

This is only talking about the kinematics.

For dynamical questions there are some subtleties.

4. Jun 15, 2014

### kelvin490

For a non-symmetric object, we can still calculate the Iω about axis of rotation. Is this value equal to the magnitude of k component? Is the k component constant no matter which point we take the angular momentum?

5. Jun 15, 2014

### mattt

As kingvitamin and I already wrote, the projection of total angular momentum vector upon the rotation axis is constant (and does not depend on the point O of the rotation axis) and equal to Iω (if the particles are describing a circle around the rotation axis with the same constant angular velocity; obviously if the angular velocity changes with time, the projection of total angular momentum vector also changes with time, but this projection does not depend on the point of the axis).