# The Physics Philes, lesson 61: Don’t Be So Rigid

Up off your bums, boys and girls. It’s physics time. Last week we started a more in depth discussion of angular momentum. This week, as promised, we will dive into angular momentum of a rigid body.

Remember when we figured out the angular momentum of a particle equals the product of the velocity, mass, distance, and the lever arm for the momentum vector? The lever arm in this situation is the perpendicular distance from the velocity vector to the origin, which is equal to the distance vector times sinΦ. (Check out the illustration in last week’s post for clarification. We can use that relationship to find the total angular momentum of a rigid body rotation about the z-axis with a particular angular speed.

Let’s pretend that we have a thin slice of a rigid body lying on an xy-plane, and that rigid body is spinning around the z-axis which goes through the origin. As each particle spins around, its instant velocity vector is perpendicular to the position vector. If we were to draw it out, it would look something like this:

(Hey, what would The Physics Philes be without a crappy paint drawing from yours truly?)

OK, so the value of sinΦ is 1 for every particle because Φ = 90 degrees. A particle with a certain mass at a certain distance from the origin has a speed equal to the distance times the angular speed. That means that we can determine the magnitude of angular momentum by swapping out the linear speed and replace it with the product of the distance from the origin and the angular speed:

Voila! But we aren’t done. If we want to figure out the total angular momentum of the slice of the body lying on the xy-plane, we just add up all the values of L for every particle, like so:

Whoa! Look at that! It’s the equation for moment of inertia! Specifically, it’s the moment of inertia for the slice about the z-axis. This same calculation holds for other slices of the same body that are parallel to the xy-plane. But for points that are not parallel to the xy-plane, it gets a little trickier…or does it? In that case, the distance vectors will have components in the z-direction in addition to the x- and y-directions. That gives the angular momentum a component that is perpendicular to the z-axis. But! If the z-axis is an axis of symmetry there are no worries. The perpendicular components for the particles on opposite sides of the axis will add up to zero, so when the body rotates its angular momentum vector lies along the symmetry axis and it’s magnitude equals the moment of inertia times the angular speed.

But we’re *still* not done! We still need to talk about the rate of change of total angular momentum. I wrote briefly last week about how the rate of change equals the torque of the net force acting on the particle. For any body – both rigid and nonrigid – the rate of change of the total angular momentum is equal to the sum of the torques of all the forces acting on the particles. Since the torques of the internal forces add up to zero (if the forces act along the line from one particle to another), the sum of the torques only includes the torques from the external forces. That means the total torque is:

This can be used for any body, even if it’s not rigid. That’s not true for what comes next. If the body is rigid and rotating about an axis of symmetry, then

and I (aka the moment of inertia) is constant. If the axis has a fixed direction, then the angular momentum and angular velocity vectors will not change direction, only magnitude. In that case, we can use:

That equation is the basic relationship for the dynamics of rigid body rotation.

So far we’ve been talking about what happens if the axis of rotation is also an axis of symmetry. That’s fine, but what happens if the axis of rotation is not an axis of symmetry? What then, smart guy? In that case, the angular momentum is, in general, not parallel to the axis so, as the body turns, the angular momentum vector traces a cone around the rotation axis. Since the angular momentum vector changes, there mus be a net external torque acting on the body even though the magnitude of the angular velocity may be constant.

Boom! That’s angular momentum of a rigid body. Let it sink in and soak into all the crevices of your wrinkly brain. (Ewww…) Next week we’ll move on to conservation of angular momentum. This stuff is getting pretty good.

*Featured image credit: Stefan Andrej Shambora*

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