# The Physics Philes, lesson 54: Axis in Motion

So far in my studies of rotational motion, I’ve focused on rotating rigid bodies where the axis of rotation is stationary. But what about when the axis of rotation is actually moving? What then? Well all your questions will be answered in the coming weeks.

I should probably first establish what I’m talking about when I say the axis rotation is moving. Picture a baton twirler. Or, if you’d rather, this mace twirler I found on YouTube:

The baton (or mace) is rotating around its center of mass. But it’s also flying through the air from one hand to the other. What we have is what physicists call combined translation and rotation. Translational motion is when the center of mass moves from one position to another. Most motion we see is some combination of translation and rotational motion. In fact, every possible motion of a rigid body can be represented as a combination of translational motion of the center of mass and rotation about an axis through the center of mass.

The kinetic energy of a rigid body has both translational and rotational motions. Don’t believe me? I’ll prove it!

For a bit of illustration – and no post would be complete without one – I drew you a diagram.

This is supposed to be a diagram of a rigid rotating body, with a typical particle denoted by m_i.. The velocity of the particle is the vector sum of the velocity of the center of mass and the velocity of the particle relative to the center of mass.

Now let’s dig back into our brain baggage and remember that kinetic energy can be expressed with the equation 1/2mv^2, which can also be expressed like this:

If we substitute that vector sum equation into this we eventually get:

The total kinetic energy is the sum of all the particles making up the body. If we express the three terms in this equation as separate sums, we get:

We can rearrange some stuff…

Okokok. Let’s see what we have. See the m in the first term? That is the total mass M of the rigid body. The second term is zero because it’s equal to the total mass times the velocity of the center of mass at the center of mass. The last term is the kinetic energy of the rotation around the center of mass. We can replace the final term with the equation for the kinetic energy of a rotating body and we get

The first term is associated with the motion of the center of mass and the second term is associated with the rotation about an axis through the center of mass. Voila! We have a relationship!

This is just an introduction. Next week we’ll dig a little bit deeper into this concept and hopefully it will start to make a little bit more real world sense.

*Featured image credit: Flickr*

## 2 Comments

“The second term is zero because it’s equal to the total mass times the velocity of the center of mass at the center of mass.”

It’s not really clear that this should be zero. This statement is saying that you’re multiplying the mass times a non-zero vector; it should be a scaling, which isn’t zero.

It’s also not clear how you arrived at that expansion. The expanded form you wrote (if I may use my own variables since I can’t easily post LaTeX: v_cm = c, and v_i = v):

c^2 + 2c•v’ + v’^2

is from

(c + v’)^2

which is not what you get when you substitute in the previous equation. Rather, I think that the middle term shouldn’t exist at all given how you’ve set it up:

(c • v)

c • (c + v’)

c • c + c • v’

And so the first is merely the kinetic energy entirely due to translation, and the latter is kinetic energy due to changing momentum of the point in question (which in this case will be caused by rotation for all of them) – but that last part of course is only redundant to what you’ve said above.

I’m going to need to mull this over… 🙂