# The Physics Philes, lesson 52: Torque, This is Your Moment (of Inertia)

Oh. My. Glob. This is week 52. Fifty-two weeks (give or take a couple) of studying physics. It’s been, at various times, frustrating, fun, and rage-inducing. But it’s mostly been a hoot and a half, and I’m nowhere near done. So! Let’s get on with it!

As promised, this week I’ll try to explain the relationship between the angular acceleration of a rigid body and the torque of a rigid body. To explain this, I’m afraid I have to poorly draw a diagram. I’m so very sorry. I’ll try to explain what it means in human words. But first, here’s that drawing:

Aahhhh what does this mean? We need to image that we have some kind of body made out of a bunch of particles, and we pretend that this body is rotating around the z-axis. Let’s say the first particle as a mass of m_1 and is distance r_1 from the axis. The net force of the particle has two components – F_1,rad and F_1, tan – as we learned from the last couple of posts. If we think waaaay back to when we discussed Newton’s second law, we know that Newton’s second law for the tangential component is

Right? Remember F=ma? That’s what we did there. But we’re not done yet. We can express tangential acceleration of our particle in terms of the angular acceleration of the body. What I means is we can plug this equation

into the previous equation to get this, multiply both sides by r and get:

Look familiar? We know from previous posts that the left side of the equation is just the magnitude of torque with respect to the rotation axis, which is equal to the component of the torque vector along the rotation axis. (Remember that F_1,rad and F_1z on our diagram do not contribute torque about the z-axis, so the left side of the equation is the total torque acting on the particle.)

Notice anything familiar about the right side of the equation? You should, because the right side (sans α, that is) is the moment of inertia of the particle about the rotational axis! So, for our little particle, we can rewrite the equation as:

We just need to do this equation for every single particle, and then add those equations up. We’d get the sum of the torques about the rotational axis equals the total moment of inertia times the angular acceleration. Because we’re working with a rigid body, the angular acceleration for all the particles will be the same. The result? The rotational analog of Newton’s second law of motion:

Cool, right? The net torque on a rigid body is equal to the body’s moment of inertia times its angular acceleration, just as Newton’s second law says that the net force is equal to a body’s mass time its acceleration.

Even though the torque of each particle is due to the net force (i.e. the vector sum of the external and internal forces), on each particle, Newton’s third law tells us that the internal forces add up to zero. For that reason, the sum of torques only includes the external forces.

It’s important to note that this only works for rigid bodies. It wouldn’t hold for anything that has different angular accelerations at different spots. Also, be sure to use rad/s^2. Trust me. It won’t work otherwise.

That’s all for now. Stay tuned next week for when I try to do some problems involving the rotational dynamics of rigid bodies. See you then!

*Featured image credit: Flickr*

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