The Physics Philes, lesson 42: Angular and Linear Movement, Long-Lost Cousins

In which angular and linear motion are related, a discus is thrown, and math is a hero.

If you’re a smart cookie, you may have anticipated what we’re going to discuss today. The equations for linear and angular motion with constant acceleration are incredibly similar. No one would fault you if you thought we might be able to relate them in some way. We can! And we’re going to do that today.

So we have a spinning thing, and we want to know the linear velocity and acceleration of one spot on that spinning thing. What are we to do? Don’t worry. Math comes to the rescue.

We know that when a thing is rotating around a fixed axis, all particles on that thing moves in a circular path. The speed of the particle is directly proportional to the angular velocity of the spinning thing. As the spinning thing spins faster, the speed of the particles goes up. No big surprise there. Do you remember how we can find the distance traveled by a spinning particle? We find the arc length, which is the radius of circle times the angle in radians. To get to the relationship between linear and angular speeds, we need to take the time derivative of s = rΘ (but r is a constant for any one particle) and take the absolute value of both sides. Now our equation looks like this:

The right side of this equation is equal to the instantaneous linear speed of the particles. The left side (sans r) is the magnitude of the instantaneous angular velocity. What does this mean? It means that we can rewrite the above equation really simply:

In other words, the linear speed of a particle on a spinning thing is equal to the radius of the spinning thing times the angular speed. The greater the distance the particle is from the center, the faster it goes. This makes sense. It needs to travel faster, because it’s traveling a greater distance in the same amount of time. The direction of the linear velocity vector is tangent to the circular path.

OK, great. But what if we need the linear acceleration? What then? Again, math is on the case.

The acceleration of a particle moving in a circle can be represented by the centripetal and tangential components. The tangential component of acceleration is the component parallel to the instantaneous velocity. It changes the magnitude of the particle’s velocity and is equal to rate of change of speed. When we take the derivative of v = rω:

Notice that α = dt is the rate of change of the angular speed. If that equation had a subscript of z paired with α and ω, that would be the rate of change for angular velocity (speed and direction).

We can derive the centripetal component of acceleration from v = rω, as well:

This is true all the time, not just when v and ω are constant. The centripetal component always points toward the axis of rotation. The equations for arch length and the centripetal component of acceleration, as well as the equation that relates linear speed to angular speed are good for any particle that has the same tangential velocity as a point in the rotating rigid body.

Now, it’s a beautiful day outside, but I hate the sun so I think I’ll stay in and work an example problem. Let’s begin.

A discus thrower moves a discus in a circle with the radius of 0.800 m. At a particular instant, the thrower is spinning around at an angular speed of 10.0 rad/s, with the angular speed increasing at 50.0 rad/s^2. What are the centripetal and tangential components of the acceleration of the discus, and what is the magnitude of the acceleration?

What do we know? We know that the radius r = 0.800 m, the angular speed ω = 10.0 rad/s, and the rate of change of angular speed α = 50.0 rad/s^2. Also, since two of our target variables are components of the acceleration vector, it might help to diagram this problem. Oh look! Here it is:

Let’s use the equations we just learned to find the centripetal and tangential components of acceleration first. Just plug the values into the equation:

Simple! Now, look back at the diagram. We need to find magnitude of acceleration a. To do that, all we need to use the Pythagorean theorem. See?

The Pythagorean theorem, as you know, says that the square root of the sum of the squares of the two legs of a right triangle is equal to the square root of the square of the hypotenuse. Or, a^2 + b^2 = c^2. Just plug in the numbers:

We did it! We figured out the centripetal and tangential components of acceleration and the magnitude of the acceleration vector! Good job!

As usual, let me know if I made any mistakes or if anything needs clarified. I’m working on something a little different for next week. It’s a surprise. (Hint: It might involve black holes. But I’ve said too much…)

Featured image credit: Franklin Park Library