# Physics! #21: University Physics 3.4

3.4 Motion in a Circle

Uniform Circular Motion

First off, let’s define what we mean.

By “circular” we just mean you’re traveling in a perfect circle. That is, you’re traveling in such a way that one side of your car is always perpendicular to some fixed point. It’s not just any curvy motion, it’s motion in a perfect circle.

By “uniform” we mean that if you check the spedometer, it’ll always return the same number. That is, you’re at “constant speed.” Note that we can’t say you’re at constant velocity because velocity is a vector. Vectors combine direction and magnitude, and while your magnitude remains constant, your direction is always changing as you go round and round the circle.

Before I get to the math, we should cover what happens if you’re not uniform or not in a circle.

Imagine you wake up in a train car with no windows. You can tell by sound that the train is moving, and you would like to know if you’re going somewhere or just looping around.

First off, you can quickly tell if you’re curving or not. When a train goes around a curve, you get pushed to the side. Remember, you can feel acceleration. You can’t feel velocity. If you don’t feel any acceleration, you’re at uniform speed going in a straight line. If you feel acceleration pushing you back, it’s because the train is speeding up. If it pushes you forward, the train is slowing down.

Think of it this way: When the train slows down, the speed change has to get communicated to your body before you slow down. It takes a little time, since you aren’t firmly attached to the train. So, if the train loses speed, it loses that speed a liiiittle before you do. So, you pitch forward. If the train gains speed, it gains before you, so you fall back.

Let’s assume you feel no backward or forward acceleration. So, at least you know you’re at uniform speed. Now, you see if you’re ever being pushed left or right. Again, same principle – if you’re moving left it’s because the train is accelerating right and it takes a minute for your body to get the message. Let’s say you feel yourself being pushed left. You focus for 10 minutes, and you seem to be accelerating the whole time. So, you guess you’re probably going in some sort of loop. Now you want to know if you’re going in a circle or in an ellipse.

Try to imagine what it feels like to go around a race track, which is close to the shape of an ellipse. You get scrunched to one side of the car really hard when you take the sharp turns. On the straightaway, you don’t get that at all. On a big enough ellipse, the long part of the track is pretty close to a straightaway. So, as you go, you’ll feel a lot of acceleration at certain points, and just a little at others.

Fortunately, planets follow ellipses, so there are lots of cool animations showing the accleration vectors, like this one.

See how you’re changing velocity most as you go around the bend?

Let’s say, in your train car, acceleration actually seems to be constant. This tells you that you’re not going around an ellipse. So, from all this, you deduce that you must be traveling at uniform velocity in a circle. Pretty cool, right?

The book’s proof is a little funky, as it depends on similar triangles. There’s a much nicer proof if you’re allowed to use polar coordinates, which naturally deal with circles, but we don’t have that tool in the kit yet. That said, Khan gives a slightly different version of the same idea here, which is a little easier to follow than the book.

By whatever means you arrive at it, the formula that pops out is:

$a_{rad}=\dfrac{v^2}{R}$

That is, if you’re driving around in a circle, your radial acceleration is equal to speed on your spedometer squared, divided by your distance from the center.

This makes sense for a few reasons. First, the units stack up right: since it’s acceleration, the right side’s units need to be in meters per second per second. Speed squared gives us meters squared per second per second. Then, we divide off the radius, which is in meters.

Second, it confirms things you know from driving. If the radius is huge, the acceleration is small. If you’re in a jet, and you start off in Texas and go along a circle’s circumference to end up in Alaska, you will probably not notice the acceleration, since the radius will be so enormous compared to your velocity. However, if you’re riding a little scooter around at 10 kmph on a circle with a 3 meter radius, you may well get thrown off. This is why you get scrunched to the side of a roller coaster car when it takes a turn, but you feel nothing when your car turns. Rollar coster turns are extremely sharp, whereas highways tend to turn slowly so your car doesn’t flip over.

Third, you’ll notice everything here is in magnitudes, not vectors. This again makes sense. Radial acceleration doesn’t point in a single direction. It always points from you to the center of the circle, but that isn’t a single direction. We also have the magnitude, not the vector, of velocity, for similar reasons.

The book goes on to give you another equation for radial acceleration:

$a_{rad}=\dfrac{2{\pi}^2R}{T^2}$

T refers to the period, which is the time it takes for one revolution. The equation is derived from the statement of velocity as “distance per period.” We know the distance is the circumference, which is given by 2πr. So, you take the v from the first equation, and switch it to “2πR/T.”

You can (and should!) check for yourself to make sure the units all work out.

Nonuniform Circular Motion

Nonuniform circular motion is motion where your speed varies. It’d be like driving around a circle and hitting the gas and brakes from time to time.

Whereas in uniform circular motion the acceleration vector always pointed to the center, now the acceleration vector points somewhere else depending on how you’re handling the gas and brake.

Here’s how to think about it. At any given time, when you’re moving around a circle, you have a radial acceleration vector that points toward the center. This is for all the reasons explained earlier. However, you should always think about it physically. When you are driving around the circle, you will feel pushed outward to some extent. That tells you that the car is accelerating toward the circle’s center.

However, now there’s another vector. If you’re speeding up at a certain point, you have to combine two acceleration vectors: One pointing inward and one pointing forward. If you’re slowing down at some point, it’s one pointing inward and one pointing backward.

The inward one is just radial acceleration pointing toward the center as defined before. The new one is your tangential acceleration, which is given by your change in speed. That is, the change in what your spedometer says over time.

“Fuck you, Zach!” you exclaim. “Just a second ago, you said radial acceleration was a magnitude, NOT a vector.” Overall, that’s true. However, at a given instant, you do have an acceleration vector that points from where you are at that instant to the center of the circle.

Once again, think about this stuff physically. If you’re zipping around a circle, you feel pushed outward. If you suddenly hit the breaks, where does your body fly? Out and to your right. Why? Because relative to you, the car is accelerating inward and backward. If you’re not buckled in, your body won’t get the info, and you’ll go flying diagonally out the window.

If you suddenly speed up around a turn, you still go outward, but you also get jerked back into your seat. That’s the way you combine these vectors.

So, the neat thing is this: If you want to know which way acceleration points as you go around a circle at nonuniform speed, all you need is a function for velocity and the length of the radius. That’s enough to get your two vectors, which you just add up.

This section closes with an important distinction between the derivative of the magnitude of velocity and the magnitude of the derivative of velocity. It’s important to understand the distinction here.

In the first case, you’re getting the speed (mag of velocity) THEN taking the derivative. So, the question you’re asking must be “how did the speed on my spedometer change at that instant.” You could be going 100 miles per hour. If you don’t change your speed, then the change in magnitude of speed is zero.

In the second case, you’re taking the derivative first, so you still care how your direction changed. THEN you’re asking the magnitude of that change. So, if you’re going 100 miles per hour in a circle, the value actually does change, since the direction changes.

Got it?

I know some of this can be a little sticky, so I highly recommend working the problems for these.

Next stop: Relative velocity!

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### 2 Responses to Physics! #21: University Physics 3.4

1. david says:

if you switch the v in the first equation to a 2pi r/T you get 4 pi^2 *r/T^2 .

• ZachWeiner says:

Oops, fixed!