The Physics Philes, lesson 12: ‘Round and ‘Round It Goes

In which cars are driven, friction is revisited, and sleds slide.

Last week I tried to explain circular motion. I spent this week trying to understand how the circular motion equations are applied. Success in this regard has been…um…nonuniform. But I think I have a pretty good grasp of the basics. Let’s see if I’m right!

First, let’s do a little review. As we know from last week, when a particle moves in a circular path at a constant speed, the particle’s acceleration is directed toward the center of the circle. (Remember, there is always acceleration toward the center because the particle is constantly changing direction.) As we saw last week, this centripetal acceleration can be expressed in terms of speed and the radius of the circle:

Centripetal acceleration can also be expressed in terms of period T, or the time it takes for a particle to make one trip around the circle:

Uniform circular motion is governed by Newton’s second law of motion. Remember Newton’s second law? Force equals mass times acceleration. Or, in math: ∑F = ma. For a particle to move in a circle, the net force on the particle must be directed toward the center. We can find this net force by substituting the a sub rad = v squared/R into Newton’s famous equation:

The particle doesn’t need to go in a full circle to find the net force. The equation is good for any path that can be regarded as part of a circular arc.

Review over! Let’s try some questions!

Let’s say we’ve got a 25.0 kg sled on some frictionless ice. The sled is tied to a pole by a 5.00 m piece of rope. Once the sled is given a little push, the sled revolves uniformly around the pole and makes five revolutions a minute. What is the force exerted on the rope?

Since the sled is moving uniformly, we are dealing with uniform circular motion, which means that the only acceleration we have to deal with the is radial (or centripetal) acceleration. To find our target variable (force F) we’ll use Newton’s second law of motion.

Here is a diagram of a sled going in a circle:

Look at that sled go! ‘Round and ’round and ’round…

But it’s not very useful in solving this problem. Here is a (more helpful) free body diagram:

Free-body diagram of our sled situation.

I included both the situational diagram and the free-body diagram because, if you’re like me, you might have a little trouble figuring out which way to orient your axes. But look, I’ve oriented the positive x-axis to point toward the middle of the circle, because that is the direction the radial acceleration must be pulling.

The question doesn’t tell us what the acceleration is, so before we use Newton’s second law we have to find the radial acceleration. Which means we’ll have to use:

The problem says that the sled makes five revolutions every minute. But to find period T we need to figure out how long it takes for the sled to make one revolution.

T = (60.0 s)/(5 rev) = 12 s

Now that we know the value for period T, we can plug in the values to the radial acceleration equation:

Now that we know radial acceleration, we can plug the numbers into the equation for Newton’s second law of motion:

So the force exerted on the rope is 34.3 N. But that question involved a frictionless surface. Let’s try one that involves friction.

Let’s say we have a car rounding a flat, unbanked curve with radius R. If the coefficient of static friction between the tires and the road is μ sub s, what is the maximum speed at which the driver can take the curve without sliding?

OK, since the car’s acceleration as it rounds the curve has a magnitude a = v^2/R, the maximum speed with correspond with the maximum radial acceleration and the maximum horizontal force on the car toward the center of its circular path. The only horizontal force acting on the care is the friction force exerted by the road. So the equations we’ll need are

Here is a free-body diagram for our situation:

It includes the static friction force, weight, and the normal force. The friction force must point toward the center of the circular path in order to cause the radial acceleration. Since the car doesn’t slid toward or away the center of the circle, the friction force is static friction: f sub max = μ sub s (n).

Acceleration is toward the center of the circular path, and there is no vertical acceleration. That means we have:

That second equation basically says that the magnitude of the normal force is equal to magnitude of the weight. The first equation says that the force needed to keep the car moving in a circular path increases as the car’s speed increases. But if we apply what we know about static friction, we know that the force available is f sub max = μ sub s (n) = μ sub s (mg). This will limit how fast the car can go around the track. If we substitute f sub max for f and v sub max for v in the net force x equation, we get:

That means that the maximum speed to the car can be expressed with this equation:

So let’s say that the coefficient of static friction is 0.96 and the radius is 230 m. The maximum speed of the car would be:

There are so many more different ways of using the uniform circular motion equations! Unfortunately, some of them I had a hard time wrapping my head around. Like conical pendulums. I have a feeling that my issues with that stems from my only very tenuous grasp of trigonometry. (I know! I’m working on it. I promise!)

Does all of this make sense to you? Did I mess anything up? Do you like my new way of typing equations? Please let me know!

Featured image credit: Mr. Fujisawa

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Mindy is an attorney and Managing Editor of Teen Skepchick. She hates the law and loves stars. You can follow her on Twitter and on Google+.


  1. September 7, 2012 at 11:41 pm —

    Uh oh, physics without trig? What have you gotten yourself into. ^_^

    • September 8, 2012 at 12:01 am —

      OK, to be fair, it’s just been a really, really long time since I’ve used trigonometry on a regular basis. It’s taking me some time to remember everything. One of these weeks I may take a math interlude and re-teach myself trig.

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