Solving Circular Motion Problems, Part 2

Guide 6-3b. Solving Circular Motion Problems, page 2

As you study the following example problems, remember the five important things about solving circular motion problems.

Circular motion is the result of a net force pulling the moving object toward the center of the circular path.
The forces that give rise to circular motion are friction, normal, weight, and tension.
Centripetal force, like net force, is never drawn on a force diagram.
Select the direction of positive displacement to be toward the center of the circular path.
The magnitude of centripetal acceleration is given by the square of the magnitude of the velocity of the object divided by the radius of its circular path.

The Level Curve

A typical way to state this problem is: How fast can you go around a curve in a car without skidding off the road? There are two variations. One is a level road and the other a banked road. The banked road without friction is actually just a variation of the horizontal swing that is solved on the previous page. The banked road with friction is a complex circular motion problem, and we don't consider it here. We do the level road problem instead. This is a situation where the force keeping the object (the car) in circular motion is static friction. The textbook does this as an example. Unfortunately, the text doesn't use the notation f_s,max to represent the maximum static friction force available. So we'll redo the example correctly here. We give overhead and side views below like we did for the last problem. Figure A looks the same as before. In Figure B, the overhead view shows the static friction pointing toward the center of the circular path. The side view shows the balancing forces.

Figure A	Figure B

We'll use f_s below with the conventional meaning that it represents any value of static friction force up to and including the maximum available.

F_net,x = f_s
F_net,_y = N - mg

Applying the second law,

ma_x = f_s
0 = N - mg.

For static friction, the inequality f_s ≤ µ_sN applies. Now N = mg from the vertical net force equation. Therefore, f_s ≤ µ_smg. For f_s, we can substitute ma_x from the horizontal net force equation. This gives

ma_x ≤ µ_smg.

We divide out the mass and substitute v²/r for a_x.

v²/r ≤ µ_sg

Solving for v,

v ≤ (µ_sgr)^½.

We carried the inequality all the way through so that we could show that the magnitude of the velocity has to be less than (µ_sgr)^½ in order for the car to remain in its circular path. An alternative to solve the problem is to use f_s,max instead of f_s and then use the equality, f_s,max = µ_sN.

Car Topping a Hill

For some reason, cars and amusement park rides make good candidates for circular motion problems. We'll do a roller coaster problem after this one. For now, let's look at the forces acting on a car as it tops a hill. We'll assume the top of the hill is a circular arc. The textbook does a similar problem for a car in a dip. You should compare that solution to the following one. The situation is represented by the odd-shaped car on the funny-looking hill to the right. The car is moving to the left. You know from experience that if you top a hill too quickly, you seem to come off of your seat. (Actually, the car is dropping below you as you tend to move in a straight line.)

Let's see what the forces are. Consider the force diagram to the left. We pick the positive direction down, because the acceleration must be toward the center of the circle. That tells us immediately that W must be greater in magnitude than N. That's why we made N a smaller vector in the diagram. The net force equation is:

F_net = W - N

Applying Newton's 2nd Law and solving for N,

ma = W - N

N = mg - ma

This tells us that the normal force is smaller than the weight by an amount ma. If you think of normal force as apparent weight, then this makes sense. The apparent weight is less than the actual weight. A situation where this effect is used to reduce apparent weight to zero is in the Vomit Comet. The plane climbs and then drops steeply. As it does so, the occupants undergo a period of apparent weightlessness.

Looping Roller Coaster

Consider the occupants of a roller coaster car upside down at the top of a loop. Their acceleration is down, yet they don't fall out. As they move to the right, the track is forcing them into a circular path. The force diagram is shown to the left. The normal force of the track and the weight both point down, so the net force equation is

F_net = W + N

Applying Newton's 2nd law, ma = mg + N, and N = ma - mg. From the way we've set up the problem, we know that N ≥ 0. Therefore, ma ≥ mg and a ≥ g. That's the condition that's necessary for the passengers not to fall out of their seats. In the case that a = g, then N = 0, and the seats exert no normal force on the passengers. The passengers are in free fall but, of course, the car is falling with them. While they feel weightless, they're still safe. How fast must the car be traveling in this case? It's easy to figure out. Substitute v²/r for the acceleration.

a ≥ g

v ²/r ≥ g

v ≥ (rg)^½

So the magnitude of the velocity must be greater than or equal to (rg)^½ in order for the occupants to remain in the car. Let's look at the case that a = 2g. Then N = 2mg - mg = mg. The normal force is equal to the actual weight. Thus, a passenger would feel just as heavy as they normally do on the ground. If they closed their eyes, they wouldn't know they were upside down. Your sensation of 'up' is actually related to how the surface you're on pushes on you.

Vertical Swing

The favorite example of this type is Tarzan swinging on a vine. What is the tension force exerted by Tarzan on the vine as it swings through the lowest point? (Tarzan might want to know if there's a chance the vine will break.) We've drawn the force diagram for Tarzan at the lowest point. The acceleration is up toward the branch around which the vine swings. This means the tension force must be greater than the weight. The net force equation is

F_net = T - W.

Applying the second law gives ma = T - mg. The tension force is then

T = m(a + g)

If we know the radius of the swing as well as Tarzan's speed at the lowest point, we can calculate the tension force using

T = m(v²/r + g).

To finish the problem, the force of the vine on Tarzan has, by Newton's 3rd Law, the same magnitude as the force of Tarzan on the vine.