AP PHYSICS C: MECHANICS › UNIT 2, FORCE AND TRANSLATIONAL DYNAMICS › TOPIC 2.10
Circular Motion
An object going round at a steady speed is accelerating the whole way, toward the center, because velocity carries a direction. No new force appears anywhere in this topic. The old ones are simply asked to point inward.
2.10.A Describe the motion of an object on a circular path, find its centripetal acceleration, and identify the real forces that produce it.
2.10.B Describe circular orbits using the relation between period and radius.
This topic covers two sheets. Click either one to open it full size, and print them together.
1. Steady speed, changing velocity
An object going around a circle at a constant speed is accelerating the entire time. That sounds like a contradiction only if you are treating speed and velocity as the same thing. Velocity carries a direction, the direction is changing every instant, so the velocity is changing, and a changing velocity is an acceleration.
The acceleration points toward the center of the circle, and its size is
ac = v² / rdirected toward the center
Note how sharply it depends on the speed. Doubling the speed around the same bend quadruples the acceleration, and therefore quadruples the force something has to supply.
A ball on a circular path. The faint copies show the velocity at other points: the same length every time, a different direction every time. Then cut the string and see where it actually goes, which is not outward.
2. There is no such thing as a centripetal force
This is the most important sentence in the topic. Centripetal describes a direction, not a kind of force. The centripetal acceleration is produced by forces that are already on the free-body diagram: a tension, a normal force, friction, gravity, or a combination of them.
So a free-body diagram never carries an arrow labeled centripetal force. If it does, something has been counted twice. The correct sequence is to draw the real forces, add them toward the center, and set the sum equal to mv²/r.
ΣFtoward the center = m v² / r
Three circular motions with three different providers. In each case the right hand side is what the circle demands, the left is what the situation can supply, and the interesting question is always what happens when the demand wins.
3. The top of the loop
At the top of a vertical circle, the weight already points toward the center. If the circle needs more inward force than the weight supplies, the track presses down as well and everything is fine. If the circle needs less, the track would have to pull the object upward, and a track cannot pull. The object leaves it.
The boundary case is the normal force reaching exactly zero, which gives the minimum speed for staying on the loop:
mg = mv² / rso vmin = √(gr)
The mass cancels, so it is the same minimum speed for a marble and for a roller coaster of the same loop radius.
4. Banking the curve
Tilt the road and the normal force is no longer vertical, so a component of it points toward the center of the bend and helps with the turn. For one particular speed the banking does the whole job and no friction is needed at all:
tan θ = v² / (gr)the speed the bank was designed for
AP Physics 1 stops there. This course does not. Below that speed the car tends to slide down the bank and friction acts up the slope; above it the car tends to slide up and friction acts down. Put those two extremes into the same two equations and you get a range of safe speeds rather than a single one, which is what a real road has. The free-body diagram is the same three arrows in every case: weight, normal force, and friction along the surface. Only the direction of the friction changes.
5. Speeding up while turning
If the speed is also changing, there is a second acceleration, tangential, along the direction of motion. The total acceleration is the vector sum of the two, and it no longer points at the center. A car accelerating out of a bend has both.
Uniform circular motion is the special case with no tangential part, and it is the only case this course asks you to handle quantitatively.
6. Going round and round: period and frequency
The period is the time for one full circuit and the frequency is the number of circuits per second. They are reciprocals, and for a steady speed the period follows from the circumference:
T = 1/fT = 2πr / v
For a satellite in a circular orbit the only force is gravity, so the gravitational force supplies the whole centripetal requirement. Setting the two expressions equal and rearranging gives a relation between the period and the radius that depends only on the mass of the central body:
T² = (4π² / GM) R³
Everything orbiting the same central body obeys it with the same constant, which is why the period of any satellite tells you the mass of the planet it is going around.
7. Reading it wrong
Adding a centripetal force to the diagram. Never. Identify the real forces and add their inward components.
Drawing an outward force. There is nothing there to exert one. If your diagram has an outward arrow, ask which object is exerting it, and the answer will not come.
Thinking a cut string sends the object outward. It goes straight, along the tangent, at the speed it had. Try the button in the first simulation.
Saying a constant speed means no acceleration. Constant velocity means no acceleration. Circular motion is the standard counterexample.
Check yourself
1. A ball on a 0.80 m string moves in a horizontal circle on a smooth table at 4.0 m/s. The ball has mass 0.50 kg. Find the centripetal acceleration and the tension, and say what happens to the tension if the speed doubles.
ac = 16/0.8 = 20 m/s² and the tension is the only inward force, so FT = 0.5 × 20 = 10 N. Doubling the speed quadruples both, to 80 m/s² and 40 N, because the speed is squared. This is why strings and tires fail suddenly rather than gradually as the speed rises.
2. A car takes a flat curve of radius 40 m with μs = 0.70. Find the maximum speed, and explain why the mass of the car does not appear.
Friction supplies the whole inward force, so μsmg = mv²/r at the limit, giving v = √(μsgr) = √(0.7 × 9.8 × 40) = 16.6 m/s. The mass cancels because it appears on both sides: a heavier car needs more inward force and gets proportionally more friction. A loaded truck and an empty one slide at the same speed on the same bend.
3. A curve of radius 50 m is banked at 12°. Find the speed at which no friction is needed, and say in which direction friction acts if the car goes faster than that.
v = √(gr tan θ) = √(9.8 × 50 × tan 12°) = √(104.2) = 10.2 m/s. Faster than that and the banking alone supplies too little inward force, so the car tends to slide up the bank and static friction acts down the slope, adding its own inward component. Slower, and friction acts up the slope instead. The three arrows on the diagram never change, only the direction of one of them.
4. A student draws a free-body diagram of a car on a flat curve and includes an outward arrow labeled centrifugal force. Explain what is wrong and what the passenger is actually feeling.
No object is exerting an outward force, so the arrow fails the two-noun test from Topic 2.2 and does not belong. The passenger is going straight, as the first law promises, while the car turns; the door then pushes them inward, and the pressure of the door is what they feel. The sensation is real and the outward force is not.
5. Find the minimum speed for a ball to stay on the inside of a vertical loop of radius 1.2 m at the top, and describe the free-body diagram at exactly that speed.
v = √(gr) = √(9.8 × 1.2) = 3.4 m/s. At exactly that speed the track exerts no force at all, so the diagram has one arrow: the weight, pointing down, which is toward the center at the top of the loop. Any slower and the required inward force is less than the weight, the track would have to pull outward, and the ball falls away from the track.
6. Two satellites orbit the same planet, one at radius R and one at 4R. Compare their periods, and say what a measurement of one period and one radius would tell you about the planet.
Period squared goes as radius cubed, so the outer one has a period 43/2 = 8 times longer. One measured pair gives the mass of the planet: rearranging T² = 4π²R³/GM gives M = 4π²R³/GT², and nothing about the satellite itself appears in it, which is why the satellite mass cancelled out at the very start.
Unit 3, Work, Energy, and Power. Unit 2 has asked what forces do to the motion at every instant. Unit 3 asks what they do over a distance, and the bookkeeping turns out to be far easier for the same problems you have just been solving arrow by arrow.