AP PHYSICS C: MECHANICS › UNIT 2, FORCE AND TRANSLATIONAL DYNAMICS › TOPIC 2.1
Systems and Center of Mass
Before a single force is drawn, one decision has to be made, and it is almost never written down: what exactly are we talking about? This topic is that decision, the one point that is allowed to stand in for whatever you chose, and the integral that finds it when the mass is spread out.
2.1.A Describe a system and its interactions, and decide when a group of objects can be treated as one object.
2.1.B Locate the center of mass of a system, both for a handful of particles and for a continuous body whose density is given as a function of position.
1. The first line you draw is not a force
Unit 1 asked what the motion was. Unit 2 asks what is causing it, and the first move is one that almost nobody writes down: deciding what it refers to. A cart, or the cart and the block on top of it. That choice is called the system, and every equation after it depends on which one you made.
Draw the boundary as a dashed loop. Forces reaching across it from outside are external, and they are the ones that change the motion of the system as a whole. Forces between two things inside the loop are internal. They exist, they can be enormous, and they contribute exactly nothing, because they always appear in pairs pointing opposite ways.
2. Two systems, one situation
Two blocks on a smooth floor, joined by a light rope, with a hand pulling one of them. Ask for the acceleration and the fastest route puts both blocks inside the boundary: the rope forces become internal, cancel, and one line of algebra finishes it. Ask instead for the tension and that route is useless, because you deliberately hid the rope. Redraw the boundary around a single block and the rope becomes external, and it is the only horizontal arrow left.
Nothing physical changed. You asked a different question and drew a different loop, which is the entire skill.
Choose which blocks you are calling the system. The dashed line is the boundary, gold arrows are forces that cross it, and the faint pair is the one that does not. Only horizontal forces are drawn: the weights and the normal forces balance and would add nothing here. Watch the acceleration readout while you switch boundaries.
3. The center of mass of a handful of particles
Treating a system as one object raises the obvious question: one object located where? At the center of mass, which is a weighted average of the positions, each position counted as many times as it has kilograms.
rcm = Σ mi ri / Σ mione such equation for each coordinate
Two shortcuts follow at once. Equal masses make the weighting do nothing, so the center of mass is the plain average of the positions. And a line of symmetry with equal mass either side of it must contain the center of mass, so a uniform rod, disc or sphere needs no arithmetic at all. Reach for symmetry before you reach for an integral: the integral will agree with it, at the cost of ten minutes.
4. When the mass is spread out
This is where AP Physics C leaves AP Physics 1 behind. A rod whose density varies along its length is not a handful of particles, so the sum becomes an integral over infinitesimal pieces of mass:
rcm = ∫ r dm / ∫ dmwith dm = λ dℓ for a rod
The whole difficulty is that dm is not a variable you can integrate over. The linear density is the derivative of the mass with respect to position, λ = dm/dℓ, and reading that as dm = λ dℓ is the move that turns an unusable integral into an ordinary one in x. Do it on both the top and the bottom. The bottom is the total mass, and you nearly always need it anyway.
Then attach limits that describe the actual object, integrate, and divide. The answer is a length, so if the units of your result are not meters you have dropped a factor somewhere and should find it before going on.
A rod of length L whose linear density is λ = λ₀(1 + kx/L). The graph is the density, the shading is the same information painted onto the rod, and the cross is the center of mass. The two readouts are the integrals themselves, in units that make λ₀ and L cancel. Push k to its limit and watch where the answer refuses to go.
5. Why that point earns its name
The center of mass is not just a convenient average. It is the one point in a complicated system that obeys a simple law. Throw a wrench across the room and every part of it does something different, while the center of mass draws a clean parabola, exactly as though the whole wrench were a single particle with a single force on it.
That is the promise the rest of Unit 2 is built on. When the second law says that a net force produces an acceleration, the acceleration it means is the acceleration of the center of mass. A diver can tuck, spin and open, and the center of mass never notices: gravity is the only external force, so its path was settled the moment she left the board.
6. Three ways this goes wrong
Integrating over dm as though it were a length. Write dm = λ dℓ first, every single time, and only then put the limits on. An integral with dm still in it and limits in meters is not an equation about anything.
Forgetting that the denominator is also an integral. For a uniform body the total mass is obvious and students carry that habit into the nonuniform case, where it is wrong. If the density varies, both halves need integrating.
Changing the system halfway through. Starting with two blocks and then quietly using the tension as though it were external gives an equation that is not so much wrong as about nothing. Name the system at the top of the page and keep it.
Check yourself
1. Two blocks joined by a rope are pulled across a smooth floor. A student says the rope tension helps accelerate the pair. What is wrong with that, and when is the tension external?
With both blocks inside the system the rope pulls forward on one and backward on the other by equal amounts, so the pair contributes nothing. It becomes external the moment the boundary is redrawn around a single block, and then it is the only horizontal force on that block.
2. A rod of length L has linear density λ = λ₀ x / L. Find its total mass and the position of its center of mass, and say what symmetry alone would have told you.
M = ∫₀ᴸ λ₀ x/L dx = λ₀L/2, and ∫₀ᴸ x λ₀ x/L dx = λ₀L²/3, so x_cm = (λ₀L²/3)/(λ₀L/2) = 2L/3. Symmetry tells you nothing here, which is the point: the rod has no line of symmetry once the density varies, so the integral is not optional. It does tell you the answer must be past the middle, which is a check worth making before you start.
3. A uniform ring lies flat on a table. Where is its center of mass, and what does that tell you about whether the center of mass has to be made of anything?
At the geometric center, in the hole, by symmetry: every element of the ring is matched by one directly opposite. There is no material there. The center of mass is a position computed from a distribution, and nothing in the definition requires it to lie on matter.
4. A gymnast leaves the floor, tucks, rotates, and opens before landing. Ignoring air resistance, describe the path of her center of mass and say why the tuck cannot change it.
A parabola, fixed at the instant she left the floor. Tucking rearranges her body with internal forces, and internal forces cannot change the motion of the system as a whole. The only external force in flight is gravity, so the center of mass keeps the parabola it was given whatever the limbs do.
5. A student writes x_cm = ∫₀ᴸ x dm / M and then integrates the top as though dm were dx, obtaining L²/2. Name both errors, and give the general check that would have caught them.
First, dm was never converted with dm = λ dℓ, so the integration variable does not match the limits. Second, M was taken as known when for a nonuniform body it is itself an integral. The check is units: ∫ x dm has units of kilogram meters, so dividing by kilograms leaves meters. The result L²/2 is in square meters, which is not a position, and that alone is enough to send you back.
Topic 2.2, Forces and Free-Body Diagrams. The system is chosen and it has a location. Next comes the picture of the arrows that cross its boundary, one straight arrow per force, and the arrows students draw that are not there at all.