AP® Physics C: Mechanics review sheet from Aim for Five (aimforfive.com/physics-c-mech/units/2/2-1)
Unit 2 · Topic 2.1
2.1 Systems and Center of Mass
Before you analyze any motion, you choose a system: the object or objects you're studying. A system's overall motion follows its center of mass, the mass-weighted average position of its parts. In Physics C you find the center of mass of a solid object by integrating, including rods whose density changes along their length. Since the 2024 course update, center of mass is taught here in Unit 2; older materials put it in Unit 4, "Systems of Particles and Linear Momentum."
Key terms
- system
- center of mass
- linear mass density
- mass element dm
- line of symmetry
Choosing a system
A system is whatever you draw an imaginary boundary around: one block, two blocks tied together, or a ball plus Earth. Everything outside is the environment.
The choice decides which forces are internal (between parts of the system) and which are external (from the environment). Only external forces change the motion of the system as a whole. For two blocks joined by a string, choosing both blocks as the system makes the tension internal, so it drops out. Choosing one block alone makes the tension external, which is how you then find it.
If the system's internal structure doesn't matter to the question, you can model it as a single object, a point with all the mass, located at the center of mass.
Center of mass of separate particles
The center of mass is a weighted average of position, where heavier parts count more:
Do the same for y (and z) separately. The center of mass of two objects lies on the line between them, closer to the heavier one, and it splits the distance in the inverse ratio of the masses.
For a symmetric object with mass spread evenly (uniform), the center of mass lies on every line of symmetry. A uniform rod's is at its midpoint, and a uniform disk's or sphere's is at its center. The center of mass doesn't have to be inside the material: a ring's is in the empty middle.
Center of mass by integration
For a continuous object, chop it into tiny pieces of mass dm and add them with an integral:
The bottom integral is just the total mass M. To integrate over position, you need to write dm in terms of dx. For a thin rod along the x-axis, the linear mass density is the mass per unit length (in kg/m), so . Then:
If λ is constant (a uniform rod), this gives L/2, as symmetry says it should. If λ grows along the rod, like λ = bx, the center of mass shifts toward the denser end. Keep λ inside the integral whenever it depends on x.
Why it matters
The center of mass is the point that moves as if all the mass were there and all the external forces acted on it. A wrench spinning as it slides across ice wobbles in a complicated way, but its center of mass moves in a straight line at constant speed. This idea comes back in Newton's second law (2.5) and in momentum (4.3).
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Three particles in a plane
Particles sit at these points: 2.0 kg at (0, 0), 3.0 kg at (4.0 m, 0) and 5.0 kg at (0, 2.0 m). Find the center of mass.
Show the solutionHide the solution
- Step 1: Total mass = 2.0 + 3.0 + 5.0 = 10.0 kg.
- Step 2: m.
- Step 3: m.
Answer: (1.2 m, 1.0 m)
- Example 2Calculator allowed
A rod that gets denser along its length
A thin rod lies along the x-axis from x = 0 to x = L = 1.5 m. Its linear mass density is , with b = 4.0 kg/m². Find the rod's mass and its center of mass.
Show the solutionHide the solution
- Step 1: Mass: kg.
- Step 2: Moment: .
- Step 3: m.
- Step 4: The trap is answering L/2 = 0.75 m out of habit. The rod is denser at the right end, so its center of mass sits to the right of the middle.
Answer: M = 4.5 kg; m from the light end.
Common mistakes
- Averaging positions without weighting by mass. The center of mass is closer to the heavier part.
- Pulling λ outside the integral when it depends on x. Only a constant λ can come out.
- Assuming the center of mass must be inside the object. For a ring or an L-shape it can be in empty space.
- Forgetting that the bottom of the center-of-mass fraction is the total mass, which also needs an integral for a nonuniform rod.
On the exam
- Expect a nonuniform rod with a given λ(x) and a request to derive its mass and center of mass. Write and both integrals with limits; the setup earns credit.
- Free-response questions often start by naming the system. State which objects are in it, because that decides which forces are internal and which are external.
Connected topics
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Check yourself
4 questions on 2.1 Systems and Center of Mass. Pick an answer to see if you got it, and why.
Three small objects lie on the x-axis: 1.0 kg at x = 0, 2.0 kg at x = 3.0 m and 3.0 kg at x = 5.0 m. Where is the center of mass of the system?
A thin rod of length L lies along the x-axis from x = 0 to x = L. Its linear mass density is , where c is a positive constant. Where is its center of mass?
A rod lies along the x-axis from x = 0 to x = 2.0 m. Its linear mass density is , where λ is in kg/m and x is in meters.
Described situation
What is the mass of the rod?
Where is the rod's center of mass?
0 of 4 answered