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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:

xcm=∑mixi∑mix_{cm} = \frac{\sum m_ix_i}{\sum m_i}

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:

xcm=∫x dm∫dmx_{cm} = \frac{\int x\,dm}{\int dm}

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 λ=dmdx\lambda = \frac{dm}{dx} is the mass per unit length (in kg/m), so dm=λ dxdm = \lambda\,dx. Then:

M=∫0Lλ dx,xcm=1M∫0Lxλ dxM = \int_0^L \lambda\,dx, \qquad x_{cm} = \frac{1}{M}\int_0^L x\lambda\,dx

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.

  1. 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 solution
    1. Step 1: Total mass = 2.0 + 3.0 + 5.0 = 10.0 kg.
    2. Step 2: xcm=(2.0)(0)+(3.0)(4.0)+(5.0)(0)10.0=1.2x_{cm} = \frac{(2.0)(0) + (3.0)(4.0) + (5.0)(0)}{10.0} = 1.2 m.
    3. Step 3: ycm=(2.0)(0)+(3.0)(0)+(5.0)(2.0)10.0=1.0y_{cm} = \frac{(2.0)(0) + (3.0)(0) + (5.0)(2.0)}{10.0} = 1.0 m.

    Answer: (1.2 m, 1.0 m)

  2. 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 λ=bx\lambda = bx, with b = 4.0 kg/m². Find the rod's mass and its center of mass.

    Show the solution
    1. Step 1: Mass: M=∫0Lbx dx=bL22=(4.0)(1.5)22=4.5M = \int_0^L bx\,dx = \frac{bL^2}{2} = \frac{(4.0)(1.5)^2}{2} = 4.5 kg.
    2. Step 2: Moment: ∫0Lx(bx) dx=bL33\int_0^L x(bx)\,dx = \frac{bL^3}{3}.
    3. Step 3: xcm=bL3/3bL2/2=2L3=1.0x_{cm} = \frac{bL^3/3}{bL^2/2} = \frac{2L}{3} = 1.0 m.
    4. 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; xcm=23L=1.0x_{cm} = \frac{2}{3}L = 1.0 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 dm=λ dxdm = \lambda\,dx 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

Videos

  • Topic 2.1 - Systems and Center of Mass

    Lessons With LondotWatch on YouTube (opens in a new tab)

  • Center of Mass by Integration (Rigid Objects with Shape)

    Flipping PhysicsWatch on YouTube (opens in a new tab)

  • AP Physics C - Center of Mass

    Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)

  • AP Physics C Mechanics - Unit 2 - Lesson 19C - Center of Mass (Distributed Mass)

    Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)

  • 17.5 Worked Example - Center of Mass of a Uniform Rod

    MIT OpenCourseWareWatch on YouTube (opens in a new tab)

  • Nonuniform Density Center of Mass

    Flipping PhysicsWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 2.1 Systems and Center of Mass. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

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?

Question 2 of 4Calculator allowed

A thin rod of length L lies along the x-axis from x = 0 to x = L. Its linear mass density is λ=cx\lambda = cx, 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 λ(x)=2+3x\lambda(x) = 2 + 3x, where λ is in kg/m and x is in meters.

Described situation

Question 3 of 4Calculator allowed

What is the mass of the rod?

Question 4 of 4Calculator allowed

Where is the rod's center of mass?

0 of 4 answered