AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/2/2-1)
Unit 2 · Topic 2.1
2.1 Systems and Center of Mass
Before you analyze forces or energy, you choose a system: the object or group of objects you're studying. Everything else is the environment. If the system's insides don't matter, you can treat the whole thing as one object located at its center of mass, the mass-weighted average position of its parts.
Key terms
- system
- environment
- object model
- center of mass
- line of symmetry
Systems and their environment
A system is whatever you draw an imaginary boundary around. A single block can be a system, and so can two blocks tied together, or a ball plus Earth. Everything outside the boundary is the environment.
The choice matters because it decides which interactions are internal (between parts of the system) and which are external (between the system and its environment). Only external forces can change how the system as a whole moves. Energy and mass can also cross the boundary, for example when a person does work on a box or a rocket throws out exhaust.
A system's properties come from the interactions inside it. A spring is springy because of forces between its atoms. Some changes outside a system can change its internal structure too, as when a hard push crushes a soda can.
When a system can be one object
If the parts of a system move together and their internal details don't affect the question, treat the system as a single object. A car rolling down a road is one object for most problems.
A smart choice of system can make a problem easier. For two blocks joined by a string, choosing both blocks as the system makes the string's tension internal, so it drops out of the equation for the system's acceleration. Choosing one block alone brings the tension back as an external force, which is how you then find it.
Sometimes the internal structure does matter. Parts of a system can move differently from each other and from the system as a whole. When a firework bursts, the pieces fly in every direction, but the system's center of mass keeps following the same smooth path it was on, as long as gravity is the only outside force.
Finding the center of mass
The center of mass is the average position of a system's mass, with heavier parts counting more. Along the x-axis:
x_cm = (m₁x₁ + m₂x₂ + m₃x₃ + …) ÷ (m₁ + m₂ + m₃ + …)
Do the same with y-coordinates to get y_cm in two dimensions. The center of mass always lies closer to the heavier mass. For two equal masses it's exactly halfway between them.
For a symmetric object with evenly spread mass, the center of mass lies on every line of symmetry, so it's at the geometric center of a uniform rod, sphere, disk or rectangle. It doesn't have to be inside the material: a ring's center of mass is at the empty middle.
To combine objects, replace each symmetric piece with a single point mass at its own center, then use the formula.
What you need to calculate
AP Physics 1 only expects center-of-mass calculations for up to five particles arranged in two dimensions, or for highly symmetric objects. For anything else, you only need to reason about roughly where it is.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Two masses on a light rod
A 2.0 kg ball sits at x = 0 and a 6.0 kg ball sits at x = 0.80 m, connected by a rod of negligible mass. Where is the system's center of mass?
Show the solutionHide the solution
- Step 1: x_cm = (m₁x₁ + m₂x₂) ÷ (m₁ + m₂) = [(2.0)(0) + (6.0)(0.80)] ÷ (2.0 + 6.0).
- Step 2: = 4.8 ÷ 8.0 = 0.60 m.
- Step 3: Check: the answer is closer to the heavier 6.0 kg ball, three times as far from the 2.0 kg ball (0.60 m) as from the 6.0 kg ball (0.20 m), matching the 3-to-1 mass ratio.
Answer: x_cm = 0.60 m from the 2.0 kg ball
- Example 2Calculator allowed
Three particles in two dimensions
Particles sit at these positions: 1.0 kg at (0, 0), 2.0 kg at (4.0 m, 0) and 1.0 kg at (0, 4.0 m). Find the center of mass.
Show the solutionHide the solution
- Step 1: Total mass = 1.0 + 2.0 + 1.0 = 4.0 kg.
- Step 2: x_cm = [(1.0)(0) + (2.0)(4.0) + (1.0)(0)] ÷ 4.0 = 8.0 ÷ 4.0 = 2.0 m.
- Step 3: y_cm = [(1.0)(0) + (2.0)(0) + (1.0)(4.0)] ÷ 4.0 = 4.0 ÷ 4.0 = 1.0 m.
Answer: (2.0 m, 1.0 m)
- Example 3Calculator allowed
A rod with a ball on the end
A uniform 3.0 kg rod is 2.0 m long and lies along the x-axis from x = 0 to x = 2.0 m. A small 1.0 kg ball is fixed to its right end. Find the center of mass of the rod–ball system.
Show the solutionHide the solution
- Step 1: The rod is uniform and symmetric, so treat it as a 3.0 kg point at its middle, x = 1.0 m.
- Step 2: The ball is a 1.0 kg point at x = 2.0 m.
- Step 3: x_cm = [(3.0)(1.0) + (1.0)(2.0)] ÷ 4.0 = 5.0 ÷ 4.0 = 1.25 m.
Answer: x_cm = 1.25 m from the left end
Common mistakes
- Averaging positions without weighting by mass. The center of mass sits closer to the heavier part.
- Forgetting to divide by the total mass of all the parts, not just the number of parts.
- Assuming the center of mass must be inside the object. For a ring or a bent shape it can be in empty space.
- Counting forces between parts of the system as if they could move the whole system. Internal forces can't change the center of mass's motion.
On the exam
- Expect short center-of-mass calculations with two to four point masses, often on a light rod or at the corners of a shape.
- Free-response questions often start with "choose a system" or define one for you. State which objects are in it, because that decides whether forces and energy transfers are internal or external.
Connected topics
Videos
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: 2.0 kg at x = 0, 3.0 kg at x = 2.0 m and 5.0 kg at x = 4.0 m. Where is the system's center of mass?
Three small objects are placed on a flat surface: 1.0 kg at (0, 0), 1.0 kg at (2.0 m, 0) and 2.0 kg at (0, 2.0 m). What are the coordinates of the system's center of mass?
A person pushes a box across a rough floor. The system is chosen to be the person and the box together. Which force is internal to this system?
Where is the center of mass of a thin, uniform circular ring?
0 of 4 answered