AP Physics 1 · Topic 2.1

Topic 2.1: Systems and Center of Mass

Unit 2: Force and Translational Dynamics18-23% of the multiple-choice section

A system is whatever set of objects you choose to analyze together. When the internal details do not matter, the whole system can be treated as one object sitting at its center of mass, the mass-weighted average position of its parts. Symmetry locates that point whenever the mass is spread evenly.

AP Physics: Unit 2 (topics 2.1 Systems and Center of Mass). AP Physics 1 Unit 2, Topic 2.1, covering learning objective 2.1.A (describe the properties and interactions of a system) and 2.1.B (describe the location of a system's center of mass with respect to the system's constituent parts). The topic's boundary statement limits center-of-mass calculations to five or fewer particles in a two-dimensional configuration or to highly symmetrical systems. The CED lists suggested skills 1.B, 2.B, 2.C, and 3.B here, and weights Unit 2 at 18 to 23 percent of the multiple-choice section.

What Topic 2.1 asks for

Topic 2.1 carries two learning objectives, and neither one is a plug-in. Objective 2.1.A asks you to describe the properties and interactions of a system. Objective 2.1.B asks you to describe the location of a system's center of mass with respect to the system's constituent parts. The first is a modeling decision; the second is the one calculation in the topic.

The CED opens with the reason both matter: system properties are determined by the interactions between objects within the system. Change what you count as inside the system and you change which forces are internal bookkeeping and which are external forces that actually move it. That choice is yours to make, and making it deliberately is the skill being tested.

Topic 2.1 opens Unit 2, Force and Translational Dynamics, which the CED weights at 18 to 23 percent of the multiple-choice section and about 22 to 27 class periods. That is the joint largest weighting in the course, shared with Unit 3. The suggested skills listed for this topic are 1.B, 2.B, 2.C, and 3.B, which sit beneath the three AP Physics 1 science practices: Creating Representations, Mathematical Routines, and Scientific Questioning and Argumentation.

Choosing the system is the first decision

The CED states the permission directly: if the properties or interactions of the constituent objects within a system are not important in modeling the behavior of the macroscopic system, the system can itself be treated as a single object. Whether that condition holds depends on the question you are answering, not on the objects.

Take a 3.0 kg block and a 1.0 kg block joined by a light string, pulled along a frictionless floor by a 6.0 N horizontal force. Ask for the acceleration and the string is irrelevant: treat the pair as one 4.0 kg object and $a = (6.0 \, \mathrm{N})/(4.0 \, \mathrm{kg}) = 1.5 \, \mathrm{m/s^2}$ . Ask for the tension in that string and the same choice fails, because tension is internal to that system and internal forces do not appear in the system equation. You have to redraw around one block.

That is the working rule. Internal forces cancel in pairs and drop out; only external forces survive. So pick the system that puts the quantity you want on the outside of the boundary. The equation doing the work here, $\vec{a}_{sys} = \vec{F}_{net}/m_{sys}$ , belongs to Topic 2.5, and the AP Physics 1 equation sheet prints it in exactly that system form.

Systems trade energy and mass with their surroundings

Two essential knowledge statements keep a system from being a sealed box. Systems may allow interactions between constituent parts of the system and the environment, which may result in the transfer of energy or mass. And individual objects within a chosen system may behave differently from each other as well as from the system as a whole.

The second is worth sitting with, because it marks the limit of the single-object model. A wheel rolling without slipping down a ramp has a center of mass that moves steadily forward while the point of the rim touching the ground is momentarily at rest. Two skaters on frictionless ice who start at rest and push apart have a center of mass that does not move at all, while both skaters certainly do. Neither case breaks the model. The model describes the system's center of mass, not the parts.

So when a question asks about the system, answer for the center of mass. When it asks about one specific block, cart, or skater, isolate that object and start again. Mixing the two is exactly the error this distinction is written to prevent.

Internal structure changes the analysis

Two more statements finish objective 2.1.A. The internal structure of a system affects the analysis of that system. And as variables external to a system are changed, the system's substructure may change.

A solid steel block and a sealed bucket of water can have the same mass and the same weight, and for a question about the net force on either one, that is all you need. Start tilting, spinning, or shaking them and the two behave nothing alike, because the water rearranges itself and the steel cannot. The internal structure you were free to ignore has become the whole problem.

The logic runs the other way too. A spring squeezed between two carts is internal structure that stores energy. Release it and the carts fly apart even though nothing outside pushed them along the track. The motion of the system's center of mass did not change, which is Topic 2.3 making its point about internal forces, but the motion of each cart changed a lot. Deciding what you are allowed to ignore is the modeling work Topic 2.1 is asking for.

Symmetry finds the center of mass for free

For systems with symmetrical mass distributions, the center of mass is located on lines of symmetry. That single statement, essential knowledge 2.1.B.1, settles most center-of-mass questions with no arithmetic at all.

A uniform sphere has symmetry through its geometric center, so that is where the center of mass sits. A uniform rod is symmetric about its perpendicular bisector, so the center of mass is at the midpoint. A uniform rectangular plate has two symmetry lines that cross at the geometric center, and the center of mass is at the crossing.

Uniform is doing real work in each of those sentences. A rod with a lead slug welded to one end is no longer symmetric about its midpoint, and its center of mass shifts toward the slug. Two cautions follow. Symmetry of shape is not enough by itself, because the mass distribution has to be symmetric as well. And the center of mass need not lie in any material at all: a uniform ring has symmetry lines that all cross at its center, where there is nothing but air, and that empty point is still the center of mass.

The center of mass equation

When symmetry does not settle it, the CED gives one equation, which the AP Physics 1 equation sheet prints:

\vec{x}_{cm} = \frac{\sum m_i \vec{x}_i}{\sum m_i}

Read it as a mass-weighted average of position. The denominator is the total mass of the system. The numerator adds one term per part, each part's mass times its position measured from whatever origin you picked. Heavier parts pull the result toward themselves, which is the built-in sanity check: the answer has to land nearer the heavy end.

Three mechanics of using it. The equation works one axis at a time, so a two-dimensional problem is two applications, one for $x$ and one for $y$ . Positions are signed coordinates on a stated axis, the same convention as Topic 1.1, so put the origin somewhere convenient and then leave it alone. And the result is a position in meters, not a mass and not a distance between parts.

The CED bounds the workload: AP Physics 1 only expects students to calculate the center of mass for systems of five or fewer particles arranged in a two-dimensional configuration or for systems that are highly symmetrical.

One object, located at the center of mass

Essential knowledge 2.1.B.3 closes the loop. A system can be modeled as a singular object that is located at the system's center of mass. That statement is the permission behind every free-body diagram you are about to draw.

Topic 2.2 spells out the consequence: forces exerted on an object or system are represented as vectors originating from the representation of the center of mass, such as a dot, and a system is treated as though all of its mass is located at the center of mass. The dot in a free-body diagram is not a cartoon of the object. It is the center of mass.

The same substitution runs through the rest of the unit. In Newton's second law the acceleration you calculate is the acceleration of the system's center of mass. In gravitation the force between two systems acts along the line connecting their centers of mass. Getting used to it now saves confusion later. Build a few diagrams in the free-body diagram builder, and read how to draw a free-body diagram for the drawing procedure itself.

How Topic 2.1 is tested

The CED lists four suggested skills for this topic: 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Read that list as a set of instructions rather than a question format. Skill 2.B is the center-of-mass calculation itself. Skill 2.C is the comparison version: same particles, one of them moved, which way does the center of mass shift and by how much. Skill 3.B is the sentence you write to defend a system choice, naming the statement you are relying on.

Rehearse the comparison version out loud, because it costs nothing and it is the easiest to skip. Then move on to Topic 2.2, where the single dot you just justified becomes the starting point of every dynamics problem, and check the full map at the Unit 2 overview.

Center of mass of three particles in a plane

Three small spheres are bolted to a light frame. Sphere 1 has mass 2.0 kg at the origin, sphere 2 has mass 3.0 kg at $(4.0 \, \mathrm{m}, 0)$ , and sphere 3 has mass 5.0 kg at $(0, 6.0 \, \mathrm{m})$ . The frame's mass is negligible. Locate the center of mass.

Fix the coordinate system first. Take the origin at sphere 1, with $x$ positive to the right and $y$ positive upward, and keep those axes for both calculations.
Total mass, the denominator both times: $\sum m_i = 2.0 \, \mathrm{kg} + 3.0 \, \mathrm{kg} + 5.0 \, \mathrm{kg} = 10.0 \, \mathrm{kg}$ .
Numerator on the $x$ axis: $\sum m_i x_i = (2.0)(0) + (3.0)(4.0) + (5.0)(0) = 12 \, \mathrm{kg \cdot m}$ .
So $x_{cm} = (12 \, \mathrm{kg \cdot m})/(10.0 \, \mathrm{kg}) = 1.2 \, \mathrm{m}$ .
Numerator on the $y$ axis: $\sum m_i y_i = (2.0)(0) + (3.0)(0) + (5.0)(6.0) = 30 \, \mathrm{kg \cdot m}$ , so $y_{cm} = (30 \, \mathrm{kg \cdot m})/(10.0 \, \mathrm{kg}) = 3.0 \, \mathrm{m}$ .
Sanity check the answer against the masses. The 5.0 kg sphere is the heaviest and sits high on the $y$ axis, so the center of mass should sit well up the $y$ axis and only a little way along $x$ . At $(1.2 \, \mathrm{m}, 3.0 \, \mathrm{m})$ it does, and the point falls inside the triangle the three spheres form, which it must for masses that are all positive.

The center of mass is at $(1.2 \, \mathrm{m}, 3.0 \, \mathrm{m})$ from sphere 1. Three particles in a two-dimensional arrangement is squarely inside the CED's limit of five or fewer particles, and the work is two separate applications of one equation, never a single combined step.

Two rods welded into an L: using symmetry first

A 1.0 kg thin uniform rod of length 0.60 m lies along the $x$ axis from the origin to $(0.60 \, \mathrm{m}, 0)$ . A 2.0 kg thin uniform rod of length 0.40 m is welded to it at the origin and runs along the $y$ axis to $(0, 0.40 \, \mathrm{m})$ . Locate the center of mass of the L-shaped object.

Reduce each rod to a point before touching the equation. Each rod is uniform, so by essential knowledge 2.1.B.1 its center of mass lies on its line of symmetry, at its midpoint.
Horizontal rod: midpoint at $(0.30 \, \mathrm{m}, 0)$ , carrying its full 1.0 kg. Vertical rod: midpoint at $(0, 0.20 \, \mathrm{m})$ , carrying its full 2.0 kg. The problem is now two particles, not two extended objects.
Total mass: $1.0 \, \mathrm{kg} + 2.0 \, \mathrm{kg} = 3.0 \, \mathrm{kg}$ .
Horizontal axis: $x_{cm} = [(1.0)(0.30) + (2.0)(0)]/3.0 = (0.30 \, \mathrm{kg \cdot m})/(3.0 \, \mathrm{kg}) = 0.10 \, \mathrm{m}$ .
Vertical axis: $y_{cm} = [(1.0)(0) + (2.0)(0.20)]/3.0 = (0.40 \, \mathrm{kg \cdot m})/(3.0 \, \mathrm{kg}) = 0.133 \, \mathrm{m}$ , which is 0.13 m to the two significant figures the data supports.
Look at where that point sits. The horizontal rod occupies $y = 0$ and the vertical rod occupies $x = 0$ , so the point $(0.10 \, \mathrm{m}, 0.13 \, \mathrm{m})$ lies in the open space between the two arms, in no material at all.

The center of mass is at $(0.10 \, \mathrm{m}, 0.13 \, \mathrm{m})$ , a point that contains none of the object. Symmetry did the first half of the job by turning two rods into two particles, and the CED equation did the second half. Note that the heavier, shorter rod pulls the answer toward the vertical arm.

Frequently asked questions

What is a system in AP Physics 1?

A system is the set of objects you decide to analyze together. Nothing in the physical situation tells you where to draw the boundary; you choose it, and that choice decides which forces count as internal and which are external. The CED adds that a system's properties are determined by the interactions between the objects inside it.

How do you calculate the center of mass?

Multiply each part's mass by its position on one axis, add those products, and divide by the total mass. That gives the coordinate of the center of mass on that axis. Repeat on the second axis for a two-dimensional layout. Positions are signed coordinates measured from an origin you choose, so fix the origin before you start.

Can the center of mass be outside the object?

Yes. A uniform ring is the clearest case: its lines of symmetry all cross at the geometric center, which is empty space, and that empty point is the center of mass. The same happens for an L-shaped or U-shaped object. The center of mass is an average position, not a piece of material.

How many masses will the AP exam ask you to combine?

The Topic 2.1 boundary statement limits calculations to systems of five or fewer particles arranged in a two-dimensional configuration, or to systems that are highly symmetrical. Anything larger is meant to be handled by symmetry or by reducing uniform pieces to points at their own midpoints first.

Why can several objects be treated as one object?

Because the CED allows it whenever the properties and interactions of the individual parts are not important for the behavior you are modeling, and because forces between the parts cancel in pairs when you add up the whole system. What survives is the external forces, and those set the acceleration of the system's center of mass.