AP Physics C: Mechanics · Topic 2.1

Topic 2.1: Systems and Center of Mass

Unit 2: Force and Translational Dynamics20-25% of the multiple-choice section

Topic 2.1 gives you two ways to locate a center of mass. For separate masses, use the weighted sum of their positions. For one continuous body whose density varies, integrate r dm over the body and divide by the total mass. The integral form is the part AP Physics 1 does not have.

AP Physics: Unit 2 (topics 2.1 Systems and Center of Mass). AP Physics C: Mechanics Unit 2, Topic 2.1. Two learning objectives. 2.1.A, describe the properties and interactions of a system, supported by 2.1.A.1 through 2.1.A.6. 2.1.B, describe the location of a system's center of mass with respect to the system's constituent parts, supported by 2.1.B.1 (symmetrical distributions put the center of mass on lines of symmetry), 2.1.B.2 (the summation form along a given axis), 2.1.B.3 (for a nonuniform solid considered as a collection of differential masses dm, the center of mass is the integral of r dm over the integral of dm), 2.1.B.3.i (linear mass density is the derivative of the rod's mass with respect to the position of the differential mass element, lambda = d/dl of m(l)), 2.1.B.3.ii (a given mass-density function can be integrated over length, area or volume to give total mass, example M_total = integral of rho(r) dV), and 2.1.B.4 (a system can be modeled as a singular object located at the system's center of mass). Topic 2.1 prints NO boundary statement in AP Physics C: Mechanics; the AP Physics 1 version of this topic does, limiting students to five or fewer particles in a two-dimensional configuration or highly symmetrical systems, and AP Physics 1 has no counterpart to 2.1.B.3. Suggested skills are 1.A, 2.B, 2.C and 3.B; skill 2.A is not among them. Three equations from this topic are printed on the AP Physics C: Mechanics sheet (the summation form, the integral form, and lambda as a derivative) and only the first appears on the AP Physics 1 sheet. The CED's sample free-response question 3, Experimental Design and Analysis, aligns to 2.1.B among six objectives. Two of the unit's six optional sample instructional activities are on Topic 2.1.

What calculus changes here, in one statement

AP Physics 1 and AP Physics C: Mechanics both call Topic 2.1 "Systems and Center of Mass", and their first learning objective is the same. The difference is one essential-knowledge statement and one boundary statement.

AP Physics C adds statement 2.1.B.3: for a nonuniform solid that can be considered as a collection of differential masses $dm$ , the solid's center of mass can be calculated using the equation

\vec{r}_{\text{cm}} = \frac{\int \vec{r}\, dm}{\int dm}

AP Physics 1 removes that statement and replaces it with a boundary statement that reads: "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." AP Physics C: Mechanics prints no boundary statement on Topic 2.1 at all.

So the algebra-based course is fenced to at most five point masses or an object you can argue about by symmetry. The calculus-based course removes the fence and hands you the tool that removes it: a rod whose thickness tapers, a disc whose density grows with radius, a wire bent into an arc. Anything you can write a density function for, you can integrate.

Statement 2.1.B.3.i then defines linear mass density as a derivative. The linear mass density of a rod or other linear rigid body is the derivative of the rod's mass with respect to the position of the differential mass element on the rigid body:

\lambda = \frac{d}{d\ell} m(\ell)

Statement 2.1.B.3.ii goes the other way. If a function of mass density is given for a solid, the total mass can be determined by integrating the mass density over the length (one dimension), area (two dimensions), or volume (three dimensions) of the solid. Its printed example is $M_{\text{total}} = \int \rho(r)\, dV$ .

That is the whole calculus addition. It is small on the page and large in what it lets a question ask.

What the CED requires of Topic 2.1

Two learning objectives, and the suggested skills the course and exam description prints beside them are 1.A, 2.B, 2.C and 3.B.

Objective 2.1.A: describe the properties and interactions of a system. Six essential-knowledge statements sit under it.

2.1.A.1: system properties are determined by the interactions between objects within the system.
2.1.A.2: 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.
2.1.A.3: systems may allow interactions between constituent parts of the system and the environment, which may result in the transfer of energy or mass.
2.1.A.4: individual objects within a chosen system may behave differently from each other as well as from the system as a whole.
2.1.A.5: the internal structure of a system affects the analysis of that system.
2.1.A.6: as variables external to a system are changed, the system's substructure may change.

Objective 2.1.B: describe the location of a system's center of mass with respect to the system's constituent parts. Four statements.

2.1.B.1: for objects or systems with symmetrical mass distributions, the center of mass is located on lines of symmetry.
2.1.B.2: the location of a system's center of mass along a given axis can be calculated using $\vec{x}_{\text{cm}} = \sum m_i \vec{x}_i / \sum m_i$ .
2.1.B.3: the integral form above, with its two sub-statements on density.
2.1.B.4: a system can be modeled as a singular object that is located at the system's center of mass.

Notice the ordering. Symmetry comes first, the discrete sum second, the integral third. That is also the order you should try them in on a real problem, because the integral is the slowest of the three and symmetry is free.

Choosing the system, and the CED's own worked discussion of it

Objective 2.1.A is not filler. The appendix of the AP Physics C: Mechanics course and exam description, "Vocabulary and Definitions of Important Ideas in AP Physics", defines a system as a collection of objects that are analyzed together, and spends two pages on why the choice matters.

Its example is a leaking box of cereal. The physicist has a decision to make: continue to include every piece of cereal as part of the system, or consider only the cereal inside the box. The appendix calls the first choice System A and the second System B, and says analyzing System A will be exceedingly complex, as the small pieces of cereal move, bounce, accelerate, and collide with each other and the environment, while analyzing System B is much simpler: the box is losing mass to the environment, but the box, bag and cereal system may be modeled as a single object that has a changing mass.

That is 2.1.A.2 and 2.1.A.3 in a concrete case, and it is worth reading before Topic 2.5, because a system whose mass changes is exactly where $\vec{F}_{\text{net}} = d\vec{p}/dt$ stops being interchangeable with $m\vec{a}$ .

The same appendix separates three model levels you will be asked to move between.

An object is a physical thing where the internal structure and properties of the thing are ignored. The appendix notes that an object has no internal structure or surface properties, and that using the object model ignores the physical size of the object itself.
A system is a collection of objects that are analyzed together, and in special cases a system can itself be reduced to a single object.
A rigid system is a system that does not change shape, but different points within the system may move in different directions and with different speeds. Its example is a rotating wheel, where the two marked points travel at different speeds, so the wheel cannot be modeled as an object if the rotation is relevant.

Statement 2.1.B.4 is the bridge: once you know where the center of mass is, a system can be modeled as a singular object located there. That is why Topic 2.1 comes before every other topic in the unit.

The three equations on the sheet, and the two that are new

The AP Physics C: Mechanics equation sheet prints three lines from this topic, in this order:

\vec{x}_{\text{cm}} = \frac{\sum m_i \vec{x}_i}{\sum m_i} \qquad \vec{r}_{\text{cm}} = \frac{\int \vec{r}\, dm}{\int dm} \qquad \lambda = \frac{d}{d\ell} m(\ell)

The AP Physics 1 sheet prints only the first of those three. The integral and the density derivative are on this sheet and nowhere on that one, which is a fair summary of the whole difference between the two versions of this topic.

The symbol list beside the mechanics table defines $\lambda$ as linear mass density. It does not define $\rho$ , even though 2.1.B.3.ii's example uses it, so if a question hands you a volume density function it will say so in words.

One more line matters here and belongs to Unit 4 rather than Unit 2, but it is printed on the same sheet and it is about the center of mass: $\vec{v}_{\text{cm}} = \sum \vec{p}_i / \sum m_i = \sum m_i \vec{v}_i / \sum m_i$ . If a question asks for the velocity of a system's center of mass, that line is already written down for you.

Three routes to a center of mass, in the order to try them

1. Symmetry. Statement 2.1.B.1 says that for objects or systems with symmetrical mass distributions, the center of mass is located on lines of symmetry. A uniform rod, disc, sphere, ring, cube or rectangular plate is done the moment you say so. Two lines of symmetry intersect at a point, and that point is the answer.

2. The discrete sum. For point masses, or for composite bodies you can chop into pieces whose individual centers you know by symmetry, use $\vec{x}_{\text{cm}} = \sum m_i \vec{x}_i / \sum m_i$ once per axis. A body with a hole cut out of it is the standard case: treat the hole as a piece of negative mass, and the sum handles it without a single integral.

3. The integral. Use it when the density varies continuously, because then no finite chop is exact. The mechanics is always the same four steps.

Pick a coordinate and write $dm$ in terms of it. For a rod, $dm = \lambda(x)\, dx$ . For a disc of surface density $\sigma(r)$ , the natural element is an annulus, $dm = \sigma(r)\, 2\pi r\, dr$ . For a solid of volume density $\rho(r)$ , $dm = \rho(r)\, dV$ .
Integrate $dm$ over the body to get the total mass. That is 2.1.B.3.ii, and it is the denominator.
Integrate $x\, dm$ over the same limits. That is the numerator.
Divide, and check that the constants in the density function cancel. They almost always do, and a symbolic answer that still contains $\lambda_0$ is usually an arithmetic slip.

The integrals you will meet are within $\int x^n\, dx = \frac{1}{n+1}x^{n+1}$ for $n \neq -1$ , which the Table of Information prints in its Calculus table.

Traps this topic sets

The center of mass is a location, not a material point. Nothing has to be there. A uniform ring's center of mass sits in the hole; the boomerang-shaped composite in a free-response figure has its center of mass off the object entirely. Statement 2.1.B.4 says the system can be modeled as a singular object located at the center of mass, which is a modeling claim, not a claim about where the atoms are.

A varying density does not move the center of mass proportionally. Doubling the density everywhere changes nothing, because the constant cancels between numerator and denominator. Only the shape of the density function matters. That cancellation is worth doing symbolically before you touch a calculator, and it is also the check that you set the integral up correctly.

Averaging positions is not averaging masses. $\sum m_i \vec{x}_i / \sum m_i$ weights each position by its mass. Students who compute $(\sum \vec{x}_i)/N$ get the centroid of the positions, which agrees with the center of mass only when all the masses are equal.

Internal forces do not move it. Statement 2.3.A.2, one topic later, says interactions between objects within a system (internal forces) do not influence the motion of a system's center of mass. Two skaters pushing off each other on frictionless ice keep their common center of mass exactly where it was. This is the fact that makes the center of mass worth defining at all.

Set the origin before you integrate, and keep it. The numbers change with the origin and the physics does not. Declare it in the first line of your work, and if the symbolic answer comes out negative, that is information about which side of your origin the answer sits on, not an error.

If you want the algebra-based version of this topic

If you are taking AP Physics 1, the page for you is AP Physics 1 Topic 2.1: Systems and Center of Mass. It covers the same system vocabulary and the same discrete sum, and it stops where the AP Physics 1 boundary statement stops. If you are taking AP Physics C: Mechanics, stay here: the integral form and the density derivative are yours and are not on that page.

	AP Physics 1 Topic 2.1	AP Physics C Topic 2.1
Objectives	2.1.A, 2.1.B	2.1.A, 2.1.B
Statements under 2.1.B	3	4
Continuous bodies	not in the course	statement 2.1.B.3
Linear mass density	not in the course	2.1.B.3.i, as a derivative
Boundary statement	five or fewer particles, or highly symmetrical	none printed
On the equation sheet	the sum only	the sum, the integral, and $\lambda$

Both courses share objective 2.1.A word for word, so the system, object and rigid-system vocabulary transfers in either direction. The other page is the better read for that vocabulary if you want it slower; this page is the one that carries the integral.

Related ideas elsewhere on this site: the center of mass glossary entry for a one-paragraph definition, and the Unit 2 hub for how this topic sits against the other nine.

How Topic 2.1 is tested

Unit 2 carries 20 to 25 percent of the multiple-choice section across about 15 to 25 class periods, the highest minimum weighting of any unit in AP Physics C: Mechanics (only Unit 3, at 15 to 25 percent, reaches the same 25 percent ceiling). Topic 2.1 is one of ten topics in it.

The suggested skills tell you the shape of the questions. Skill 1.A is create diagrams, tables, charts, or schematics to represent physical situations. Skill 2.B is calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway. Skill 2.C is compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario. Skill 3.B is apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Notice which skill is absent: 2.A, derive a symbolic expression, is not among Topic 2.1's suggested skills, even though it carries 25 to 30 percent of the multiple-choice section overall. Topic 2.1 questions tend to ask you to compute or compare a location rather than to produce a general expression.

The CED's own sample free-response set aligns question 3, the Experimental Design and Analysis question worth 10 points, to six learning objectives, and 2.1.B is the first of them. Its other suggested skills for that question are 3.A, 2.D, 1.B, 2.B and 3.C.

Two of the six optional sample instructional activities the CED lists for Unit 2 are on Topic 2.1: one has students use two bathroom scales and a long wooden plank to determine the location of their own center of mass and then how far it moves as they raise their arms, and the other has pairs of students trade free-body diagrams and suggest situations that would produce them.

If you want practice on the surrounding material, the forces and Newton's laws practice set is the closest fit on this site, though it is written for the algebra-based treatment.

A rod whose density grows as the square of position

A thin rod of length $L = 0.80$ m lies along the $x$ -axis with one end at the origin. Its linear mass density is $\lambda(x) = \lambda_0 (x/L)^2$ with $\lambda_0 = 1.2$ kg/m. Find (a) the total mass, (b) the position of the center of mass, and (c) how the answer to (b) would change if $\lambda_0$ were doubled.

Set the axis before anything else: $x$ measured from the light end at the origin, positive toward the heavy end at $x = L$ .
Write the mass element. Statement 2.1.B.3.i defines $\lambda$ as $dm/d\ell$ , so $dm = \lambda(x)\, dx = \lambda_0 (x/L)^2\, dx$ .
(a) Statement 2.1.B.3.ii says the total mass is the density integrated over the length: $M = \int_0^L \lambda_0 \dfrac{x^2}{L^2}\, dx = \dfrac{\lambda_0}{L^2}\cdot\dfrac{L^3}{3} = \dfrac{\lambda_0 L}{3}$ . The integral used is $\int x^n\, dx = \frac{1}{n+1}x^{n+1}$ , printed in the Calculus table of the Table of Information.
Numerically, $M = \dfrac{(1.2)(0.80)}{3} = 0.32$ kg.
Sanity check the size. A uniform rod at the maximum density $\lambda_0 = 1.2$ kg/m would mass $1.2 \times 0.80 = 0.96$ kg, and $0.32$ kg is one third of that, which is what a quadratic profile should give.
(b) Use $\vec{r}_{\text{cm}} = \int \vec{r}\, dm / \int dm$ from 2.1.B.3. The numerator is $\int_0^L x \lambda_0 \dfrac{x^2}{L^2}\, dx = \dfrac{\lambda_0}{L^2}\cdot\dfrac{L^4}{4} = \dfrac{\lambda_0 L^2}{4}$ .
Divide before substituting numbers: $x_{\text{cm}} = \dfrac{\lambda_0 L^2 / 4}{\lambda_0 L / 3} = \dfrac{3}{4}L$ . The density constant cancels, which is the check that the setup is right.
Numerically, $x_{\text{cm}} = 0.75(0.80) = 0.60$ m from the light end. The arithmetic route agrees: numerator $= (1.2)(0.80)^2/4 = 0.192$ kg m, and $0.192 / 0.32 = 0.60$ m.
(c) Doubling $\lambda_0$ doubles the total mass to 0.64 kg and leaves the center of mass at 0.60 m. The constant cancelled in step 7, so only the shape of $\lambda(x)$ can move the center of mass, never its overall scale.
Compare with a uniform rod, whose center of mass is at $L/2 = 0.40$ m by the symmetry argument of 2.1.B.1. The quadratic profile pushes it 0.20 m further toward the heavy end.

(a) $M = \lambda_0 L / 3 = 0.32$ kg. (b) $x_{\text{cm}} = \frac{3}{4}L = 0.60$ m from the light end. (c) It does not change: $\lambda_0$ cancels between the numerator and the denominator, so scaling the density uniformly moves nothing.

A disc with a hole, done without an integral

A uniform disc of radius $R = 0.24$ m has a circular hole of radius $R/2$ cut out of it. The center of the hole is a distance $R/2$ from the center of the disc. Locate the center of mass of what is left.

Put the origin at the center of the original full disc, with the positive $x$ -axis pointing toward the center of the hole. The problem is symmetric about that axis, so by 2.1.B.1 the center of mass lies on it and $y_{\text{cm}} = 0$ .
This is a case where the integral is the wrong tool. The material is uniform, so the discrete sum of 2.1.B.2 handles it if you treat the hole as a piece of negative mass. Full disc = remainder + hole, so remainder = full disc minus hole.
Write the masses in terms of the surface density $\sigma$ : the full disc is $M_{\text{full}} = \sigma \pi R^2$ and the removed piece is $m_{\text{hole}} = \sigma \pi (R/2)^2 = M_{\text{full}}/4$ . So the remainder has mass $M_{\text{rem}} = \tfrac{3}{4} M_{\text{full}}$ .
Apply 2.1.B.2 to the full disc, whose center of mass is at the origin by symmetry: $M_{\text{full}} (0) = M_{\text{rem}} x_{\text{cm}} + m_{\text{hole}} \left(\dfrac{R}{2}\right)$ .
Substitute the mass fractions: $0 = \tfrac{3}{4} M_{\text{full}}\, x_{\text{cm}} + \tfrac{1}{4} M_{\text{full}} \left(\dfrac{R}{2}\right)$ , so $x_{\text{cm}} = -\dfrac{1}{3}\cdot\dfrac{R}{2} = -\dfrac{R}{6}$ .
Numerically, $x_{\text{cm}} = -0.24/6 = -0.040$ m. The minus sign says the center of mass has moved 4.0 cm away from the hole, which is the direction it must move.
Check with explicit numbers. Take $\sigma = 8.0\ \mathrm{kg/m^2}$ : $M_{\text{full}} = 8.0\pi(0.24)^2 = 1.4476$ kg, $m_{\text{hole}} = 8.0\pi(0.12)^2 = 0.36191$ kg, $M_{\text{rem}} = 1.08573$ kg. Then $x_{\text{cm}} = -(0.36191)(0.12)/1.08573 = -0.0400$ m, and $\sigma$ has cancelled as it had to.
Note where the answer lives. The point $x = -0.040$ m is inside the remaining material here, but push the hole out toward the rim and the center of mass of the remainder can end up inside the hole, where there is no matter at all. That is allowed: 2.1.B.4 makes the center of mass a modeling location, not a physical particle.

$x_{\text{cm}} = -R/6 = -0.040$ m, that is, 4.0 cm from the center of the original disc on the side away from the hole, on the line joining the two centers.

Frequently asked questions

What is the center of mass formula for a continuous object in AP Physics C?

Essential knowledge 2.1.B.3 of the AP Physics C: Mechanics course and exam description states that for a nonuniform solid that can be considered as a collection of differential masses dm, the solid's center of mass equals the integral of r dm divided by the integral of dm. Both integrals run over the whole body. In one dimension you write dm as lambda(x) dx, where lambda is the linear mass density, and the integral in the denominator is just the total mass. This equation is printed on the AP Physics C: Mechanics equation sheet. It does not appear on the AP Physics 1 sheet, which prints only the summation form for discrete masses.

What is the difference between AP Physics 1 and AP Physics C for center of mass?

Both courses share learning objective 2.1.A on systems word for word, and both print the summation formula for the center of mass of discrete masses. AP Physics C: Mechanics adds essential knowledge 2.1.B.3, the integral of r dm over the integral of dm for a continuous nonuniform body, plus 2.1.B.3.i defining linear mass density as the derivative of mass with respect to position and 2.1.B.3.ii on getting total mass by integrating a density function over a length, area or volume. AP Physics 1 instead carries a boundary statement limiting students to systems of five or fewer particles in a two-dimensional configuration, or highly symmetrical systems. AP Physics C: Mechanics prints no boundary statement on this topic.

How do you find the total mass of a rod with variable density?

Integrate the density function over the length of the rod. Essential knowledge 2.1.B.3.ii of the AP Physics C: Mechanics framework says that if a function of mass density is given for a solid, the total mass can be determined by integrating the mass density over the length in one dimension, the area in two dimensions, or the volume in three dimensions, and gives the example that total mass equals the integral of rho(r) dV. For a rod along the x-axis with linear mass density lambda(x), the total mass is the integral of lambda(x) dx from one end to the other. That same integral is the denominator of the center of mass expression, so you have already done half the work.

Is linear mass density on the AP Physics C equation sheet?

Yes. The AP Physics C: Mechanics Table of Information prints lambda equals the derivative of m with respect to l, in the Mechanics equations table, and its symbol list defines lambda as linear mass density. It sits directly below the integral form of the center of mass. Neither line appears on the AP Physics 1 equation sheet. Note that the volume density symbol rho is used in essential knowledge 2.1.B.3.ii but is not defined in the mechanics symbol list, so a question involving a volume density will describe it in words.

Can the center of mass be located where there is no mass?

Yes, and it often is. A uniform ring has its center of mass at the geometric center, which is empty space, and a horseshoe or an L-shaped bracket has its center of mass outside the material. Essential knowledge 2.1.B.4 of the AP Physics C: Mechanics framework says a system can be modeled as a singular object that is located at the system's center of mass, which is a statement about how you may model the system, not a claim that any matter sits at that point. The center of mass is the point that moves as though the net external force acted on the whole mass concentrated there.

Do internal forces move a system's center of mass?

No. Essential knowledge 2.3.A.2 of the AP Physics C: Mechanics framework states that interactions between objects within a system, that is, internal forces, do not influence the motion of a system's center of mass. Two ice skaters pushing apart on a frictionless surface keep their common center of mass exactly where it was, because the forces they exert on each other are a Newton's third law pair internal to the system. Only a net external force can change the velocity of a system's center of mass, which is what essential knowledge 2.5.A.3 says in Topic 2.5.

When should you use the integral instead of the summation for center of mass?

Use symmetry first, the summation second, and the integral only when neither works. Essential knowledge 2.1.B.1 says the center of mass of a symmetrical mass distribution lies on the lines of symmetry, which settles uniform rods, discs and spheres immediately. The summation handles separate point masses and composite bodies you can split into pieces whose centers you already know, including bodies with holes, where the hole is treated as negative mass. The integral of 2.1.B.3 is needed only when the density varies continuously, because then no finite set of pieces is exact.