Systems and Center of Mass (AP Physics C: Mechanics 2.1)

Toss a hammer end over end across a room. The head and the end of the handle trace loops, but one point between them, much closer to the heavy head, follows the same smooth parabola a thrown ball would. That point is the hammer's center of mass. Before you can apply any of Newton's laws, you have to decide what object you are analyzing and where its center of mass sits.

A system is whatever you draw a boundary around

A system is the object, or collection of objects, that you choose to analyze. Everything outside it is the environment. The choice is yours, and it decides how forces are sorted: a force that one part of the system exerts on another part is internal, and a force exerted by something in the environment is external. Topic 2.3 shows why internal forces cannot move the system as a whole.

What a system does comes from how its parts interact. A rubber ball bounces because its molecules push back when the ball is squashed; a steel beam holds its shape because neighboring atoms pull on each other. Even the system's mass is a property of the collection: it is the sum of the parts' masses, and how that mass is spread out sets where the center of mass is.

Whether you need any of that detail depends on the question you are asking.

A $1500\ \text{kg}$ car braking on a highway: every part moves together, so the car can be one object, a single mass with forces on it.
The same car in a crash test: the crumple zone collapses while the passenger cabin stays rigid. The internal structure is now the whole question, and the car has to be modeled as separate parts.

The working rule: if the parts move together and the question doesn't ask about the forces between them, model the system as a single object. If it asks what one part does, or what one part does to another, the parts have to appear in your analysis.

Three more facts about systems come up constantly.

Parts can move differently from the whole. Two skaters standing still on ice push off each other. One glides left, the other glides right, and the center of mass of the two-skater system stays exactly where it was. Neither skater moves the way the system does.

Mass and energy can cross the boundary. A rocket that burns fuel throws mass out of its nozzle, so a system containing only the rocket loses mass every second. A system containing the rocket and all of its exhaust keeps a constant mass. Energy crosses boundaries too: a crate sliding to a stop on a rough floor loses kinetic energy, which ends up as thermal energy in the crate and floor. Which of these transfers you have to track depends on where you put the boundary.

Internal structure matters, and it can change. A rope can only pull, while a rigid rod can push or pull, so the same "connector" gives different equations depending on what it is made of. And the structure of a system can change when you change something outside it. Push gently on the bottom block of a two-block stack and the stack moves as one object. Push harder and the top block slips: you now have two objects with two different accelerations. Pull hard enough on a chain and a link breaks, turning one system into two.

Center of mass of separate particles

For particles spread along the $x$ -axis, the center of mass is the mass-weighted average of their positions:

x_{cm} = \frac{\sum m_i x_i}{\sum m_i}.

The same formula applies to each axis separately, so in two or three dimensions you find $x_{cm}$ , $y_{cm}$ and $z_{cm}$ one at a time, or write it as a vector, $\vec{r}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$ .

Take a $2.0\ \text{kg}$ mass at $x = 0$ and a $6.0\ \text{kg}$ mass at $x = 4.0\ \text{m}$ :

x_{cm} = \frac{(2.0)(0) + (6.0)(4.0)}{2.0 + 6.0} = 3.0\ \text{m}.

The center of mass is $3.0\ \text{m}$ from the light mass and $1.0\ \text{m}$ from the heavy one. For two masses, the distances are always in the inverse ratio of the masses ( $2.0 \times 3.0 = 6.0 \times 1.0$ ), so the center of mass sits between them, closer to the heavier one. Use that as a quick check on any answer.

Why weight by mass? Differentiate the definition with respect to time: $\vec{v}_{cm} = \frac{\sum m_i \vec{v}_i}{M}$ , so $M\vec{v}_{cm}$ equals the total momentum of all the particles. That is what lets the whole system's motion be described by one point carrying the total mass $M$ .

Symmetry finds it without arithmetic. If the mass is arranged symmetrically about a line, every bit of mass on one side has a twin at the mirror-image spot on the other side. Their contributions to the position measured across that line cancel in the sum, so the center of mass must lie on the line. A uniform rod's center of mass is at its midpoint; a uniform rectangle's, disk's, ring's or sphere's is at its geometric center. Two lines of symmetry pin it at their intersection.

The center of mass doesn't have to be inside the material. A ring's is at its empty center. A uniform horseshoe's lies on its line of symmetry, in the empty space between the arms, but not at the center of the circle it was bent from. Measured along the symmetry axis, more of the metal lies on the curved side of that center (the bend) than on the open side (the gap between the arm tips), so the average position is pulled from the circle's center toward the bend.

Composite objects: replace each piece by a single particle with that piece's mass, placed at that piece's own center of mass, then use the sum. Worked example 2 does this.

Continuous objects: integrate over mass elements

A rod or a plate is not a list of particles, so the sum becomes an integral. Slice the object into tiny pieces of mass $dm$ , each at position $x$ :

The density is what makes the integral doable: it converts "a piece of mass" into "a piece of length". For a uniform rod of length $L$ from $x = 0$ to $x = L$ , $\lambda = \frac{M}{L}$ is constant and

x_{cm} = \frac{1}{M}\int_0^L x\,\frac{M}{L}\,dx = \frac{1}{L}\cdot\frac{L^2}{2} = \frac{L}{2},

which agrees with symmetry. Now let the density grow toward one end, $\lambda = kx$ with $k$ a constant in $\text{kg/m}^2$ :

M = \int_0^L kx\,dx = \frac{kL^2}{2}, \qquad x_{cm} = \frac{1}{M}\int_0^L x\,(kx)\,dx = \frac{kL^3/3}{kL^2/2} = \frac{2L}{3}.

More mass near $x = L$ pulls the center of mass past the midpoint, toward the dense end.

In two or three dimensions, the density is per area, $\sigma = \frac{dm}{dA}$ in $\text{kg/m}^2$ , or per volume, $\rho = \frac{dm}{dV}$ in $\text{kg/m}^3$ , and the total mass is $\int \sigma\,dA$ or $\int \rho\,dV$ . The trick is to slice the object into strips on which the density is constant, so the integral has one variable. A rectangular plate of length $L$ and width $w$ whose density depends only on distance from one edge, $\sigma = \sigma_0\frac{x}{L}$ , slices into strips of width $dx$ and area $dA = w\,dx$ :

M = \int_0^L \sigma_0 \frac{x}{L}\,w\,dx = \frac{\sigma_0 w L}{2}.

Treating a system as one object

Once you know where the center of mass is, the system's overall motion follows one rule: the center of mass moves as if all the mass were concentrated there and every external force acted there. Topic 2.5 writes this as Newton's second law for the system, $\vec{a}_{cm} = \frac{\sum \vec{F}_{ext}}{M}$ , and Topic 2.3 shows why internal forces drop out (they come in equal and opposite pairs, so they add to zero).

That is why the skaters' push leaves their center of mass at rest: the push is internal. It is also why a firework that bursts in mid-air scatters its pieces in every direction while the center of mass of all the pieces keeps following the original parabola, at least until the first fragment hits the ground and the ground's force, an external one, joins in. And it is why a free-body diagram (Topic 2.2) draws an object as a single dot at its center of mass.

The single-object model throws away two things: how the parts move relative to each other, and rotation about the center of mass, which Unit 5 handles. When a question asks about either, go back to the parts.

Worked examples

Worked exampleThree particles in a plane

A $1.0\ \text{kg}$ particle is at the origin, a $2.0\ \text{kg}$ particle is at $(3.0\ \text{m},\ 0)$ , and a $3.0\ \text{kg}$ particle is at $(0,\ 2.0\ \text{m})$ . Find the center of mass of the three-particle system.

Show solution

Work one axis at a time. The total mass is $M = 1.0 + 2.0 + 3.0 = 6.0\ \text{kg}$ .

x_{cm} = \frac{(1.0)(0) + (2.0)(3.0) + (3.0)(0)}{6.0} = 1.0\ \text{m}

y_{cm} = \frac{(1.0)(0) + (2.0)(0) + (3.0)(2.0)}{6.0} = 1.0\ \text{m}

The center of mass is at $(1.0\ \text{m},\ 1.0\ \text{m})$ . The particle at the origin adds nothing to either numerator, but its $1.0\ \text{kg}$ still belongs in the denominator: it pulls the center of mass toward the origin. The answer lies inside the triangle the particles form, where a weighted average has to be, and it is not on any particle.

Worked exampleA rod with a ball on the end

A uniform rod of length $1.20\ \text{m}$ and mass $0.60\ \text{kg}$ has a small $0.90\ \text{kg}$ ball fixed to its right end. How far from the rod's left end is the center of mass of the rod-and-ball system?

Show solution

Treat each piece as a particle at its own center of mass. By symmetry, the uniform rod acts like $0.60\ \text{kg}$ at its midpoint, $0.60\ \text{m}$ from the left end. The ball is a particle at $1.20\ \text{m}$ . Measuring from the left end:

x_{cm} = \frac{(0.60)(0.60) + (0.90)(1.20)}{0.60 + 0.90} = \frac{0.36 + 1.08}{1.50} = 0.96\ \text{m}.

Check it: $0.96\ \text{m}$ is between the rod's center ( $0.60\ \text{m}$ ) and the ball ( $1.20\ \text{m}$ ), and closer to the ball, which is the heavier piece. Replacing the rod by a point at its center is what saves you an integral here, and it works only because the rod is uniform.

Worked exampleA rod that gets denser along its length

A thin rod lies on the $x$ -axis from $x = 0$ to $x = L$ . Its linear mass density is $\lambda(x) = \lambda_0\left(1 + \frac{2x}{L}\right)$ , where $\lambda_0$ is a positive constant.

(a) Derive an expression for the rod's total mass $M$ in terms of $\lambda_0$ and $L$ .

(b) Derive an expression for the position $x_{cm}$ of the rod's center of mass in terms of $L$ .

(c) A student says the center of mass must be at $x = \frac{L}{2}$ because the rod is a straight, symmetric shape. Explain what is wrong with this claim, using the density function.

(d) For $L = 0.90\ \text{m}$ and $\lambda_0 = 0.40\ \text{kg/m}$ , calculate $M$ and $x_{cm}$ .

Show solution

(a) Start from the definition of density, $dm = \lambda\,dx$ , and add up every element over the length of the rod:

M = \int_0^L \lambda_0\left(1 + \frac{2x}{L}\right)dx = \lambda_0\left[x + \frac{x^2}{L}\right]_0^L = \lambda_0(L + L) = 2\lambda_0 L.

Units check: $\lambda_0 L$ is $(\text{kg/m})(\text{m}) = \text{kg}$ .

(b) Weight each element's position by its mass, then divide by the total mass from (a):

\int_0^L x\,dm = \int_0^L \lambda_0\left(x + \frac{2x^2}{L}\right)dx = \lambda_0\left(\frac{L^2}{2} + \frac{2L^2}{3}\right) = \frac{7}{6}\lambda_0 L^2

x_{cm} = \frac{\frac{7}{6}\lambda_0 L^2}{2\lambda_0 L} = \frac{7L}{12}.

Showing $dm = \lambda\,dx$ , the integral with its limits, and the division by $M$ as separate lines is what earns the derivation points; a bare $\frac{7L}{12}$ shows none of the physics.

(c) Symmetry of shape is not enough: the mass has to be distributed symmetrically. Here the density at $x = L$ is $\lambda_0(1 + 2) = 3\lambda_0$ , three times the density at $x = 0$ , so more mass lies in the right half of the rod than the left. The weighted average of position is pulled to the right of the midpoint, and indeed $\frac{7L}{12} > \frac{6L}{12}$ .

(d) $M = 2(0.40\ \text{kg/m})(0.90\ \text{m}) = 0.72\ \text{kg}$ , and $x_{cm} = \frac{7(0.90\ \text{m})}{12} = 0.525\ \text{m} \approx 0.53\ \text{m}$ from the light end.

Where students lose points

Check yourself

Question 1
A firework shell is launched on a parabolic path and bursts at the top of its flight into many fragments that fly off in different directions. Air resistance is negligible. Before any fragment lands, the center of mass of all the fragments
1. A
  stops at the point of the burst and falls straight down
2. B
  continues along the parabola the shell was following
3. C
  moves along a straight line in the direction the shell was moving
4. D
  follows the path of the most massive fragment
Show answer
B
The burst is an internal interaction, so it cannot change the motion of the center of mass. The only external force is still gravity, so the center of mass keeps following the same parabola. Choice A treats the explosion as if it removed the system's velocity, but internal forces add to zero.
Question 2
A $1.0\ \text{kg}$ particle sits at $x = 8.0\ \text{m}$ . Where on the $x$ -axis must a $3.0\ \text{kg}$ particle be placed so that the center of mass of the two is at the origin?
Show answer
At $x = -\frac{8}{3}\ \text{m} \approx -2.7\ \text{m}$ .
Setting $x_{cm} = 0$ requires $(1.0)(8.0) + (3.0)x = 0$ , so $x = -2.7\ \text{m}$ . The heavier particle ends up a third as far from the center of mass as the lighter one, on the opposite side.
Question 3
A rod along the $x$ -axis from $x = 0$ to $x = L$ has linear mass density $\lambda = Cx^3$ , where $C$ is a positive constant. Where is its center of mass?
1. A
  $x = \frac{L}{2}$
2. B
  $x = \frac{2L}{3}$
3. C
  $x = \frac{3L}{4}$
4. D
  $x = \frac{4L}{5}$
Show answer
D
$M = \int_0^L Cx^3\,dx = \frac{CL^4}{4}$ and $\int_0^L x\,Cx^3\,dx = \frac{CL^5}{5}$ , so $x_{cm} = \frac{CL^5/5}{CL^4/4} = \frac{4L}{5}$ . Choice B is the answer for $\lambda \propto x$ ; a density that rises more steeply puts even more of the mass near $x = L$ .
Question 4
A thin uniform wire is bent into a semicircle. Which statement about the wire's center of mass is correct?
1. A
  It is at the midpoint of the arc, on the wire.
2. B
  It is at the center of the full circle the arc belongs to.
3. C
  It is on the semicircle's line of symmetry, in the empty space between the arc and the center of the circle.
4. D
  It cannot be located without knowing the wire's mass.
Show answer
C
The wire is symmetric about the line through the arc's midpoint and the circle's center, so the center of mass lies on that line. Every piece of wire is on the arc side of the center, so the weighted average is pulled away from the center (ruling out B), but it is an average of points on a curve, so it is not on the wire itself (ruling out A). The mass cancels out of the ratio, so D is wrong.

Systems and Center of Mass

A system is whatever you draw a boundary around

Center of mass of separate particles

Continuous objects: integrate over mass elements

Treating a system as one object

Worked examples

Where students lose points

Check yourself

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