Center of Mass & Systems of Particles

One point that moves simply

Throw a wrench across a room. It tumbles, end over end, and no part of it traces anything you could call a nice curve. The handle loops, the head loops the other way, and the whole thing looks hopeless to describe. But paint a dot in exactly the right place on that wrench and film it again: the dot flies a clean parabola, the same parabola a marble thrown with the same push would follow.

That place is the center of mass. It is not a physical part of the object. It is a bookkeeping point, an average position, and its whole value is that it is the one point of a complicated system that obeys $F = ma$ as if the system were a single particle sitting there. In the last lesson on collisions you used conservation of momentum to skip past the messy contact forces. The center of mass is the same trick made geometric: you skip past the messy internal structure and keep one point.

Averaging positions, weighted by mass

Start with the simplest possible system: two masses on a line. Put $m_1 = 2.0$ kg at $x = 0$ and $m_2 = 3.0$ kg at $x = 5.0$ m. Where is the "average" position? Not the midpoint at 2.5 m, because the 3.0 kg mass ought to count for more. Imagine the two sitting on a weightless plank, and ask where you'd have to put a knife-edge so the plank balances. Balance means the turning effects match: $m_1 d_1 = m_2 d_2$, where $d_1$ and $d_2$ are the distances from the knife-edge to each mass.

If the balance point is at $x$, then $d_1 = x$ and $d_2 = 5.0 - x$, so $2.0x = 3.0(5.0 - x)$, giving $5.0x = 15$ and $x = 3.0$ m. Closer to the heavy one, as it should be.

Now write that in a way that scales. Rearranged, $m_1 x + m_2 x = m_1 x_1 + m_2 x_2$, so

$x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}.$

Nothing about the derivation cared that there were two masses. For any number of them, in any number of dimensions,

$\vec R_{\text{cm}} = \frac{\sum_i m_i \vec r_i}{\sum_i m_i} = \frac{1}{M}\sum_i m_i \vec r_i,$

where $m_i$ is the mass of the $i$th piece, $\vec r_i$ is its position vector, and $M = \sum_i m_i$ is the total mass. In components this is three separate one-dimensional averages: $x_{\text{cm}}$ uses only the $x$ values, $y_{\text{cm}}$ only the $y$ values. They never mix. That fact alone turns most two-dimensional center-of-mass problems into two easy one-dimensional ones.

Two things to notice about the formula. Each mass is weighted by its share of the total, $m_i/M$, and those shares add to one, so $\vec R_{\text{cm}}$ always lands somewhere between the extremes: never outside the spread of the particles. And the origin is yours to choose. Put it wherever makes the arithmetic easiest; the physical point you find won't move.

Worked example 1. Masses of 1.0, 2.0 and 3.0 kg sit at $x = 0$, 2.0 m and 6.0 m. Find the center of mass.

Total mass first: $M = 1.0 + 2.0 + 3.0 = 6.0$ kg. Then the weighted sum of positions:

$\sum m_i x_i = (1.0)(0) + (2.0)(2.0) + (3.0)(6.0) = 0 + 4.0 + 18 = 22\ \text{kg·m}.$

So $x_{\text{cm}} = 22/6.0 = 3.67$ m. It sits between 2.0 and 6.0, pulled toward the heavy end, which is the sanity check to run every time.

Solid objects: symmetry first, then pieces

A real object has mass spread continuously, so the sum becomes an integral, $\vec R_{\text{cm}} = \frac{1}{M}\int \vec r \, dm$. You will need that sometimes. Reach for it last, because two shortcuts handle most problems.

Symmetry. If an object is uniform and has a plane of symmetry, its center of mass lies on that plane: every bit of mass on one side is matched by an equal bit mirrored on the other, and the two contributions to the average cancel around the plane. A uniform rod of length 2.4 m has its center of mass at 1.2 m. A uniform disk, sphere, or rectangular plate has it at the geometric center. No integral, no arithmetic.

Composite bodies. Cut the object into pieces whose centers you already know, then treat each piece as a point mass at its own center. This works because the definition is a sum, and you can group the terms of a sum however you like. For a uniform sheet, mass is proportional to area, so you can use areas in place of masses and let the density cancel.

Worked example 2. A uniform L-shaped plate is a $4.0 \times 1.0$ m rectangle lying along the $x$-axis with its lower-left corner at the origin, plus a $1.0 \times 3.0$ m rectangle stacked on its left end, occupying $x$ from 0 to 1.0 m and $y$ from 1.0 to 4.0 m. Find the center of mass.

Piece A: area $4.0 \times 1.0 = 4.0$ m². Its center is at the middle of that rectangle, $(2.0, 0.5)$ m.

Piece B: area $1.0 \times 3.0 = 3.0$ m². It runs from $x = 0$ to 1.0 and $y = 1.0$ to 4.0, so its center is at $(0.5, 2.5)$ m.

Total area 7.0 m². Now the two averages, using area as the weight:

$x_{\text{cm}} = \frac{(4.0)(2.0) + (3.0)(0.5)}{7.0} = \frac{8.0 + 1.5}{7.0} = \frac{9.5}{7.0} = 1.36\ \text{m}$

$y_{\text{cm}} = \frac{(4.0)(0.5) + (3.0)(2.5)}{7.0} = \frac{2.0 + 7.5}{7.0} = \frac{9.5}{7.0} = 1.36\ \text{m}$

Equal, because this L is symmetric about the line $y = x$. Spotting that in advance would have halved the work.

When you have to integrate

Two cases force an integral: a curved uniform shape, and a rod whose density varies.

For the varying rod, slice it. A slice at position $x$ of width $dx$ has mass $dm = \lambda(x)\,dx$, where $\lambda$ is the linear density in kg per metre. The total mass is $M = \int \lambda\,dx$ and the weighted sum is $\int x\lambda\,dx$, so

$x_{\text{cm}} = \frac{\int_0^L x\,\lambda(x)\,dx}{\int_0^L \lambda(x)\,dx}.$

Worked example 3. A rod of length 4.0 m has $\lambda(x) = \lambda_0(1 + x/L)$ with $L = 4.0$ m. Find its center of mass.

The constant $\lambda_0$ appears in both integrals, so it will cancel: drop it now and carry it in your head. Denominator:

$\int_0^L \left(1 + \frac{x}{L}\right)dx = \left[x + \frac{x^2}{2L}\right]_0^L = L + \frac{L}{2} = \frac{3L}{2}.$

Numerator:

$\int_0^L x\left(1 + \frac{x}{L}\right)dx = \int_0^L \left(x + \frac{x^2}{L}\right)dx = \frac{L^2}{2} + \frac{L^2}{3} = \frac{5L^2}{6}.$

Divide: $x_{\text{cm}} = \frac{5L^2/6}{3L/2} = \frac{5L^2}{6}\cdot\frac{2}{3L} = \frac{5L}{9}$. With $L = 4.0$ m that is $20/9 = 2.22$ m. Past the midpoint, because the far end is denser. If instead $\lambda = 2x$ on a rod of length 3.0 m, the same method gives numerator $\int_0^3 2x^2 dx = 18$ and denominator $\int_0^3 2x\,dx = 9$, so $x_{\text{cm}} = 2.0$ m: exactly $2L/3$, which is where any $\lambda \propto x$ rod balances.

For curved uniform solids the integrals are standard and worth memorising rather than rederiving under time pressure. Measured from the flat face:

• Half disk (semicircular plate) of radius $R$: $4R/3\pi \approx 0.424R$.
• Solid hemisphere of radius $R$: $3R/8 = 0.375R$.
• Solid cone of height $h$: $h/4$ above the base.
• Hemispherical shell (a bowl): $R/2$.
• Semicircular wire (not a plate): $2R/\pi \approx 0.637R$.

The pattern is that the more mass sits near the wide base, the lower the center of mass, and a solid always sits lower than the shell of the same shape.

Why this point and not some other

Everything above is geometry. Here is the physics that makes the point worth finding. Differentiate the definition with respect to time. The masses are constant, so

$M\vec V_{\text{cm}} = \sum_i m_i \vec v_i = \vec P_{\text{total}}.$

The total momentum of any system is the total mass times the velocity of its center of mass. A 12 kg system whose center of mass moves at 2.5 m/s carries $12 \times 2.5 = 30$ kg·m/s of momentum, regardless of how the parts are moving individually. Differentiate once more:

$M\vec A_{\text{cm}} = \sum_i m_i \vec a_i = \sum_i \vec F_i = \vec F_{\text{ext}}.$

That last step is the important one. The forces on the pieces come in two kinds: internal (piece pushing on piece) and external (everything else). Internal forces come in third-law pairs, equal and opposite, so they cancel exactly in the sum. Only external forces survive:

$\vec F_{\text{ext}} = M\vec A_{\text{cm}}.$

So the center of mass accelerates as though the whole mass were concentrated there and only outside forces acted. Nothing you do inside a system can accelerate its center of mass. That is why you cannot lift yourself by your own belt, and why an astronaut drifting free must throw a tool to get anywhere.

The corollary problems actually use: with no net external force, the center of mass moves at constant velocity, and if it started at rest it stays put. Walk from one end of a floating raft to the other and the raft slides backward by exactly enough to hold $\vec R_{\text{cm}}$ fixed.

The exploding projectile

This is the classic application and it appears constantly. Fire a shell on a parabola and let it burst at the top. The explosive forces are internal, so by the argument above they do nothing to the center of mass. Gravity is still the only external force and still acts on the total mass. The center of mass therefore carries on along the original parabola as if nothing had happened, right up until a fragment hits the ground and the ground starts pushing.

Worked example 4. A 6.0 kg firework at the top of its flight, 100 m downrange, splits into a 2.0 kg piece and a 4.0 kg piece. The 2.0 kg piece lands at 60 m. The unexploded firework would have landed at 200 m. Where does the 4.0 kg piece land?

Write down what you know before anything else: the center of mass lands at 200 m. It must, because gravity has not changed. Assume both fragments land at about the same time, so that at the moment of landing the definition applies to the two landing spots:

$M x_{\text{cm}} = m_1 x_1 + m_2 x_2$ $(6.0)(200) = (2.0)(60) + (4.0)x_2$ $1200 = 120 + 4.0 x_2$ $x_2 = \frac{1080}{4.0} = 270\ \text{m}.$

The heavy piece overshoots the undisturbed landing point, which is the whole point of the problem: the light piece was thrown backward, so the heavy one had to be thrown forward.

Toppling: where the center of mass sits over the support

An object tips when its center of mass passes beyond the edge of whatever supports it. Gravity acts at the center of mass, so once that point is past the edge, the weight's turning effect swings the object off rather than back onto the support.

Worked example 5. Two identical uniform blocks of length $L = 0.30$ m are stacked at the edge of a table, each pushed out as far as it can go. What is the biggest total overhang?

Work from the top down, because the top block only has to worry about the block under it. Its center is at its midpoint, so it can hang out at most $L/2$ past the front edge of the lower block.

Now the pair. The lower block's center is at its own midpoint; the top block's center is $L/2$ further out. Their combined center, equal masses, is halfway between: $L/4$ ahead of the lower block's midpoint. The pair tips when that combined center passes the table edge, so the lower block's midpoint can be at most $L/4$ back from the edge: meaning the lower block sticks out $L/2 - L/4 = L/4$ past the table.

Total overhang $= L/2 + L/4 = 3L/4 = 0.75 \times 0.30 = 0.225$ m.

A related favourite is the chain over a pulley. A 2.0 m, 5.0 kg uniform chain hangs with 1.4 m on one side and 0.60 m on the other. Each side is a uniform rod, so each has its own center at its own midpoint: 0.70 m and 0.30 m below the pulley. Mass is proportional to length, so the sides weigh $5.0 \times 1.4/2.0 = 3.5$ kg and $5.0 \times 0.6/2.0 = 1.5$ kg. Then

$y_{\text{cm}} = \frac{(3.5)(0.70) + (1.5)(0.30)}{5.0} = \frac{2.45 + 0.45}{5.0} = \frac{2.90}{5.0} = 0.58\ \text{m}$

below the pulley. Composite bodies again: nothing new, just a careful splitting.

How to recognise a center-of-mass problem

Four signals, and you will see at least one in nearly every problem of this type.

Something splits, or joins, or rearranges, with no outside push. An exploding shell, a person walking on a raft, a cannon recoiling on ice, two skaters pushing off. The phrase "frictionless" or "in space" is the tell. Find the center of mass before and after, set them equal, and you have your equation.

The problem asks for a balance point, a pivot, or a tipping condition. Seesaws, overhanging blocks, a ladder about to slip off a ledge. You are being asked whether the center of mass is over the support.

The object is made of standard shapes glued together, or has a hole. Split into pieces, use areas or volumes as weights, and treat the hole as negative.

The problem hands you a density that depends on position. $\lambda = 2x$, or $\rho = \rho_0(1 + r/R)$. That is an explicit instruction to integrate; set up the two integrals and let the constants cancel.

One habit is worth building. Before writing anything, draw the axis and mark your origin, then guess where the answer should land. The center of mass is always inside the spread of the mass and always closer to the heavy part. A guess that the answer is near 3 m will catch an arithmetic slip that produces 30 m.

From here the course turns to rotation. You have the point a body translates about; next you need the angles it turns through, which starts with angular kinematics. The center of mass comes straight back when you find that a rolling object's kinetic energy splits neatly into the motion of its center of mass plus the spin around it.