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The center of mass of particles and of continuous bodies, $\int x\,dm/M$ for rods of changing density, and its motion $M\vec{a}_{\text{cm}} = \vec{F}_{\text{ext}}$ through walks, recoils and explosions.
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By the end of this lesson you will be able to locate the center of mass of particles and continuous bodies, and use $M\vec{a}_{\text{cm}} = \vec{F}_{\text{ext}}$ to predict how systems move.
From Physics 1 you know the center of mass of a few particles as a weighted average, and that a thrown object's center of mass follows a parabola. The last lesson showed that forces change momentum. From calculus you can integrate over a length. This lesson finds centers of mass of continuous bodies and shows why the center of mass is the one point whose motion depends only on external forces.
| Term | What it means |
|---|---|
| Center of mass | The mass-weighted average position, $\vec{r}_{\text{cm}} = \sum m_i\vec{r}_i/M$. |
| Linear density | $\lambda = dm/dx$, mass per unit length of a rod, in kg/m. |
| First moment | $\int x\,dm$, the sum of each piece's mass times its position. |
| Internal force | A force between parts of the system; they cancel in pairs by Newton's third law. |
| External force | A force on the system from something outside it. |
| Total momentum | $\vec{P} = \sum m_i\vec{v}_i = M\vec{v}_{\text{cm}}$. |
| Symmetry axis | A line about which a body's mass is balanced, on which the center of mass must lie. |
For particles of masses $m_i$ at positions $\vec{r}_i$, the center of mass is
$$\vec{r}_{\text{cm}} = \frac{\sum m_i\vec{r}_i}{M}, \qquad M = \sum m_i.$$
For a continuous body the sum becomes an integral. For a rod along $x$ with linear density $\lambda(x)$, each slice has $dm = \lambda\,dx$, and
$$x_{\text{cm}} = \frac{1}{M}\int x\,dm = \frac{\int x\lambda\,dx}{\int\lambda\,dx}.$$
The center of mass matters because of how it moves. Summing Newton's second law over every particle, the internal forces cancel in pairs by the third law, leaving
$$M\vec{a}_{\text{cm}} = \vec{F}_{\text{ext}}.$$
However a body spins, flexes or breaks apart, its center of mass moves like a single particle of mass $M$ pushed by the net external force. Its velocity is $\vec{v}_{\text{cm}} = \vec{P}/M$, so when no net external force acts, the center of mass moves at constant velocity, or stays put.
Another way: picture
Throw a wrench spinning across a room. Every point of it wobbles through a complicated path, except one: the center of mass traces a smooth parabola, as if the whole wrench were a pebble. The spinning, the wobbling and even a break in midair are internal matters; gravity alone decides the path of that one point.
Another way: steps
For two masses on a line, the center of mass divides the separation in the inverse ratio of the masses: a $3$ kg and a $1$ kg mass $4$ m apart balance $1$ m from the heavier. Each coordinate is found separately, so particles in a plane need two weighted averages, one for $x$ and one for $y$.
Symmetry saves most of the work for real bodies. A uniform sphere, cube or cylinder has its center of mass at its geometric center; a uniform rod at its midpoint; an isosceles triangle somewhere on its symmetry axis. Any body with a mirror plane has its center of mass in that plane. Before integrating anything, use every symmetry to reduce the question to the coordinates that remain unknown.
A baseball bat, a tapered antenna or a rod made of two materials has a density that changes along its length. Slice it into pieces of length $dx$ and mass $\lambda(x)\,dx$. The total mass is $\int\lambda\,dx$, and the first moment is $\int x\lambda\,dx$.
For $\lambda = kx$ on a rod of length $L$, the mass is $\tfrac{1}{2}kL^2$ and the moment $\tfrac{1}{3}kL^3$, so $x_{\text{cm}} = \tfrac{2}{3}L$: pulled toward the dense end, and independent of $k$, because only the shape of the density matters. For $\lambda = kx^2$ the center moves further, to $\tfrac{3}{4}L$. A bat's center of mass sits about two-thirds of the way from the handle, which is why it feels heavy at the barrel.
Write Newton's second law for each particle: $m_i\vec{a}_i = \vec{F}_i^{\text{ext}} + \sum_j\vec{F}_{ij}$, where $\vec{F}_{ij}$ is the force particle $j$ exerts on $i$. Add all the equations. Every internal force appears twice, as $\vec{F}_{ij}$ and $\vec{F}_{ji}$, which are equal and opposite, so they cancel. The left side adds to $M\vec{a}_{\text{cm}}$, by the definition of the center of mass.
The result, $M\vec{a}_{\text{cm}} = \vec{F}_{\text{ext}}$, is what lets us treat a car, a planet or a person as a point. The forces holding them together, however large, never affect the path of the center of mass. Only external forces do, and often only gravity is external.
If the net external force is zero, the center of mass has constant velocity, and if the system starts at rest, the center of mass stays put. A person walking along a floating raft pushes the raft backward through friction, an internal force of the person-plus-raft system. The raft slides back just enough that the center of mass does not move.
The same reasoning explains a rowboat drifting backward when someone walks to the bow, an astronaut on a spacewalk pushing off a satellite, and a gun recoiling when fired. In each case the mass-weighted displacements add to zero. Water drag and friction with the ground are external, and over time they break the rule, which is why a raft eventually drifts back.
When a firework shell bursts, the forces that tear it apart are internal. The center of mass of the fragments continues on the shell's parabola as if nothing had happened, until the first fragment hits the ground and the ground supplies an external force.
This makes predictions easy. If a shell splits into two equal pieces that land at the same time, and one lands short of where the shell would have, the other lands just as far beyond. If the pieces are unequal, the landing points are weighted: the heavy piece lands closer to the original point. Artillery engineers, and forensic investigators studying debris, use exactly this reasoning.
Checking an answer. The center of mass must lie within the smallest box that contains all the mass. It must move toward the heavier parts. It must lie on every symmetry axis. And for a system at rest with no external force, the weighted sum of all displacements must be zero.
The center of mass need not be inside the material. A ring's is at its empty center; a boomerang's lies in the air between its arms. A high jumper arching backward over the bar in the Fosbury flop bends so far that her center of mass passes under the bar while every part of her body passes over it.
That is why the technique, introduced by Dick Fosbury of Oregon State at the 1968 Olympics, let jumpers clear higher bars for the same takeoff speed: the height the center of mass must reach is set by the launch, and bending lets the body clear a bar above that height. The center of mass is an average, not a place in the body, and its trajectory obeys the equations while the body arranges itself around it.
For an irregular flat shape, a center of mass is easy to find by hanging. Suspend the shape from a pin at one point and let it settle: its center of mass hangs directly below the pin, so draw a vertical line through the pin. Hang it from a second point and draw another. The center of mass is where the lines cross.
Engineers find the center of mass of aircraft this way in principle, weighing each wheel on a separate scale and taking the weighted average of the wheel positions. Every airliner is weighed and its center of mass computed before each flight, since an aircraft whose center of mass is too far back cannot be controlled. Loading cargo and seating passengers are planned to keep it within limits.
The fireworks over the National Mall on the Fourth of July are launched as shells from mortars. Each shell climbs on a parabola, and at the top a timed fuse sets off a burst charge that scatters its stars. The burst forces are internal to the shell, so the center of mass of all the glowing fragments keeps following the shell's parabola until they start to burn out or reach the ground.
That is why a well-made shell's burst looks centered on the point where the shell was going: the stars spread symmetrically about a center that moves on smoothly. Display designers use the principle to predict where debris will fall, keeping crowds outside the fallout zone. If a shell splits into two equal pieces that land together, one short of the expected point, the other lands the same distance beyond it.
Before every flight, loadmasters on an Air Force C-17 or a FedEx freighter compute the airplane's center of mass. Each pallet's mass and position along the cargo floor enter a weighted average with the aircraft's own mass and the fuel in each tank. The result must fall within a narrow range, a few percent of the wing's width, for the plane to be stable and controllable.
A center of mass too far forward makes the plane hard to lift at takeoff; too far back, and it can pitch up out of control. As fuel burns, the center of mass shifts, and the flight computer tracks it. Every load plan is the formula $\sum m_ix_i/M$ applied to dozens of items, one of the most consequential weighted averages computed every day.
Because we locate the center of mass by pointing at an object, it is easy to assume it is always inside the material. It is an average of positions, not a piece of the body. For a ring, a horseshoe, a boomerang or a high jumper arched over a bar, it lies in empty space, and the body arranges itself around it.
A second error is to think internal forces can move the center of mass. A person on frictionless ice cannot shift their center of mass by flailing their arms; only an external push, such as friction from the ground, can.
Masses of $2.0$, $3.0$ and $5.0$ kg sit at $(0, 0)$, $(4.0, 0)$ and $(2.0, 3.0)$ m. Find the total mass.
$M = 2.0 + 3.0 + 5.0 = 10\ \text{kg}$
Add the masses.
Find the weighted sum of $x$.
$\sum m_ix_i = 0 + 12 + 10 = 22\ \text{kg m}$
Each mass times its $x$.
Divide for $x_{\text{cm}}$.
$x_{\text{cm}} = \dfrac{22}{10} = 2.2\ \text{m}$
Weighted average.
Find the weighted sum of $y$.
$\sum m_iy_i = 0 + 0 + 15 = 15\ \text{kg m}$
Only the third mass is off the axis.
Divide for $y_{\text{cm}}$.
$y_{\text{cm}} = \dfrac{15}{10} = 1.5\ \text{m}$
Inside the triangle, as it must be.
A rod of length $0.90$ m has $\lambda = 4.0x$ kg/m. Integrate for its mass.
$M = \displaystyle\int_0^{0.90}4.0x\,dx = 2.0 \times 0.81 = 1.62\ \text{kg}$
$\int x\,dx = x^2/2$.
Integrate for its first moment.
$\displaystyle\int x\,dm = \int_0^{0.90}4.0x^2\,dx = \tfrac{4.0}{3} \times 0.729 = 0.972\ \text{kg m}$
$\int x^2\,dx = x^3/3$.
Divide the moment by the mass.
$x_{\text{cm}} = \dfrac{0.972}{1.62} = 0.60\ \text{m}$
The center of mass.
Compare with two-thirds of the length.
$\tfrac{2}{3} \times 0.90 = 0.60\ \text{m}$
The general result for $\lambda \propto x$.
Compare with a uniform rod.
$\text{uniform: } 0.45\ \text{m}$
The heavy end pulls the center $0.15$ m toward itself.
A $60$ kg person stands at the left end of a $140$ kg raft $5.0$ m long, both at rest. Note what cannot change.
$x_{\text{cm}} = \text{constant}$
No external horizontal force if drag is ignored.
Write the balance of displacements.
$60\,\Delta x_p + 140\,\Delta x_r = 0$
Mass-weighted displacements cancel.
Relate the person's displacement to the raft's.
$\Delta x_p = 5.0 + \Delta x_r$
She walks $5.0$ m along a raft that itself moves.
Substitute and solve for the raft.
$60(5.0 + \Delta x_r) + 140\,\Delta x_r = 0 \Rightarrow \Delta x_r = -1.5\ \text{m}$
The raft moves left.
Find the person's displacement over the water.
$\Delta x_p = 5.0 - 1.5 = 3.5\ \text{m}$
Less than the length of the raft.
Check the center of mass.
$60 \times 3.5 + 140 \times (-1.5) = 210 - 210 = 0$
It has not moved.
Find the weighted sum.
$\sum m_ix_i = 4.0 + 36 = 40\ \text{kg m}$
Each mass times its position.
Divide by the total mass.
$x_{\text{cm}} = \dfrac{40}{10}$
Weighted average.
Evaluate the position.
A rod $60$ cm long has a density that grows in proportion to the distance from its left end, $\lambda = kx$. Where is its center of mass?
Complete the worked solution: particles of $2$ kg at $(6, 0)$, $2$ kg at $(4, 5)$ and $6$ kg at $(0, 6)$, in meters. Find the total mass, and the $x$ and $y$ coordinates of the center of mass.
Add the three masses.
$M =$ m
The total mass.
Weight each $x$ by its mass, add, and divide by the total.
$x_{\text{cm}} =$ x
The third particle is on the $y$ axis and adds nothing.
Weight each $y$ by its mass, add, and divide by the total.
$y_{\text{cm}} =$ y
The first particle is on the $x$ axis and adds nothing.
Check that the point lies among the particles.
$\text{inside the triangle the three particles make}$
A weighted average of positions.
Match each statement about the center of mass to its expression.
| $\sum m_ix_i/M$ | $\int x\,dm/M$ | $M\vec{a}_{\text{cm}} = \vec{F}_{\text{ext}}$ | $\vec{P}/M$ | |
|---|---|---|---|---|
| particles | ||||
| continuous body | ||||
| motion of the center of mass | ||||
| velocity of the center of mass |
A rod along $0 \le x \le 3$ m has density $\lambda = 3x$ kg/m. Fill in its mass in kg, its first moment $\int x\,dm$ in kg m, and the position of its center of mass in m.
| value | |
|---|---|
| mass (kg) | |
| first moment (kg m) | |
| center of mass (m) |
Masses of $8$ kg and $2$ kg are joined by a light rod. Measuring from the $8$ kg mass, write the position of the center of mass as a formula in the separation $d$.
Answer:
A $68$ kg person stands at one end of a $132$ kg raft floating at rest on a calm lake, and walks $4$ m to the other end. Ignoring water drag, how far does the raft move, in meters?
Answer: m the raft moves
At a Fourth of July show over the National Mall, a shell would have landed $66$ m downrange. At the top of its flight it bursts into two equal pieces that move horizontally, so both reach the ground together. One lands $34$ m downrange. Ignoring air resistance, where does the other land, in meters downrange?
Answer: m downrange for the second piece
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Masses of $2$ kg and $8$ kg are joined by a light rod. Measuring from the $2$ kg mass, write the position of the center of mass as a formula in the separation $d$.
Answer:
You can find and use the center of mass. Explain to someone why a firework's sparks spread around a point that keeps moving on the shell's path.
19. Your turn: masses of $4.0$ kg at $x = 1.0$ m and $6.0$ kg at $x = 6.0$ m. Find the center of mass., step 3
$x_{\text{cm}} = 4.0\ \text{m}$
Nearer the heavier mass.