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Center of Mass

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Center of Mass - Complete Interactive Lesson

Part 1: COM Definition (Discrete)

Center of Mass — Definition (Discrete Systems)

Part 1 of 7

The center of mass (COM) is the mass-weighted average position of a system. For a collection of $N$ discrete particles:

$\vec{r}_{\text{cm}} = \frac{\sum_{i=1}^{N} m_i \vec{r}_i}{\sum_{i=1}^{N} m_i} = \frac{1}{M}\sum_{i=1}^{N} m_i \vec{r}_i$

In component form:

$x_{\text{cm}} = \frac{\sum m_i x_i}{M}, \quad y_{\text{cm}} = \frac{\sum m_i y_i}{M}, \quad z_{\text{cm}} = \frac{\sum m_i z_i}{M}$

where $M = \sum m_i$ is the total mass.

Key Properties

The COM is a unique point for any mass distribution
It does not need to lie within the physical body (e.g., a ring)
For a uniform-density symmetric object, COM lies at the geometric center

Two-Dimensional Systems

For particles in the $xy$ -plane, compute each component separately.

Worked Example

Three masses:

$m_1 = 2$ kg at $(0, 0)$

Using Symmetry

If a mass distribution has a line of symmetry, the COM lies on that line. If it has two perpendicular lines of symmetry, the COM is at their intersection.

Negative Mass Trick

To find the COM of an object with a hole, treat it as:

$\vec{r}_{\text{cm}} = \frac{M_{\text{full}}\vec{r}_{\text{full}} - M_{\text{hole}}\vec{r}_{\text{hole}}}{M_{\text{full}} - M_{\text{hole}}}$

Summary

Concept	Formula
COM (1D)	$x_{\text{cm}} = \frac{\sum m_i x_i}{M}$

Part 2: COM (Continuous Bodies)

Center of Mass — Continuous Bodies via Integration

Part 2 of 7

For a continuous mass distribution, sums become integrals:

$\vec{r}_{\text{cm}} = \frac{1}{M}\int \vec{r}\,dm$

Part 3: COM Velocity & Momentum

COM Velocity and Momentum

Part 3 of 7

Differentiating the COM position gives the COM velocity:

$\vec{v}_{\text{cm}} = \frac{d\vec{r}_{\text{cm}}}{dt} = \frac{1}{M}\sum_{i} m_i \vec{v}_i = \frac{\vec{p}_{\text{total}}}{M}$

Part 4: COM Reference Frame

COM Reference Frame

Part 4 of 7

The center-of-mass frame (also called the zero-momentum frame) is the reference frame in which the total momentum is zero:

$\vec{p}_{\text{total}}^{\,\prime} = \sum m_i \vec{v}_i^{\,\prime} = 0$

Part 5: COM Under External Forces

COM Motion Under External Forces

Part 5 of 7

Newton's second law for the center of mass:

$\vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}} = M\frac{d\vec{v}_{\text{cm}}}{dt} = \frac{d\vec{p}_{\text{total}}}{dt}$

Part 6: Problem-Solving Workshop

Center of Mass — Problem-Solving Workshop

Part 6 of 7

Strategy Guide

Step	Action
1	Identify the system and all masses
2	Choose coordinates (use symmetry!)
3	Determine if the problem is discrete or continuous
4	For $\vec{F}_{\text{ext}} = 0$ : use

Part 7: Review & Applications

Center of Mass — Review & Applications

Part 7 of 7 — Comprehensive Assessment

Formula Reference

Formula	Expression
Discrete COM	$\vec{r}_{\text{cm}} = \frac{\sum m_i \vec{r}_i}{M}$

m_2 = 3

kg at

(4, 0)

m_3 = 5

kg at

(2, 3)

$x_{\text{cm}} = \frac{2(0) + 3(4) + 5(2)}{2+3+5} = \frac{0+12+10}{10} = 2.2 \text{ m}$

$y_{\text{cm}} = \frac{2(0) + 3(0) + 5(3)}{10} = \frac{15}{10} = 1.5 \text{ m}$

So $\vec{r}_{\text{cm}} = (2.2, 1.5)$ m.

Mfull−MholeMfullrfull−Mholerhole

This is equivalent to adding a "negative mass" at the hole's position.

Example: Disk with Off-Center Hole

A uniform disk of mass $M$ and radius $R$ has a circular hole of radius $R/2$ cut from it, centered at $x = R/2$ from the disk center.

Let $\sigma$ be surface mass density. $M_{\text{full}} = \sigma \pi R^2$ , $M_{\text{hole}} = \sigma \pi (R/2)^2 = M/4$ .

$x_{\text{cm}} = \frac{M(0) - (M/4)(R/2)}{M - M/4} = \frac{-MR/8}{3M/4} = -\frac{R}{6}$

The COM shifts away from the hole.

Next: Part 2 — Center of mass for continuous mass distributions using integration.

The key is expressing $dm$ in terms of geometry:

Geometry	Mass element
1D (rod, wire)	$dm = \lambda(x)\,dx$
2D (plate, disk)	$dm = \sigma(x,y)\,dA$
3D (solid)	$dm = \rho(x,y,z)\,dV$

where $\lambda$ , $\sigma$ , $\rho$ are linear, surface, and volume mass densities.

Non-Uniform Density

Worked Example: Rod with $\lambda(x) = \alpha x$

A rod extends from $x = 0$ to $x = L$ with linear density $\lambda(x) = \alpha x$ .

Step 1: Find total mass

$M = \int_0^L \alpha x\,dx = \alpha \frac{L^2}{2}$

Step 2: Find $x_{\text{cm}}$

$x_{\text{cm}} = \frac{1}{M}\int_0^L x \cdot \alpha x\,dx = \frac{1}{\alpha L^2/2}\cdot \alpha \frac{L^3}{3} = \frac{2L}{3}$

The COM is shifted toward the denser (heavier) end, which makes physical sense.

Two-Dimensional Bodies

Semicircular Wire (Uniform)

A uniform semicircular wire of radius $R$ and mass $M$ lies in the upper half-plane.

By symmetry, $x_{\text{cm}} = 0$ . For $y_{\text{cm}}$ , parameterize with angle $\theta$ :

$dm = \lambda R\,d\theta, \quad y = R\sin\theta$

$y_{\text{cm}} = \frac{1}{M}\int_0^{\pi} R\sin\theta \cdot \lambda R\,d\theta = \frac{\lambda R^2}{M}[-\cos\theta]_0^{\pi} = \frac{\lambda R^2 \cdot 2}{\lambda \pi R} = \frac{2R}{\pi}$

Semicircular Disk (Uniform)

For a solid semicircular disk, use area element $dA = r\,dr\,d\theta$ :

$y_{\text{cm}} = \frac{1}{M}\int_0^{\pi}\int_0^R (r\sin\theta)\sigma \cdot r\,dr\,d\theta = \frac{4R}{3\pi}$

Shape	$y_{\text{cm}}$
Semicircular wire	$2R/\pi$
Semicircular disk	$4R/(3\pi)$

Summary

Method	Expression
General	$\vec{r}_{\text{cm}} = \frac{1}{M}\int \vec{r}\,dm$
1D rod	$x_{\text{cm}} = \frac{\int x\,\lambda(x)\,dx}{\int \lambda(x)\,dx}$
2D plate	Use $dm = \sigma\,dA$ with appropriate coordinates
Key results	Semicircle wire $2R/\pi$ ; hemisphere $3R/8$

Next: Part 3 — COM velocity and momentum.

Therefore the total momentum of the system equals:

$\vec{p}_{\text{total}} = M\vec{v}_{\text{cm}}$

This is a profound result: the total momentum of a system is the same as if all the mass were concentrated at the COM moving with $\vec{v}_{\text{cm}}$ .

$\vec{a}_{\text{cm}} = \frac{d\vec{v}_{\text{cm}}}{dt} = \frac{1}{M}\sum m_i \vec{a}_i = \frac{\vec{F}_{\text{ext}}}{M}$

Internal forces cancel in pairs (Newton's third law), so only external forces determine COM motion.

Explosions and Internal Forces

When a body explodes or breaks apart, no external forces act during the explosion. Therefore:

$\vec{v}_{\text{cm, before}} = \vec{v}_{\text{cm, after}}$

Worked Example

A 10 kg projectile moving at $20$ m/s horizontally explodes into two pieces. A 4 kg piece comes to rest. Find the velocity of the 6 kg piece.

Solution:

$p_{\text{before}} = 10 \times 20 = 200 \text{ kg}\cdot\text{m/s}$

$p_{\text{after}} = 4(0) + 6v_2 = 200$

$v_2 = \frac{200}{6} = \frac{100}{3} \approx 33.3 \text{ m/s}$

The COM continues at $v_{\text{cm}} = 20$ m/s throughout.

Recoil Problems

A classic application: a person standing on a frictionless surface throws an object.

Worked Example

A 60 kg person on a frictionless frozen lake throws a 5 kg ball at $10$ m/s (relative to ground). Both start at rest.

$0 = 60 v_p + 5(10) \implies v_p = -\frac{50}{60} = -\frac{5}{6} \approx -0.83 \text{ m/s}$

The person recoils in the opposite direction.

Continuous Mass Loss

If mass is ejected continuously (foreshadowing rockets), the momentum equation becomes differential:

$d\vec{p} = \vec{v}\,dm$

This leads to the variable-mass equation we'll study in Topic 8.

Summary

Concept	Key Equation
COM velocity	$\vec{v}_{\text{cm}} = \frac{\sum m_i \vec{v}_i}{M}$
Total momentum	$\vec{p}_{\text{total}} = M\vec{v}_{\text{cm}}$
No external forces	$\vec{v}_{\text{cm}} = \text{const}$
Explosions	COM velocity unchanged
Recoil	$\sum \vec{p}_i = 0$ if starting from rest

Next: Part 4 — The center of mass reference frame.

To transform from the lab frame to the COM frame, subtract $\vec{v}_{\text{cm}}$ :

$\vec{v}_i^{\,\prime} = \vec{v}_i - \vec{v}_{\text{cm}}$

Why Use the COM Frame?

Total momentum is always zero — simplifies collision analysis
Kinetic energy splits into COM motion + internal motion
Elastic collisions are symmetric in the COM frame

Kinetic Energy Decomposition

The total KE in the lab frame separates as:

$K_{\text{total}} = \underbrace{\frac{1}{2}M v_{\text{cm}}^2}_{K_{\text{cm}}} + \underbrace{\sum \frac{1}{2}m_i v_i'^2}_{K_{\text{int}}}$

where $v_i' = |\vec{v}_i - \vec{v}_{\text{cm}}|$ .

Interpretation

$K_{\text{cm}}$ : energy of the system's bulk motion
$K_{\text{int}}$ : energy of internal (relative) motion
In a perfectly inelastic collision, $K_{\text{int}} \to 0$ (all internal KE is lost)

Worked Example

A 2 kg ball at $6$ m/s collides with a 4 kg ball at rest.

$v_{\text{cm}} = \frac{2(6) + 4(0)}{6} = 2 \text{ m/s}$

$K_{\text{cm}} = \frac{1}{2}(6)(2)^2 = 12 \text{ J}$

$K_{\text{int}} = \frac{1}{2}(2)(6-2)^2 + \frac{1}{2}(4)(0-2)^2 = 16 + 8 = 24 \text{ J}$

$K_{\text{total}} = 12 + 24 = 36 \text{ J} = \frac{1}{2}(2)(6)^2 \checkmark$

Elastic Collisions in the COM Frame

In the COM frame, an elastic collision is beautifully simple: each particle reverses its velocity.

$v_1'^{\text{after}} = -v_1'^{\text{before}}, \quad v_2'^{\text{after}} = -v_2'^{\text{before}}$

Transforming back to the lab frame:

$v_1^{\text{after}} = v_1'^{\text{after}} + v_{\text{cm}} = -v_1' + v_{\text{cm}}$

This gives the familiar results:

$v_1^{\text{after}} = \frac{m_1 - m_2}{m_1 + m_2}v_1 + \frac{2m_2}{m_1+m_2}v_2$

$v_2^{\text{after}} = \frac{2m_1}{m_1+m_2}v_1 + \frac{m_2 - m_1}{m_1+m_2}v_2$

Summary

Concept	Key Result
COM frame transform	$\vec{v}_i' = \vec{v}_i - \vec{v}_{\text{cm}}$
Zero momentum	$\sum m_i \vec{v}_i' = 0$ always
KE decomposition	$K = K_{\text{cm}} + K_{\text{int}}$
Elastic (COM frame)	Velocities reverse
Perfectly inelastic	All $K_{\text{int}}$ is lost
Reduced mass	$\mu = m_1 m_2/(m_1+m_2)$

Next: Part 5 — COM motion under external forces.

This is the most powerful consequence of the COM concept:

No matter how complex the internal interactions, the COM moves as if it were a single point particle of mass $M$ subject to the net external force.

A wrench tossed in the air: the COM follows a parabola even though the wrench rotates
A firework in flight: the COM continues on the parabolic trajectory after explosion
A binary star system: the COM follows the gravitational trajectory of the total mass

Projectile Breakup

Worked Example

A projectile is launched at $45°$ with speed $v_0$ . At the top of its trajectory, it breaks into two equal pieces. One piece falls straight down. Where does the other piece land?

Solution:

Step 1: Range of intact projectile: $R = v_0^2 \sin(90°)/g = v_0^2/g$

Step 2: At the peak, the projectile is at $x = R/2$ , $y = R/4$ , with velocity $(v_0/\sqrt{2}, 0)$ .

Step 3: The COM must continue the original parabolic path and land at $x = R$ .

Step 4: Piece 1 ( $m/2$ ) falls straight down from $x = R/2$ . By COM condition at landing time:

$R = \frac{(m/2)(R/2) + (m/2)(x_2)}{m}$

$x_2 = 2R - R/2 = \frac{3R}{2} = \frac{3v_0^2}{2g}$

The second piece lands at $3R/2$ from the launch point — 50% farther than the original range.

COM of an Atwood Machine

Consider an Atwood machine with masses $m_1 > m_2$ , connected by a massless string over a frictionless pulley.

The acceleration: $a = \frac{(m_1 - m_2)g}{m_1 + m_2}$

COM acceleration: Mass $m_1$ accelerates down, $m_2$ accelerates up, both with magnitude $a$ .

$a_{\text{cm}} = \frac{m_1 a_{\downarrow} + m_2 a_{\uparrow}}{m_1 + m_2} = \frac{m_1(-a) + m_2(a)}{m_1+m_2}$

Wait — actually $m_1$ goes down ( $-a$ ) and $m_2$ goes up ( $+a$ )... but the net external force is downward while the string/pulley system is internal.

$a_{\text{cm}} = \frac{(m_1 - m_2)g - 0}{m_1 + m_2}$

Actually: the constraint forces (tension, normal from pulley) contribute externally via the pulley support. The COM accelerates downward at:

$a_{\text{cm}} = \frac{(m_1 - m_2)^2 g}{(m_1+m_2)^2}$

Summary

Scenario	COM Behavior
Only gravity	$\vec{a}_{\text{cm}} = \vec{g}$ (parabolic path)
No external forces	$\vec{v}_{\text{cm}} = \text{const}$
Breakup/explosion	COM continues original trajectory
Person on boat	COM stays fixed; boat shifts

Next: Part 6 — Problem-solving workshop.

\vec{v}_{\text{cm}} = \text{const}

Problem 1: Two-Dimensional Integration

Find the COM of a quarter-disk of radius $R$ and uniform surface density $\sigma$ in the first quadrant.

Solution:

By symmetry, $x_{\text{cm}} = y_{\text{cm}}$ .

$M = \sigma \cdot \frac{\pi R^2}{4}$

Using polar coordinates with $x = r\cos\theta$ :

$x_{\text{cm}} = \frac{\sigma}{M}\int_0^{\pi/2}\int_0^R (r\cos\theta)\,r\,dr\,d\theta$

$= \frac{\sigma}{M} \cdot \frac{R^3}{3} \cdot [\sin\theta]_0^{\pi/2} = \frac{\sigma R^3/3}{\sigma \pi R^2/4} = \frac{4R}{3\pi}$

$\vec{r}_{\text{cm}} = \left(\frac{4R}{3\pi}, \frac{4R}{3\pi}\right)$

Problem 2: Collision + COM

A 3 kg block moving at $4$ m/s to the right collides elastically with a 1 kg block at rest.

In the COM frame:

$v_{\text{cm}} = \frac{3(4) + 1(0)}{4} = 3 \text{ m/s}$

COM frame velocities:

$v_1' = 4 - 3 = 1$ m/s (right)
$v_2' = 0 - 3 = -3$ m/s (left)

Check: $3(1) + 1(-3) = 0$ ✓

After elastic collision (reverse in COM frame):

$v_1'^{\text{after}} = -1$ m/s
$v_2'^{\text{after}} = +3$ m/s

Back to lab frame:

$v_1^{\text{after}} = -1 + 3 = 2$ m/s
$v_2^{\text{after}} = 3 + 3 = 6$ m/s

Verify: $3(2) + 1(6) = 12 = 3(4) + 1(0)$ ✓ and KE is conserved ✓

Workshop Takeaways

Problem Type	Key Technique
Non-uniform rod	$x_{\text{cm}} = \frac{\int x\lambda(x)dx}{\int \lambda(x)dx}$
2D shapes	Use polar coords for circular regions
Collisions	Transform to COM frame, reverse, transform back
Missing piece	Negative-mass subtraction
Solids of revolution	Use disk/shell slicing

Next: Part 7 — Comprehensive review & applications.

Shape	COM
Uniform rod	$L/2$ from end
Solid cone	$h/4$ from base
Semicircle wire	$2R/\pi$ from center
Solid hemisphere	$3R/8$ from base

AP-Style Free Response

A uniform solid disk of mass $M$ and radius $R$ has a hole of radius $R/3$ drilled through it. The hole is centered at distance $R/3$ from the disk's center. Find the COM of the remaining piece.

Solution:

Let the disk center be at the origin and the hole center at $x = R/3$ .

Area of full disk: $\pi R^2$ . Area of hole: $\pi(R/3)^2 = \pi R^2/9$ .

Mass of full disk: $M_{\text{full}} = M \cdot \frac{\pi R^2}{\pi R^2 - \pi R^2/9} = M \cdot \frac{9}{8}$

Mass of hole: $M_{\text{hole}} = \frac{M}{8} \cdot \frac{9}{8}$ ... Let me redo this more carefully.

Let $\sigma = M/(\pi R^2 - \pi R^2/9) = M/(8\pi R^2/9) = 9M/(8\pi R^2)$

$M_{\text{full}} = \sigma \pi R^2 = 9M/8$

$M_{\text{hole}} = \sigma \pi R^2/9 = M/8$

$x_{\text{cm}} = \frac{M_{\text{full}}(0) - M_{\text{hole}}(R/3)}{M_{\text{full}} - M_{\text{hole}}} = \frac{-(M/8)(R/3)}{9M/8 - M/8} = \frac{-MR/24}{M} = -\frac{R}{24}$

The COM shifts $R/24$ away from the hole.

🎉 Topic Complete — Center of Mass

You've mastered:

Part	Topic	Status
1	Discrete COM definition	✅
2	Continuous bodies (integration)	✅
3	COM velocity & momentum	✅
4	COM reference frame	✅
5	COM motion under external forces	✅
6	Problem-solving workshop	✅
7	Review & applications	✅

Key Insight: The center of mass reduces complex multi-body problems to single-particle dynamics. Master the COM frame and you'll cut through collision and explosion problems with ease.

Mfull−MholeMfull(0)−Mhole(R/3)

9M/8−M/8−(M/8)(R/3)=

Center of Mass

Example: Disk with Off-Center Hole

Non-Uniform Density

Worked Example: Rod with $\lambda(x) = \alpha x$

Two-Dimensional Bodies

Semicircular Wire (Uniform)

Semicircular Disk (Uniform)

Summary

COM Acceleration

Explosions and Internal Forces

Worked Example

Recoil Problems

Worked Example

Continuous Mass Loss

Summary

Why Use the COM Frame?

Kinetic Energy Decomposition

Interpretation

Worked Example

Elastic Collisions in the COM Frame

Summary

Applications

Projectile Breakup

Worked Example

COM of an Atwood Machine

Summary

Problem 1: Two-Dimensional Integration

Problem 2: Collision + COM

Workshop Takeaways

Key COM Positions

AP-Style Free Response

🎉 Topic Complete — Center of Mass

Center of Mass

Center of Mass - Complete Interactive Lesson

Part 1: COM Definition (Discrete)

Center of Mass — Definition (Discrete Systems)

Key Properties

Two-Dimensional Systems

Worked Example

Using Symmetry

Negative Mass Trick

Summary

Part 2: COM (Continuous Bodies)

Center of Mass — Continuous Bodies via Integration

Part 3: COM Velocity & Momentum

COM Velocity and Momentum

Part 4: COM Reference Frame

COM Reference Frame

Part 5: COM Under External Forces

COM Motion Under External Forces

Part 6: Problem-Solving Workshop

Center of Mass — Problem-Solving Workshop

Strategy Guide

Part 7: Review & Applications

Center of Mass — Review & Applications

Formula Reference

Example: Disk with Off-Center Hole

Non-Uniform Density

Worked Example: Rod with λ(x)=αx\lambda(x) = \alpha xλ(x)=αx

Two-Dimensional Bodies

Semicircular Wire (Uniform)

Semicircular Disk (Uniform)

Summary

COM Acceleration

Explosions and Internal Forces

Worked Example

Recoil Problems

Worked Example

Continuous Mass Loss

Summary

Why Use the COM Frame?

Kinetic Energy Decomposition

Interpretation

Worked Example

Elastic Collisions in the COM Frame

Summary

Applications

Projectile Breakup

Worked Example

COM of an Atwood Machine

Summary

Problem 1: Two-Dimensional Integration

Problem 2: Collision + COM

Workshop Takeaways

Key COM Positions

AP-Style Free Response

🎉 Topic Complete — Center of Mass

Worked Example: Rod with $\lambda(x) = \alpha x$