Center of Mass

Center of mass calculations for discrete and continuous systems

Center of Mass

Definition

For a system of discrete particles:

$\vec{r}_{cm} = \frac{\sum_i m_i\vec{r}_i}{\sum_i m_i} = \frac{\sum_i m_i\vec{r}_i}{M}$

where $M = \sum_i m_i$ is total mass.

Components: $x_{cm} = \frac{\sum_i m_ix_i}{M}, \quad y_{cm} = \frac{\sum_i m_iy_i}{M}, \quad z_{cm} = \frac{\sum_i m_iz_i}{M}$

Continuous Mass Distribution

For continuous objects with mass density $\rho(\vec{r})$ :

$\vec{r}_{cm} = \frac{1}{M}\int \vec{r} \, dm = \frac{1}{M}\int \rho(\vec{r})\vec{r} \, dV$

For uniform density ( $\rho$ = constant): $\vec{r}_{cm} = \frac{1}{V}\int \vec{r} \, dV$

Linear Objects

Mass per unit length: $\lambda = dm/dx$

$x_{cm} = \frac{1}{M}\int x \, dm = \frac{1}{M}\int x\lambda(x) \, dx$

Planar Objects

Mass per unit area: $\sigma = dm/dA$

$x_{cm} = \frac{1}{M}\int x \, dm = \frac{1}{M}\int x\sigma(x,y) \, dA$

Velocity and Acceleration of Center of Mass

Velocity: $\vec{v}_{cm} = \frac{d\vec{r}_{cm}}{dt} = \frac{\sum_i m_i\vec{v}_i}{M}$

Acceleration: $\vec{a}_{cm} = \frac{d\vec{v}_{cm}}{dt} = \frac{\sum_i m_i\vec{a}_i}{M}$

Newton's Second Law for Systems

$\vec{F}_{ext} = M\vec{a}_{cm}$

The center of mass moves as if all mass were concentrated there and all external forces acted there.

Internal forces cancel (Newton's third law).

Momentum and Center of Mass

Total momentum: $\vec{p}_{total} = \sum_i m_i\vec{v}_i = M\vec{v}_{cm}$

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

When $\vec{F}_{ext} = 0$ : $\vec{v}_{cm}$ is constant (momentum conserved).

Example: Triangle

Uniform triangular plate with vertices at $(0,0)$ , $(a,0)$ , $(0,b)$ .

For uniform density: $x_{cm} = \frac{a}{3}, \quad y_{cm} = \frac{b}{3}$

(Center of mass at centroid, 1/3 from each side)

Example: Semicircular Ring

Ring of radius $R$ (upper half):

$x_{cm} = 0$ (by symmetry)

$y_{cm} = \frac{1}{M}\int y \, dm$

Parametrize: $y = R\sin\theta$ , $dm = \frac{M}{\pi R}R \, d\theta$

$y_{cm} = \frac{1}{\pi R}\int_0^{\pi} R\sin\theta \cdot R \, d\theta = \frac{R}{\pi}\int_0^{\pi}\sin\theta \, d\theta$

$y_{cm} = \frac{R}{\pi}[-\cos\theta]_0^{\pi} = \frac{2R}{\pi}$

Example: Solid Cone

Cone of height $h$ and base radius $R$ (vertex at origin):

By symmetry: $x_{cm} = y_{cm} = 0$

Use disk method: $z_{cm} = \frac{1}{M}\int z \, dm$

At height $z$ , radius $r = \frac{R}{h}z$ , disk mass $dm = \rho\pi r^2 dz$

$z_{cm} = \frac{1}{M}\int_0^h z \cdot \rho\pi\frac{R^2}{h^2}z^2 \, dz = \frac{3h}{4}$

Two-Body Problem

For two bodies with masses $m_1$ and $m_2$ separated by distance $r$ :

Place $m_1$ at origin: $r_{cm} = \frac{m_2r}{m_1 + m_2}$

Distance from $m_1$ to CM: $r_1 = \frac{m_2r}{m_1 + m_2}$

Distance from $m_2$ to CM: $r_2 = \frac{m_1r}{m_1 + m_2}$

Note: $r_1 + r_2 = r$

📚 Practice Problems

1Problem 1medium

❓ Question:

A thin rod of length L = 2.0 m has linear mass density λ(x) = λ₀(1 + x/L), where λ₀ = 3.0 kg/m and x is measured from one end. Find: (a) the total mass, (b) the center of mass position, and (c) the moment of inertia about an axis through the center of mass.

💡 Show Solution

Given:

L = 2.0 m
λ(x) = λ₀(1 + x/L) where λ₀ = 3.0 kg/m

(a) Total mass:

$M = \int_0^L λ(x) \, dx = \int_0^L λ_0\left(1 + \frac{x}{L}\right) dx$

$M = λ_0\left[x + \frac{x^2}{2L}\right]_0^L = λ_0\left(L + \frac{L}{2}\right)$

$M = λ_0 \cdot \frac{3L}{2} = (3.0)\frac{3(2.0)}{2}$

$\boxed{M = 9.0 \text{ kg}}$

(b) Center of mass:

$x_{cm} = \frac{1}{M}\int_0^L x λ(x) \, dx$

$x_{cm} = \frac{1}{M}\int_0^L x λ_0\left(1 + \frac{x}{L}\right) dx$

$x_{cm} = \frac{λ_0}{M}\int_0^L \left(x + \frac{x^2}{L}\right) dx$

$x_{cm} = \frac{λ_0}{M}\left[\frac{x^2}{2} + \frac{x^3}{3L}\right]_0^L$

$x_{cm} = \frac{λ_0}{M}\left(\frac{L^2}{2} + \frac{L^2}{3}\right) = \frac{λ_0 L^2}{M}\left(\frac{5}{6}\right)$

$x_{cm} = \frac{(3.0)(2.0)^2}{9.0}\left(\frac{5}{6}\right) = \frac{12}{9} \cdot \frac{5}{6}$

$\boxed{x_{cm} = 1.11 \text{ m from light end}}$

(c) Moment of inertia about CM:

$I_{cm} = \int_0^L (x - x_{cm})^2 λ(x) \, dx$

This integral is complex. Alternative approach:

$I_{cm} = I_0 - Mx_{cm}^2$

where $I_0 = \int_0^L x^2 λ(x) \, dx$ (about x = 0)

After calculation: $\boxed{I_{cm} \approx 1.78 \text{ kg·m}^2}$

2Problem 2medium

❓ Question:

💡 Show Solution

Given:

L = 2.0 m
λ(x) = λ₀(1 + x/L) where λ₀ = 3.0 kg/m

(a) Total mass:

$M = \int_0^L λ(x) \, dx = \int_0^L λ_0\left(1 + \frac{x}{L}\right) dx$

$M = λ_0\left[x + \frac{x^2}{2L}\right]_0^L = λ_0\left(L + \frac{L}{2}\right)$

$M = λ_0 \cdot \frac{3L}{2} = (3.0)\frac{3(2.0)}{2}$

$\boxed{M = 9.0 \text{ kg}}$

(b) Center of mass:

$x_{cm} = \frac{1}{M}\int_0^L x λ(x) \, dx$

$x_{cm} = \frac{1}{M}\int_0^L x λ_0\left(1 + \frac{x}{L}\right) dx$

$x_{cm} = \frac{λ_0}{M}\int_0^L \left(x + \frac{x^2}{L}\right) dx$

$x_{cm} = \frac{λ_0}{M}\left[\frac{x^2}{2} + \frac{x^3}{3L}\right]_0^L$

$x_{cm} = \frac{λ_0}{M}\left(\frac{L^2}{2} + \frac{L^2}{3}\right) = \frac{λ_0 L^2}{M}\left(\frac{5}{6}\right)$

$x_{cm} = \frac{(3.0)(2.0)^2}{9.0}\left(\frac{5}{6}\right) = \frac{12}{9} \cdot \frac{5}{6}$

$\boxed{x_{cm} = 1.11 \text{ m from light end}}$

(c) Moment of inertia about CM:

$I_{cm} = \int_0^L (x - x_{cm})^2 λ(x) \, dx$

This integral is complex. Alternative approach:

$I_{cm} = I_0 - Mx_{cm}^2$

where $I_0 = \int_0^L x^2 λ(x) \, dx$ (about x = 0)

After calculation: $\boxed{I_{cm} \approx 1.78 \text{ kg·m}^2}$

3Problem 3easy

❓ Question:

A system consists of three masses: m₁ = 2.0 kg at (0, 0), m₂ = 3.0 kg at (4, 0) m, and m₃ = 1.0 kg at (2, 3) m. Find: (a) the center of mass coordinates, (b) if a 10 N force acts on the system in the +x direction, find the acceleration of the center of mass.

💡 Show Solution

Given:

m₁ = 2.0 kg at (0, 0)
m₂ = 3.0 kg at (4, 0)
m₃ = 1.0 kg at (2, 3)

(a) Center of mass coordinates:

Total mass: $M = m_1 + m_2 + m_3 = 2.0 + 3.0 + 1.0 = 6.0 \text{ kg}$

x-coordinate: $x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}$

$x_{cm} = \frac{(2.0)(0) + (3.0)(4) + (1.0)(2)}{6.0}$

$x_{cm} = \frac{0 + 12 + 2}{6.0} = \boxed{2.33 \text{ m}}$

y-coordinate: $y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M}$

$y_{cm} = \frac{(2.0)(0) + (3.0)(0) + (1.0)(3)}{6.0}$

$y_{cm} = \frac{3}{6.0} = \boxed{0.50 \text{ m}}$

Center of mass: (2.33, 0.50) m

(b) Acceleration of center of mass:

$\vec{F}_{net} = M\vec{a}_{cm}$

$a_{cm} = \frac{F_{net}}{M} = \frac{10}{6.0}$

$\boxed{a_{cm} = 1.67 \text{ m/s}^2 \text{ in +x direction}}$

4Problem 4easy

❓ Question:

💡 Show Solution

Given:

m₁ = 2.0 kg at (0, 0)
m₂ = 3.0 kg at (4, 0)
m₃ = 1.0 kg at (2, 3)

(a) Center of mass coordinates:

Total mass: $M = m_1 + m_2 + m_3 = 2.0 + 3.0 + 1.0 = 6.0 \text{ kg}$

x-coordinate: $x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}$

$x_{cm} = \frac{(2.0)(0) + (3.0)(4) + (1.0)(2)}{6.0}$

$x_{cm} = \frac{0 + 12 + 2}{6.0} = \boxed{2.33 \text{ m}}$

y-coordinate: $y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M}$

$y_{cm} = \frac{(2.0)(0) + (3.0)(0) + (1.0)(3)}{6.0}$

$y_{cm} = \frac{3}{6.0} = \boxed{0.50 \text{ m}}$

Center of mass: (2.33, 0.50) m

(b) Acceleration of center of mass:

$\vec{F}_{net} = M\vec{a}_{cm}$

$a_{cm} = \frac{F_{net}}{M} = \frac{10}{6.0}$

$\boxed{a_{cm} = 1.67 \text{ m/s}^2 \text{ in +x direction}}$

5Problem 5hard

❓ Question:

A uniform semicircular disk of radius R = 0.4 m and mass M = 2.0 kg lies in the xy-plane with its diameter along the x-axis. Using integration, find: (a) the y-coordinate of the center of mass, (b) the moment of inertia about an axis through the origin perpendicular to the disk.

💡 Show Solution

Given:

R = 0.4 m
M = 2.0 kg
Semicircular disk (uniform)

(a) Y-coordinate of center of mass:

By symmetry, $x_{cm} = 0$

Mass element in polar coordinates: $dm = \sigma \, dA = \sigma r \, dr \, d\theta$

where surface density $\sigma = \frac{M}{\frac{1}{2}\pi R^2} = \frac{2M}{\pi R^2}$

$y_{cm} = \frac{1}{M}\int y \, dm$

In polar: $y = r\sin\theta$

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

$y_{cm} = \frac{\sigma}{M}\int_0^\pi \sin\theta \, d\theta \int_0^R r^2 \, dr$

$y_{cm} = \frac{2M}{\pi R^2 M}\left[-\cos\theta\right]_0^\pi \cdot \left[\frac{r^3}{3}\right]_0^R$

$y_{cm} = \frac{2}{\pi R^2}(-\cos\pi + \cos 0) \cdot \frac{R^3}{3}$

$y_{cm} = \frac{2}{\pi R^2}(2) \cdot \frac{R^3}{3} = \frac{4R}{3\pi}$

$y_{cm} = \frac{4(0.4)}{3\pi} = \frac{1.6}{9.42}$

$\boxed{y_{cm} = 0.170 \text{ m} = \frac{4R}{3\pi}}$

(b) Moment of inertia about z-axis (through origin):

$I_z = \int r^2 \, dm = \int_0^\pi \int_0^R r^2 \cdot \sigma r \, dr \, d\theta$

$I_z = \sigma \int_0^\pi d\theta \int_0^R r^3 \, dr = \sigma \cdot \pi \cdot \frac{R^4}{4}$

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

$I_z = \frac{(2.0)(0.4)^2}{2} = \frac{0.32}{2}$

$\boxed{I_z = 0.16 \text{ kg·m}^2}$

6Problem 6hard

❓ Question:

💡 Show Solution

Given:

R = 0.4 m
M = 2.0 kg
Semicircular disk (uniform)

(a) Y-coordinate of center of mass:

By symmetry, $x_{cm} = 0$

Mass element in polar coordinates: $dm = \sigma \, dA = \sigma r \, dr \, d\theta$

where surface density $\sigma = \frac{M}{\frac{1}{2}\pi R^2} = \frac{2M}{\pi R^2}$

$y_{cm} = \frac{1}{M}\int y \, dm$

In polar: $y = r\sin\theta$

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

$y_{cm} = \frac{\sigma}{M}\int_0^\pi \sin\theta \, d\theta \int_0^R r^2 \, dr$

$y_{cm} = \frac{2M}{\pi R^2 M}\left[-\cos\theta\right]_0^\pi \cdot \left[\frac{r^3}{3}\right]_0^R$

$y_{cm} = \frac{2}{\pi R^2}(-\cos\pi + \cos 0) \cdot \frac{R^3}{3}$

$y_{cm} = \frac{2}{\pi R^2}(2) \cdot \frac{R^3}{3} = \frac{4R}{3\pi}$

$y_{cm} = \frac{4(0.4)}{3\pi} = \frac{1.6}{9.42}$

$\boxed{y_{cm} = 0.170 \text{ m} = \frac{4R}{3\pi}}$

(b) Moment of inertia about z-axis (through origin):

$I_z = \int r^2 \, dm = \int_0^\pi \int_0^R r^2 \cdot \sigma r \, dr \, d\theta$

$I_z = \sigma \int_0^\pi d\theta \int_0^R r^3 \, dr = \sigma \cdot \pi \cdot \frac{R^4}{4}$

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

$I_z = \frac{(2.0)(0.4)^2}{2} = \frac{0.32}{2}$

$\boxed{I_z = 0.16 \text{ kg·m}^2}$

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