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: $r_1 + r_2 = r$

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

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

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

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