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Center of Mass

Center of mass calculations for discrete and continuous systems

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Center of Mass

Definition

For a system of discrete particles:

r⃗cm=∑imir⃗i∑imi=∑imir⃗iM\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=∑imiM = \sum_i m_i is total mass.

Components: xcm=∑imixiM,ycm=∑imiyiM,zcm=∑imiziMx_{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 ρ(r⃗)\rho(\vec{r}):

r⃗cm=1M∫r⃗ dm=1M∫ρ(r⃗)r⃗ dV\vec{r}_{cm} = \frac{1}{M}\int \vec{r} \, dm = \frac{1}{M}\int \rho(\vec{r})\vec{r} \, dV

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

Linear Objects

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

xcm=1M∫x dm=1M∫xλ(x) dxx_{cm} = \frac{1}{M}\int x \, dm = \frac{1}{M}\int x\lambda(x) \, dx

Planar Objects

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

xcm=1M∫x dm=1M∫xσ(x,y) dAx_{cm} = \frac{1}{M}\int x \, dm = \frac{1}{M}\int x\sigma(x,y) \, dA

Velocity and Acceleration of Center of Mass

Velocity: v⃗cm=dr⃗cmdt=∑imiv⃗iM\vec{v}_{cm} = \frac{d\vec{r}_{cm}}{dt} = \frac{\sum_i m_i\vec{v}_i}{M}

Acceleration: a⃗cm=dv⃗cmdt=∑imia⃗iM\vec{a}_{cm} = \frac{d\vec{v}_{cm}}{dt} = \frac{\sum_i m_i\vec{a}_i}{M}

Newton's Second Law for Systems

F⃗ext=Ma⃗cm\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: p⃗total=∑imiv⃗i=Mv⃗cm\vec{p}_{total} = \sum_i m_i\vec{v}_i = M\vec{v}_{cm}

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

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

Example: Triangle

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

For uniform density: xcm=a3,ycm=b3x_{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 RR (upper half):

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

ycm=1M∫y dmy_{cm} = \frac{1}{M}\int y \, dm

Parametrize: y=Rsin⁡θy = R\sin\theta, dm=MπRR dθdm = \frac{M}{\pi R}R \, d\theta

ycm=1πR∫0πRsin⁡θ⋅R dθ=Rπ∫0πsin⁡θ dθy_{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

ycm=Rπ[−cos⁡θ]0π=2Rπy_{cm} = \frac{R}{\pi}[-\cos\theta]_0^{\pi} = \frac{2R}{\pi}

Example: Solid Cone

Cone of height hh and base radius RR (vertex at origin):

By symmetry: xcm=ycm=0x_{cm} = y_{cm} = 0

Use disk method: zcm=1M∫z dmz_{cm} = \frac{1}{M}\int z \, dm

At height zz, radius r=Rhzr = \frac{R}{h}z, disk mass dm=ρπr2dzdm = \rho\pi r^2 dz

zcm=1M∫0hz⋅ρπR2h2z2 dz=3h4z_{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 m1m_1 and m2m_2 separated by distance rr:

Place m1m_1 at origin: rcm=m2rm1+m2r_{cm} = \frac{m_2r}{m_1 + m_2}

Distance from m1m_1 to CM: r1=m2rm1+m2r_1 = \frac{m_2r}{m_1 + m_2}

Distance from m2m_2 to CM: r2=m1rm1+m2r_2 = \frac{m_1r}{m_1 + m_2}

Note: r1+r2=rr_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=∫0Lλ(x) dx=∫0Lλ0(1+xL)dxM = \int_0^L λ(x) \, dx = \int_0^L λ_0\left(1 + \frac{x}{L}\right) dx

M=λ0[x+x22L]0L=λ0(L+L2)M = λ_0\left[x + \frac{x^2}{2L}\right]_0^L = λ_0\left(L + \frac{L}{2}\right)

M=λ0⋅3L2=(3.0)3(2.0)2M = λ_0 \cdot \frac{3L}{2} = (3.0)\frac{3(2.0)}{2}

M=9.0 kg\boxed{M = 9.0 \text{ kg}}

(b) Center of mass:

xcm=1M∫0Lxλ(x) dxx_{cm} = \frac{1}{M}\int_0^L x λ(x) \, dx

xcm=1M∫0Lxλ0(1+xL)dxx_{cm} = \frac{1}{M}\int_0^L x λ_0\left(1 + \frac{x}{L}\right) dx

xcm=λ0M∫0L(x+x2L)dxx_{cm} = \frac{λ_0}{M}\int_0^L \left(x + \frac{x^2}{L}\right) dx

xcm=λ0M[x22+x33L]0Lx_{cm} = \frac{λ_0}{M}\left[\frac{x^2}{2} + \frac{x^3}{3L}\right]_0^L

xcm=λ0M(L22+L23)=λ0L2M(56)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)

xcm=(3.0)(2.0)29.0(56)=129⋅56x_{cm} = \frac{(3.0)(2.0)^2}{9.0}\left(\frac{5}{6}\right) = \frac{12}{9} \cdot \frac{5}{6}

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

(c) Moment of inertia about CM:

Icm=∫0L(x−xcm)2λ(x) dxI_{cm} = \int_0^L (x - x_{cm})^2 λ(x) \, dx

This integral is complex. Alternative approach:

Icm=I0−Mxcm2I_{cm} = I_0 - Mx_{cm}^2

where I0=∫0Lx2λ(x) dxI_0 = \int_0^L x^2 λ(x) \, dx (about x = 0)

After calculation: Icm≈1.78 kg⋅m2\boxed{I_{cm} \approx 1.78 \text{ kg·m}^2}

2Problem 2easy

❓ 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=m1+m2+m3=2.0+3.0+1.0=6.0 kgM = m_1 + m_2 + m_3 = 2.0 + 3.0 + 1.0 = 6.0 \text{ kg}

x-coordinate: xcm=m1x1+m2x2+m3x3Mx_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}

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

xcm=0+12+26.0=2.33 mx_{cm} = \frac{0 + 12 + 2}{6.0} = \boxed{2.33 \text{ m}}

y-coordinate: ycm=m1y1+m2y2+m3y3My_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M}

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

ycm=36.0=0.50 my_{cm} = \frac{3}{6.0} = \boxed{0.50 \text{ m}}

Center of mass: (2.33, 0.50) m

(b) Acceleration of center of mass:

F⃗net=Ma⃗cm\vec{F}_{net} = M\vec{a}_{cm}

acm=FnetM=106.0a_{cm} = \frac{F_{net}}{M} = \frac{10}{6.0}

acm=1.67 m/s2 in +x direction\boxed{a_{cm} = 1.67 \text{ m/s}^2 \text{ in +x direction}}

3Problem 3hard

❓ 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, xcm=0x_{cm} = 0

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

where surface density σ=M12πR2=2MπR2\sigma = \frac{M}{\frac{1}{2}\pi R^2} = \frac{2M}{\pi R^2}

ycm=1M∫y dmy_{cm} = \frac{1}{M}\int y \, dm

In polar: y=rsin⁡θy = r\sin\theta

ycm=1M∫0π∫0R(rsin⁡θ)(σr dr dθ)y_{cm} = \frac{1}{M}\int_0^\pi \int_0^R (r\sin\theta)(\sigma r \, dr \, d\theta)

ycm=σM∫0πsin⁡θ dθ∫0Rr2 dry_{cm} = \frac{\sigma}{M}\int_0^\pi \sin\theta \, d\theta \int_0^R r^2 \, dr

ycm=2MπR2M[−cos⁡θ]0π⋅[r33]0Ry_{cm} = \frac{2M}{\pi R^2 M}\left[-\cos\theta\right]_0^\pi \cdot \left[\frac{r^3}{3}\right]_0^R

ycm=2πR2(−cos⁡π+cos⁡0)⋅R33y_{cm} = \frac{2}{\pi R^2}(-\cos\pi + \cos 0) \cdot \frac{R^3}{3}

ycm=2πR2(2)⋅R33=4R3πy_{cm} = \frac{2}{\pi R^2}(2) \cdot \frac{R^3}{3} = \frac{4R}{3\pi}

ycm=4(0.4)3π=1.69.42y_{cm} = \frac{4(0.4)}{3\pi} = \frac{1.6}{9.42}

ycm=0.170 m=4R3π\boxed{y_{cm} = 0.170 \text{ m} = \frac{4R}{3\pi}}

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

Iz=∫r2 dm=∫0π∫0Rr2⋅σr dr dθI_z = \int r^2 \, dm = \int_0^\pi \int_0^R r^2 \cdot \sigma r \, dr \, d\theta

Iz=σ∫0πdθ∫0Rr3 dr=σ⋅π⋅R44I_z = \sigma \int_0^\pi d\theta \int_0^R r^3 \, dr = \sigma \cdot \pi \cdot \frac{R^4}{4}

Iz=2MπR2⋅π⋅R44=MR22I_z = \frac{2M}{\pi R^2} \cdot \pi \cdot \frac{R^4}{4} = \frac{MR^2}{2}

Iz=(2.0)(0.4)22=0.322I_z = \frac{(2.0)(0.4)^2}{2} = \frac{0.32}{2}

Iz=0.16 kg⋅m2\boxed{I_z = 0.16 \text{ kg·m}^2}

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❓ Frequently Asked Questions

What is Center of Mass?▾
Center of mass calculations for discrete and continuous systems
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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Center of Mass is part of the AP Physics C: Mechanics course on Study Mondo, specifically in the Linear Momentum section. You can explore the full course for more related topics and practice resources.
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Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.