Physics · Senior secondary (Class 11-12, JEE / NEET, AP, A-level, IB)

Continuous Mass Distribution ka Centre of Mass

Jab ek body mein masscontinuously distributedhota hai, jaise ki ek rod ya disk, tab uska Centre of Mass (COM) nikalne ke liye humintegrationka use karte hain. Yeh discrete particles ke COM se thoda alag hai.

1Centre of Mass kya hai?

Definition

Centre of Mass (COM) — Centre of Mass ek hypothetical point hai jahan par body ka saara mass concentrated mana ja sakta hai, aur external force lagne par body aise move karti hai jaise saara mass us point par ho.

Discrete particles ke liye COM nikalna easy hai, lekin continuous bodies ke liye humeininfinitesimal mass elementsconsider karne padte hain.

Discrete vs. Continuous Mass Distribution

  • Mass elements

    Discrete Mass
    Individual particles (m1,m2,...m_1, m_2, ...)
    Continuous Mass
    Infinitesimal elements (dmdm)

  • Position

    Discrete Mass
    Specific points (r1,r2,...r_1, r_2, ...)
    Continuous Mass
    Varies continuously

  • Calculation

    Discrete Mass
    Summation (Σ\Sigma)
    Continuous Mass
    Integration (∫\int)

2COM ke liye General Formula

Continuous mass distribution ke liye, COM ke coordinates (XCM,YCM,ZCM)(X_{CM}, Y_{CM}, Z_{CM}) ko integral form mein define kiya jaata hai. Yeh formula har type ki continuous body par apply hota hai.

Centre of Mass for Continuous Body

XCM=∫x dm∫dm=∫x dmMX_{CM} = \frac{\int x \, dm}{\int dm} = \frac{\int x \, dm}{M}
  • XCMX_{CM} = x-coordinate of COM
  • xx = x-coordinate of mass element $dm$
  • dmdm = infinitesimal mass element
  • MM = total mass of the body

Similarly, YCM=∫y dmMY_{CM} = \frac{\int y \, dm}{M} aur ZCM=∫z dmMZ_{CM} = \frac{\int z \, dm}{M} hote hain. Integral ∫dm\int dm total mass MM ke equal hota hai.

Exam tip: Symmetry ka use karo! Agar body kisi axis ke about symmetric hai, toh COM us axis par hi hoga. Isse calculation simplify ho jaati hai.

Continuous mass distribution ke liye Centre of Mass (COM) ka conceptcalculuspar based hai. Jab body ko discrete particles mein divide nahi kiya ja sakta, tab hum infinitesimal mass elements (dmdm) ka use karte hain aur unke positions ko integrate karte hain.

Definition

Infinitesimal Mass Element ($dm$) — Ek bahut chhota, point-like mass jo continuous body ka hissa hota hai, jiska volume ya length zero ko approach karta hai.

Derivation — COM for Continuous Mass Distribution

  1. XCM=∑mixi∑miX_{CM} = \frac{\sum m_i x_i}{\sum m_i}

    Discrete particles ke liye COM ka formula

  2. As Δmi→0,∑Δmixi→∫x dm\text{As } \Delta m_i \to 0, \sum \Delta m_i x_i \to \int x \, dm

    Continuous limit mein summation integral ban jaata hai

  3. ∑Δmi→∫dm=M\sum \Delta m_i \to \int dm = M

    Total mass M, sabhi infinitesimal masses ka sum

  4. XCM=∫x dm∫dm=∫x dmMX_{CM} = \frac{\int x \, dm}{\int dm} = \frac{\int x \, dm}{M}

    Continuous mass distribution ke liye COM formula

Mass Element in terms of Density

dm=ρ dVordm=λ dLordm=σ dAdm = \rho \, dV \quad \text{or} \quad dm = \lambda \, dL \quad \text{or} \quad dm = \sigma \, dA
  • dmdm = infinitesimal mass element
  • ρ\rho = volume mass density (mass per unit volume)
  • dVdV = infinitesimal volume element
  • λ\lambda = linear mass density (mass per unit length)
  • dLdL = infinitesimal length element
  • σ\sigma = surface mass density (mass per unit area)
  • dAdA = infinitesimal area element

Exam question · JEE Main style

Ek uniform rod hai jiska length LL hai aur mass MM hai. Is rod ka Centre of Mass (COM) kahan hoga, agar rod ko x-axis par origin se LL tak rakha gaya hai?
  1. A.L4\frac{L}{4}
  2. B.L2\frac{L}{2}
  3. C.2L3\frac{2L}{3}
  4. D.LL

Exam question · NEET style

Ek non-uniform thin rod hai jiska length LL hai. Iski linear mass density λ(x)=kx\lambda(x) = kx hai, jahan kk ek constant hai aur xx rod ke ek end se distance hai. Is rod ka Centre of Mass (COM) us end se kitni doori par hoga?
  1. A.L2\frac{L}{2}
  2. B.L3\frac{L}{3}
  3. C.2L3\frac{2L}{3}
  4. D.3L4\frac{3L}{4}
Common mistake: Common mistake hai dmdm ko sirf dxdx ya dVdV likh dena. Hamesha dm=ρ dVdm = \rho \, dV (ya λ dL\lambda \, dL, σ dA\sigma \, dA) use karna yaad rakho, jahan density term (ρ,λ,σ\rho, \lambda, \sigma) constant bhi ho sakti hai ya variable bhi.

3Mass Element ($dm$) ko Kaise Choose Karein?

Integration karne ke liye, humein dmdm ko body ke geometry aur mass density ke terms mein express karna hota hai. Yeh step sabse crucial hai.

dm Choose Karne ka Process

  1. 1

    Identify Geometry

    Body ki shape (rod, disk, sphere) dekho.

  2. 2

    Define Density

    Linear (λ\lambda), surface (σ\sigma), ya volume (ρ\rho) density use karo.

  3. 3

    Choose Element

    Ek infinitesimal element (length dxdx, area dAdA, volume dVdV) choose karo.

  4. 4

    Express dm

    dm=λdxdm = \lambda dx, dm=σdAdm = \sigma dA, ya dm=ρdVdm = \rho dV likho.

  5. 5

    Set Limits

    Integration ke limits body ki extent ke according set karo.

Rod ka Mass Element: Rod, x, dx, dm Full size

Rod ka Mass Element

Ek uniform rod ke liye, $dm = \lambda dx$ hota hai, jahan $\lambda$ linear mass density hai.

  1. 1Rod — Total length L
  2. 2x — Position of mass element
  3. 3dx — Infinitesimal length of element
  4. 4dm — Mass of this element

4Example: Uniform Rod ka COM

Chalo ek uniform rod ka COM nikalte hain. Uniform ka matlab hai ki mass density constant hai.

Real-life example · Uniform Rod ka COM

Ek uniform rod hai jiski length LL hai aur total mass MM hai. Rod x-axis par x=0x=0 se x=Lx=L tak rakhi hai. Iski linear mass density λ=M/L\lambda = M/L hai. Ek small element dxdx at position xx ka mass dm=λdxdm = \lambda dx hoga. XCM=∫0Lx dm∫0Ldm=∫0Lx(λdx)∫0L(λdx)=λ∫0Lxdxλ∫0Ldx=[x2/2]0L[x]0L=L2/2L=L/2X_{CM} = \frac{\int_0^L x \, dm}{\int_0^L dm} = \frac{\int_0^L x (\lambda dx)}{\int_0^L (\lambda dx)} = \frac{\lambda \int_0^L x dx}{\lambda \int_0^L dx} = \frac{[x^2/2]_0^L}{[x]_0^L} = \frac{L^2/2}{L} = L/2. So, uniform rod ka COM uske center par (L/2L/2) hota hai.

Did you know? Did you know? Agar Earth ka mass uniformly distributed hota, toh uska COM exactly uske geometrical center par hota. Lekin Earth ka core denser hai, isliye uska COM thoda sa center se shift hota hai!

5Exam Corner

Important for exams

  • Continuous mass ke liye COM integration se nikalta hai.
  • XCM=∫x dmMX_{CM} = \frac{\int x \, dm}{M} (similarly for Y and Z).
  • dmdm ko λdx\lambda dx, σdA\sigma dA, ya ρdV\rho dV mein express karna seekho.
  • Symmetry ka use karke calculations simplify karo.
  • Standard shapes (rod, ring, disk, sphere) ke COM positions yaad rakho.
  • Non-uniform density problems mein λ,σ,ρ\lambda, \sigma, \rho position ke function hote hain.

Exam question · JEE Main style

Ek thin rod ki length LL hai. Iski linear mass density λ=kx\lambda = kx hai, jahan kk constant hai aur xx rod ke ek end se distance hai. Rod ka Centre of Mass (COM) us end se kitni doori par hoga?
  1. A.L/2L/2
  2. B.2L/32L/3
  3. C.3L/43L/4
  4. D.L/3L/3
Remember it: COM for Continuous: 'I Can Integrate Mass' - Integrate x dmx \, dm over total mass MM.

Remember

  • Continuous mass distribution ka COMintegrationse nikalta hai.
  • Formula hai XCM=∫x dmMX_{CM} = \frac{\int x \, dm}{M}.
  • dmdm kolinear, surface, ya volume densityke terms mein express karte hain.
  • Symmetryse COM calculations bahut simplify ho jaati hain.
  • Non-uniform density mein, densityposition ka functionhoti hai.

Check yourself

  1. Question 1

    Ek uniform circular disk ka radius RR hai. Iska Centre of Mass kahan hoga?

  2. Question 2

    Ek thin semicircular ring ka radius RR hai. Iska Centre of Mass (y-axis par, agar diameter x-axis par ho) kya hoga?

Read it? Now test it. Timed exam questions on this topic, then a report on what to revise.

Take a test

Want a lesson on another topic? Type any topic or add a photo of your notes — get a lesson like this in seconds.

Learn a topic

More Physics explained