Centre of Mass: Quick Notes

Imagine tracking a whole bunch of particles by following just one special point that represents the “average” location of their mass. That point is the centre of mass (CM). Here’s how to find it and why it helps.

1. Two-Particle System

  • Place the two masses on an x-axis at positions x1 and x2.
  • The CM lies at
    \[ X=\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}} \]
  • If the masses are equal (m1=m2), the CM sits exactly halfway between them:
    \[ X=\frac{x_{1}+x_{2}}{2} \]

2. Many Particles on a Line

For n particles scattered along the same line, the CM coordinate is

\[ X=\frac{\displaystyle\sum_{i=1}^{n} m_{i}x_{i}}{\displaystyle\sum_{i=1}^{n} m_{i}} \]

3. Particles in a Plane

Place three masses at \((x_{1},y_{1}),\,(x_{2},y_{2}),\,(x_{3},y_{3})\). The CM coordinates come from two neat averages:

\[ \begin{aligned} X&=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}}{m_{1}+m_{2}+m_{3}},\\[4pt] Y&=\frac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}}{m_{1}+m_{2}+m_{3}}. \end{aligned} \]

(For equal masses, the CM lands on the centroid of the triangle.)

4. Particles in 3-D

Let M be the total mass \(\bigl(M=\sum m_{i}\bigr)\). Then

\[ X=\frac{\sum m_{i}x_{i}}{M},\qquad Y=\frac{\sum m_{i}y_{i}}{M},\qquad Z=\frac{\sum m_{i}z_{i}}{M}. \]

Vector form (slick and compact):

\[ \mathbf R=\frac{1}{M}\sum m_{i}\mathbf r_{i}. \]

5. Continuous Bodies

When particles are packed so tightly that you’d rather treat the object as a solid “blob,” replace the sums with integrals:

\[ \mathbf R=\frac{1}{M}\int \mathbf r\,dm. \]

Choosing the CM as the origin makes life easy because then

\[ \int x\,dm=\int y\,dm=\int z\,dm=0. \]

6. Finding the CM Fast with Symmetry

  • Uniform thin rod: CM at the geometric center.
  • Uniform ring, disc, or sphere: CM at the geometric center.
  • Triangular lamina: CM at the intersection of the three medians (the usual centroid).
  • Uniform L-shaped plate: Break it into rectangles or squares, find each small CM, then average with the formula in Section 3.

7. Motion of the Centre of Mass

Treat every particle’s position vector as \(\mathbf r_{i}\) and the CM position as \(\mathbf R\). Multiply equation after equation by time derivatives and you get:

  1. Position: \(M\mathbf R = \sum m_{i}\mathbf r_{i}\)
  2. Velocity: \(M\mathbf V = \sum m_{i}\mathbf v_{i}\)
  3. Acceleration: \(M\mathbf A = \sum m_{i}\mathbf a_{i}\)

Because internal forces cancel in pairs, the grand finale is

\[ M\mathbf A=\mathbf F_{\text{ext}}, \]

which reads: The CM accelerates exactly as if all the mass were packed at that point and only external forces acted on it.

8. Worked Examples (Highlights)

  • Equilateral-triangle masses (100 g, 150 g, 200 g): The CM sits closer to the heavier corner than to the lighter ones. The explicit numbers follow directly from Section 3’s formula.
  • Triangular lamina: Its CM is right at the centroid—the point where the three medians meet.
  • Uniform L-plate (three 1 kg squares): Using coordinates, the CM comes out at \(\bigl(X=5/6\text{ m},\;Y=5/6\text{ m}\bigr)\), lying on the diagonal line that splits the “L.”

High-Yield Ideas for NEET

  1. Apply the weighted-average formulas for discrete masses quickly—NEET loves straightforward number crunches.
  2. Exploit symmetry to place the CM of uniform rods, rings, discs, and spheres without calculation.
  3. Remember \(M\mathbf A=\mathbf F_{\text{ext}}\); questions often ask about how a system moves when internal forces (like explosions) act.
  4. Be comfortable with vector notation \(\mathbf R=\tfrac{1}{M}\sum m_{i}\mathbf r_{i}\); it makes multi-dimensional problems manageable.
  5. Break complex shapes into simpler pieces (method of composite bodies) to find their CM—a favorite trick in exams.

You now have a solid grip on centre-of-mass basics. Tackle a few practice problems and watch how quickly these ideas click!