Centre of mass — definition for system of particles: Exercises

Level 1 — Recognition

Exercise 1.1

Two particles: $m_{1} = 4$ kg at $x_{1} = 0$ , $m_{2} = 4$ kg at $x_{2} = 8$ m. Find $x_{c m}$ .

Recall Solution 1.1

WHAT we do: plug into the formula. $x_{c m} = \frac{4 ( 0 ) + 4 ( 8 )}{4 + 4} = \frac{32}{8} = 4 m$ WHY the answer makes sense: the masses are equal, so the weights cancel and the COM is the plain midpoint of $0$ and $8$ , which is $4$ m. ✔️

Exercise 1.2

State, without full computation: for $m_{1} = 1$ kg at $x = 0$ and $m_{2} = 9$ kg at $x = 10$ m, is $x_{c m}$ closer to $0$ or to $10$ ?

Recall Solution 1.2

WHAT & WHY: the COM always leans toward the heavier mass. Here $9$ kg $≫ 1$ kg, so $x_{c m}$ sits close to $x = 10$ m. (Full number, if you want it: $x_{c m} = \frac{1 ( 0 ) + 9 ( 10 )}{10} = 9$ m — indeed close to $10$ .)

Level 2 — Application

Exercise 2.1

$m_{1} = 2$ kg at $x_{1} = 1$ m, $m_{2} = 3$ kg at $x_{2} = 4$ m, $m_{3} = 5$ kg at $x_{3} = 6$ m. Find $x_{c m}$ .

Recall Solution 2.1

Total mass: $M = 2 + 3 + 5 = 10$ kg. Numerator (the moment, units kg·m): $2 (1) + 3 (4) + 5 (6) = 2 + 12 + 30 = 44$ kg·m. $x_{c m} = \frac{44}{10} = 4.4 m$ WHY this looks right: $5$ kg (the heaviest) sits at $6$ m and drags the average up past the middle particle at $4$ m — so $4.4$ m leaning toward the heavy side is exactly what we expect. ✔️

Exercise 2.2

Three particles in a plane: $m_{1} = 1$ kg at $(0, 0)$ , $m_{2} = 2$ kg at $(4, 0)$ , $m_{3} = 1$ kg at $(0, 4)$ . Find $(x_{c m}, y_{c m})$ .

Recall Solution 2.2

WHY split by axis: the vector equation is just three independent scalar averages (parent §3). Handle $x$ and $y$ separately. $M = 1 + 2 + 1 = 4$ kg. $x_{c m} = \frac{1 ( 0 ) + 2 ( 4 ) + 1 ( 0 )}{4} = \frac{8}{4} = 2 m$ $y_{c m} = \frac{1 ( 0 ) + 2 ( 0 ) + 1 ( 4 )}{4} = \frac{4}{4} = 1 m$ COM at $(2, 1)$ m — inside the triangle, pulled toward the $x$ -axis edge because the heavier $2$ kg mass sits there. See figure. ✔️

Figure — Centre of mass — definition for system of particles

Level 3 — Analysis

Exercise 3.1

Two particles: $m_{1} = 2$ kg at $x_{1} = 0$ , and a mass $m_{2}$ at $x_{2} = 6$ m. For what $m_{2}$ is $x_{c m} = 4$ m?

Recall Solution 3.1

WHAT we do: treat $m_{2}$ as the unknown and set up the equation. $4 = \frac{2 ( 0 ) + m _{2} ( 6 )}{2 + m _{2}}$ Solve — multiply both sides by $(2 + m_{2})$ : $4 (2 + m_{2}) = 6 m_{2} \Rightarrow 8 + 4 m_{2} = 6 m_{2} \Rightarrow 8 = 2 m_{2} \Rightarrow m_{2} = 4 kg$ WHY sensible: $x_{c m} = 4$ m is closer to $x_{2} = 6$ than to $x_{1} = 0$ , so $m_{2}$ must be the heavier one — and indeed $4 > 2$ kg. ✔️

Exercise 3.2

A COM sits at the origin. Particle A: $3$ kg at $(- 2, 0)$ m. Particle B: $2$ kg at $(x_{B}, y_{B})$ . Find $(x_{B}, y_{B})$ .

Recall Solution 3.2

WHY this works: if the COM is at the origin, then $\sum m_{i} x_{i} = 0$ and $\sum m_{i} y_{i} = 0$ (the numerators vanish, because $x_{c m} = y_{c m} = 0$ and $M \neq = 0$ ). x-equation: $3 (- 2) + 2 x_{B} = 0 \Rightarrow - 6 + 2 x_{B} = 0 \Rightarrow x_{B} = 3$ m. y-equation: $3 (0) + 2 y_{B} = 0 \Rightarrow y_{B} = 0$ m. So B is at $(3, 0)$ m. WHY it looks right: the two masses must sit on opposite sides of the origin along the same line for the origin to balance them; the lighter $2$ kg mass sits farther out ( $3$ m vs $2$ m) to compensate for being lighter. ✔️

Level 4 — Synthesis

Exercise 4.1

A $1.0$ m uniform rod of mass $2$ kg has its COM at its middle, $x = 0.5$ m. A point mass of $1$ kg is glued to its right end ( $x = 1.0$ m). Where is the COM of the rod+mass system?

Recall Solution 4.1

KEY idea (synthesis): an extended body may be replaced by a single point mass at its own COM. So treat the rod as " $2$ kg at $x = 0.5$ m." (This is the bridge to Centre of Mass of Continuous Bodies.) $x_{c m} = \frac{2 ( 0.5 ) + 1 ( 1.0 )}{2 + 1} = \frac{1.0 + 1.0}{3} = \frac{2.0}{3} \approx 0.667 m$ WHY it looks right: gluing mass on the right end shifts the balance point right of the rod's own centre ( $0.5$ m), toward the added mass. $0.667$ m is exactly that rightward nudge. ✔️

Exercise 4.2

Four equal masses $m$ sit at three corners of a square of side $a$ : $(0, 0)$ , $(a, 0)$ , $(0, a)$ — and the fourth corner $(a, a)$ is empty. Find the COM.

Recall Solution 4.2

WHY equal masses simplify: with all masses $= m$ , the $m$ 's cancel and $x_{c m}$ becomes the plain average of the $x$ -coordinates (and same for $y$ ). Note there are only three particles. $x_{c m} = \frac{m ( 0 ) + m ( a ) + m ( 0 )}{3 m} = \frac{a}{3}, y_{c m} = \frac{m ( 0 ) + m ( 0 ) + m ( a )}{3 m} = \frac{a}{3}$ COM at $(\frac{a}{3}, \frac{a}{3})$ . WHY interesting: the COM sits in empty space — no particle is there. It lands closer to the corner $(0, 0)$ which is "shared" by two of the occupied edges. ✔️

Level 5 — Mastery

Exercise 5.1 (COM velocity, connecting to momentum)

Two particles move on a line: $m_{1} = 2$ kg with velocity $v_{1} = + 3$ m/s, $m_{2} = 1$ kg with velocity $v_{2} = - 6$ m/s. Find the COM velocity $v_{c m}$ , and state what it tells you about the total momentum.

Recall Solution 5.1

WHY the formula generalises: differentiate $x_{c m} = \frac{\sum m _{i} x _{i}}{M}$ in time — the masses are constant, so positions become velocities: $v_{c m} = \frac{\sum _{i} m _{i} v _{i}}{M}$ $v_{c m} = \frac{2 ( 3 ) + 1 ( - 6 )}{2 + 1} = \frac{6 - 6}{3} = 0 m/s$ WHAT it means: the COM is stationary. Since total momentum $P = M v_{c m} = 3 \times 0 = 0$ , the system's momentum is zero — the two particles' momenta exactly cancel. This is the launching pad for Conservation of Linear Momentum and analysing Collisions — Elastic and Inelastic in the COM frame. ✔️

Exercise 5.2 (Mastery: the "remove-a-piece" method)

A uniform square plate of side $2$ m and mass $4$ kg is centred at the origin (its COM is at $(0, 0)$ ). A small square hole is cut out; the removed piece has mass $1$ kg and its own centre was at $(0.5, 0)$ . Find the COM of the remaining plate.

Recall Solution 5.2

KEY synthesis idea — subtraction (negative mass trick): the full plate = (remaining plate) + (removed piece). Momentum-of-position bookkeeping means moments add: $M_{f u l l} x_{f u l l} = M_{r e m} x_{r e m} + M_{c u t} x_{c u t}$ Solve for the remaining part's $x_{r e m}$ : $x_{r e m} = \frac{M _{f u l l} x _{f u l l} - M _{c u t} x _{c u t}}{M _{r e m}}$ Numbers: $M_{f u l l} = 4$ , $x_{f u l l} = 0$ ; $M_{c u t} = 1$ , $x_{c u t} = 0.5$ ; $M_{r e m} = 4 - 1 = 3$ kg. $x_{r e m} = \frac{4 ( 0 ) - 1 ( 0.5 )}{3} = \frac{- 0.5}{3} \approx - 0.167 m$ By symmetry $y_{r e m} = 0$ . So the remaining plate's COM is at $(- \frac{1}{6}, 0) \approx (- 0.167, 0)$ m. WHY it looks right: we cut mass from the right ( $x = + 0.5$ ), so the leftover balances slightly to the left of centre. The negative answer confirms the shift is leftward. ✔️

Active Recall

Recall Did you spot the pattern across levels?

L1–L2: plug in — keep the weights $m_{i}$ , split axes.
L3: invert — solve for an unknown mass or position by setting the moment equation.
L4: compose — replace an extended body by a point at its own COM.
L5: differentiate ( $v_{c m}$ ) and subtract (negative-mass hole). One formula, five moves.

Ladder of moves

recognise, apply, invert, compose, differentiate/subtract

Why does a cut-out piece carry a minus sign?

Because full = remaining + removed, so the removed moment is subtracted to recover the remaining part.

What is

v_{c m}

in terms of momentum?

v_{c m} = \frac{\sum m _{i} v _{i}}{M} = \frac{P}{M}

Connections

Newton's Second Law for a System of Particles — Exercise 5.1 is the doorway to $F_{e x t} = M a_{c m}$ .
Conservation of Linear Momentum — a zero COM velocity means zero total momentum.
Collisions — Elastic and Inelastic — the COM frame (Exercise 5.1) simplifies collision algebra.
Centre of Mass of Continuous Bodies — Exercises 4.1 & 5.2 use the "point at its own COM" bridge.
Weighted Average and Moments — every solution is a weighted average / moment balance.