Classical-Mechanics · Unit 12 · Video 3 · Interactive Practice

Where a System Really Balances: Center of Mass from Two Particles to a Rod

IKey Formulas

Formula	Name	What it takes
$\vec{R}_{\text{cm}} = \dfrac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$	Two particles	Both masses and both positions
$\vec{R}_{\text{cm}} = \dfrac{1}{m_{\text{sys}}}\displaystyle\sum_{i=1}^{N} m_i\vec{r}_i$	$N$ particles	$m_{\text{sys}} = \sum_i m_i$ ; holds component by component
$\vec{R}_{\text{cm}} = \dfrac{\int_{\text{body}} \vec{r}\,'\,dm}{\int_{\text{body}} dm}$	Continuous body	The denominator is just $M$
$dm = \lambda(x')\,dx'$	Linear mass density	Trades mass for a length element

Key Insight: All three are the same mass-weighted average of position. The only hard part of the continuous case is that the integration variable is mass, not length — so you must trade $dm$ for $\lambda(x')\,dx'$ before the integral can be evaluated.

IITwo Particles, One Weighted Average

Each position is weighted by its own mass, so the balance point drifts toward the heavier particle.

m₁ 2 kg m₂ 6 kg

IIIThe Earth–Moon Balance Point

The Moon pulls the balance point off the Earth's center — far enough to reach the surface?

Guess: R_cm in 10⁶ m 3.00

💡 The balance point only breaks the surface if the Moon is about $1.4$ times heavier: setting $m_m r_{em}/(m_e + m_m) = R_E$ gives $m_m = m_e R_E/(r_{em} - R_E) = 1.01 \times 10^{23}\ \text{kg}$ .

IVFrom a Density Law to the Balance Point

Piling mass toward one end drags the balance point with it; only the density inside the integral changes.

λ ∝ x′ⁿ, exponent n 2 swept to x′ 0.60 L

VQuiz Questions

Problem 1 · Two Masses on a Line

Given: $m_1 = 3.0\ \text{kg}$ at $x_1 = 2.0\ \text{m}$ and $m_2 = 5.0\ \text{kg}$ at $x_2 = 10.0\ \text{m}$ on the $x$ -axis — find $x_{\text{cm}}$ .

✅ Correct! The heavier particle claims

5/8

of the weight, so the balance point sits

5.0

m from

m_1

and only

3.0

m from

m_2

❌ That is the geometric midpoint.

(x_1 + x_2)/2

is the center of mass only when

m_1 = m_2

; here

m_2

is heavier, so the point must slide toward it.

❌ The weights are attached to the wrong particles. You computed

(m_2x_1 + m_1x_2)/(m_1+m_2)

— each position must be multiplied by its own mass.

❌ You stopped at the numerator.

56\ \text{kg}\cdot\text{m}

is the mass-weighted sum; dividing by the total mass

8.0

kg is what turns it back into a position.

❌ Not quite. Use

x_{\text{cm}} = (m_1x_1 + m_2x_2)/(m_1+m_2)

with numerator

3.0(2.0) + 5.0(10.0) = 56

Show solution

Both particles lie on one axis, so the vector definition reduces to its $x$ -component:

x_{\text{cm}} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} = \frac{(3.0)(2.0) + (5.0)(10.0)}{3.0 + 5.0} = \frac{6.0 + 50.0}{8.0} = \frac{56}{8.0} = 7.0\ \text{m}

Check the two limits. Equal masses would give the midpoint $6.0$ m; letting $m_1 \to 0$ would give $10.0$ m. The answer $7.0$ m lies between the midpoint and the heavy particle, exactly where a weighted average belongs.

Problem 2 · Moving the Origin

Given: the Earth–Moon system with $m_e = 5.98 \times 10^{24}\ \text{kg}$ , $m_m = 7.34 \times 10^{22}\ \text{kg}$ and $r_{em} = 3.84 \times 10^{8}\ \text{m}$ . With the origin at the Earth's center the answer was $4.66 \times 10^{6}\ \text{m}$ . Now put the origin at the Moon's center, with $\hat{\imath}$ pointing toward the Earth — find $R_{\text{cm}}$ in this frame.

✅ Correct! The same physical point, described from a new origin:

4.66 \times 10^{6} + 3.79 \times 10^{8} = 3.84 \times 10^{8}\ \text{m} = r_{em}

❌ The point does not move, but its coordinate does.

4.66 \times 10^{6}\ \text{m}

is its distance from the Earth's center; from the Moon it is almost the whole separation away.

❌ That is the midpoint $r_{em}/2$ . The midpoint would be the answer only for two equal masses; here the Earth outweighs the Moon by more than

80

times.

❌ Check the direction. The center of mass lies between the two bodies, so the two distances must add to

r_{em}

, not overshoot it.

❌ Not quite. With the origin on the Moon, the Moon's term dies instead:

R_{\text{cm}} = m_e r_{em}/(m_e + m_m)

Show solution

The origin is ours to choose, and the choice kills whichever term sits at the origin. Placing it on the Moon makes $\vec{r}_m = \vec{0}$ , so

\vec{R}_{\text{cm}} = \frac{m_e\vec{r}_e + m_m\vec{r}_m}{m_e + m_m} = \frac{m_e r_{em}}{m_e + m_m}\,\hat{\imath}

= \frac{(5.98 \times 10^{24})(3.84 \times 10^{8})}{6.05 \times 10^{24}} = \frac{2.30 \times 10^{33}}{6.05 \times 10^{24}} = 3.79 \times 10^{8}\ \text{m}

Consistency check: $3.79 \times 10^{8} + 4.66 \times 10^{6} = 3.84 \times 10^{8}\ \text{m}$ , the full separation. The balance point never moved — only the ruler did. Notice too that the Earth-based calculation is the easier one, because the small mass carries the surviving term.

Problem 3 · A Rod Whose Density Grows Linearly

Given: a thin rod of length $L$ and total mass $M$ lying from $x' = 0$ to $x' = L$ , with linear mass density $\lambda(x') = \lambda_0 x'/L$ — find the constant $\lambda_0$ and the center of mass.

What is $\lambda_0$ ?

Where is $R_{\text{cm}}$ ?

✅ Correct! A linear density piles less mass to the right than a quadratic one, so the balance point stops at

2L/3

, short of the

3L/4

of the video's rod.

❌ That is the uniform rod's density.

M/L

is the average density;

\lambda_0

here is the peak value at

x' = L

, and it must be larger than the average.

❌ That is the $\lambda \propto x'^2$ answer.

\int_0^L x'^2\,dx' = L^3/3

gives

\lambda_0 = 3M/L

; for

\lambda \propto x'

the integral is

L^2/2

instead.

❌ Not quite. Fix

\lambda_0

by demanding that the density integrate to the total mass:

\int_0^L \lambda_0 x'/L\,dx' = M

❌ That is the uniform rod. The mass here is not spread evenly, so the balance point cannot stay at the midpoint.

❌ Right number, wrong end.

L/3

is the distance from the heavy end; measured from

x' = 0

as asked, the center of mass is at

L - L/3 = 2L/3

❌ That belongs to the quadratic rod.

3L/4

comes from

\lambda \propto x'^2

; a gentler linear pile-up cannot drag the balance point that far.

❌ Not quite. Evaluate

R_{\text{cm}} = \frac{1}{M}\int_0^L x'\lambda(x')\,dx'

, where the integrand is proportional to

x'^2

Show solution

Step 1 — fix $\lambda_0$ from the total mass.

M = \int_{\text{body}} dm = \int_0^L \frac{\lambda_0 x'}{L}\,dx' = \frac{\lambda_0}{L}\cdot\frac{L^2}{2} = \frac{\lambda_0 L}{2} \quad\Longrightarrow\quad \lambda_0 = \frac{2M}{L}

The units check: mass over length, kilograms per meter.

Step 2 — integrate for the center of mass.

R_{\text{cm}} = \frac{1}{M}\int_0^L x'\,\frac{\lambda_0 x'}{L}\,dx' = \frac{\lambda_0}{ML}\int_0^L x'^2\,dx' = \frac{\lambda_0}{ML}\cdot\frac{L^3}{3}

= \frac{1}{ML}\cdot\frac{2M}{L}\cdot\frac{L^3}{3} = \frac{2L}{3}

Sanity check. For $\lambda \propto x'^n$ the same two integrals give $R_{\text{cm}} = \frac{n+1}{n+2}L$ : the uniform rod ( $n = 0$ ) balances at $L/2$ , this rod ( $n = 1$ ) at $2L/3$ , and the video's rod ( $n = 2$ ) at $3L/4$ . Steeper density, later balance point.

Problem 4 · A Rod Plus a Bead

Given: a uniform rod of mass $M$ and length $L$ lying from $x = 0$ to $x = L$ , with a point particle of mass $M$ glued to its right end at $x = L$ — find the center of mass of the combined system.

✅ Correct! And it is the same

3L/4

as the rod with

\lambda \propto x'^2

— two very different bodies can share a balance point, because only the mass distribution decides.

❌ That is the bare rod. The glued particle carries as much mass as the entire rod, so it must pull the balance point to the right of

L/2

❌ You placed the rod's mass at its far end. A body enters as its total mass sitting at its own center of mass — the rod acts as

M

L/2

, not at

L

❌ Wrong total mass. Dividing by

M

instead of

2M

puts the answer at

3L/2

, off the end of the rod entirely — a weighted average can never leave the segment joining the parts.

❌ Not quite. Replace the rod by a single particle of mass

M

x = L/2

, then apply the two-particle definition.

Show solution

Step 1 — collapse the rod. The uniform rod's own integral was done in the video: it balances at its midpoint, so for this purpose it is a particle of mass $M$ at $x = L/2$ .

Step 2 — two particles.

x_{\text{cm}} = \frac{M\left(\frac{L}{2}\right) + M(L)}{M + M} = \frac{\frac{3}{2}ML}{2M} = \frac{3L}{4}

Step 3 — read the answer. The bead drags the balance point from $L/2$ out to $3L/4$ , exactly the distance the quadratic density $\lambda \propto x'^2$ managed by smearing the extra mass along the rod instead of concentrating it at the end.

The general rule this uses: in $\sum_i m_i\vec{r}_i$ , each "particle" may be an entire extended body, entered as its total mass located at its own center of mass. That is what makes multi-part systems tractable without ever redoing the integral.

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