Classical-Mechanics · Unit 20 · Video 6 · Interactive Practice

The Physical Pendulum: Moment of Inertia, Center of Mass, and Swing Speed

IKey Formulas

Formula	Name	What it takes
$I_S = I_1 + I_{cm} + M_2 L_2^2$	Moment of inertia about the pivot	Moments about one common axis add; the parallel axis theorem carries the disc's $I_{cm}$ out to $S$
$L_{cm} = \dfrac{M_1 \frac{L_1}{2} + M_2 L_2}{M_1 + M_2}$	Centre of mass of rod and disc	Each part's mass times the position of its own centre, divided by the total mass
$(M_1 + M_2)\, g\, L_{cm}\,\big(1 - \cos\theta_i\big) = \tfrac{1}{2} I_S \omega_f^2$	Energy balance, release to bottom	Released from rest, the pivot does no work, no friction
$\omega_f = \sqrt{\dfrac{2\left(M_1 \frac{L_1}{2} + M_2 L_2\right) g \big(1 - \cos\theta_i\big)}{I_1 + I_{cm} + M_2 L_2^2}}$	Angular speed at the bottom	Both earlier results, in one expression

Key Insight: The potential energy of a rigid body may be computed as if all of its mass sat at the centre of mass; its kinetic energy may not. That is why $L_{cm}$ and $I_S$ stand on opposite sides of the same equation — $I_S$ depends on how the mass is spread about the pivot, not only on where its average sits.

IIVisualization 1 — Where the Mass Balances

The combined centre of mass is the mass-weighted average of the rod's midpoint and the disc's centre.

Disc mass M₂ (kg) 1.0

IIIVisualization 2 — Shifting the Disc's Axis Out to the Pivot

The disc turns about $S$ , not about its own centre, so $I_{cm}$ alone is not its contribution.

Disc centre L₂ (m) 0.80

IVVisualization 3 — Falling into the Bottom of the Swing

Released from rest, the pendulum trades the potential energy of its centre of mass for $\tfrac{1}{2} I_S \omega^2$ .

Release angle θᵢ (deg) 30 Swing 30°

💡 At $\theta_i = 0$ the pendulum was never displaced, and the formula returns $\omega_f = 0$ : no height was given up, so there is nothing to convert.

VQuiz Questions

Problem 1 · Moment of Inertia About the Pivot

Given: a uniform rod with $I_1 = 0.667\ \mathrm{kg\,m^2}$ about the pivot $S$ , and a disc of mass $M_2 = 1.00\ \mathrm{kg}$ with $I_{cm} = 0.020\ \mathrm{kg\,m^2}$ about the axis through its own centre, rigidly fixed to the rod with that centre $L_2 = 0.80\ \mathrm{m}$ from $S$ — find $I_S$ , the moment of inertia of the whole pendulum about the pivot. (The rod has length $L_1 = 1.00\ \mathrm{m}$ .)

✅ Correct! The disc's contribution about

S

I_{cm} + M_2 L_2^2 = 0.020 + 0.640 = 0.660\ \mathrm{kg\,m^2}

, and the rod's

I_1

was already defined about the pivot.

❌ That is $I_1 + I_{cm}$ . Leaving out

M_2 L_2^2

treats the disc as if its centre sat at the pivot, contributing nothing for its distance from

S

❌ Close — one term is missing.

I_1 + M_2L_2^2 = 1.307

drops

I_{cm}

; the parallel axis theorem adds

M_2 L_2^2

I_{cm}

, it does not replace it.

❌ Check which distance you squared. The shift is the separation of the two axes,

L_2 = 0.80\ \mathrm{m}

, not the rod's length

L_1 = 1.00\ \mathrm{m}

❌ Not quite. Moments of inertia about one common axis add:

I_S = I_1 + \big(I_{cm} + M_2 L_2^2\big)

— all three terms.

Show solution

Moments of inertia about the same axis add, so treat the rod and the disc separately and add their contributions about $S$ .

Rod: $I_1$ is already given about the pivot, so it needs no adjustment.

Disc: $I_{cm}$ is about the axis through the disc's own centre. That axis is parallel to the pivot axis (both perpendicular to the plane of the swing) and separated from it by $L_2$ , so the parallel axis theorem applies:

I_{2,S} = I_{cm} + M_2 L_2^2 = 0.020 + (1.00)(0.80)^2 = 0.020 + 0.640 = 0.660\ \mathrm{kg\,m^2}

Total:

I_S = I_1 + I_{cm} + M_2 L_2^2 = 0.667 + 0.020 + 0.640 = 1.327\ \mathrm{kg\,m^2}

The shift term alone is $0.640$ , about $48\%$ of the total — the disc's mass sitting $0.80\ \mathrm{m}$ from the pivot matters far more than how that mass is spread about its own centre.

Problem 2 · The Combined Centre of Mass

Given: a uniform rod of mass $M_1 = 2.00\ \mathrm{kg}$ and length $L_1 = 1.00\ \mathrm{m}$ hanging from the pivot, with a disc of mass $M_2 = 1.00\ \mathrm{kg}$ whose centre lies $L_2 = 0.80\ \mathrm{m}$ from the pivot along the rod — find $L_{cm}$ , the distance from the pivot to the centre of mass of the two together.

✅ Correct! The average lands between

0.500\ \mathrm{m}

and

0.800\ \mathrm{m}

, and nearer the rod because the rod is twice as heavy.

❌ That is the plain midpoint of $0.500$ and $0.800$ . The midpoint is correct only when the two masses are equal; here

M_1 = 2M_2

, so the average is pulled toward the rod's centre.

❌ The weights are attached to the wrong positions. The rod's own centre at

0.500\ \mathrm{m}

carries

M_1

, and the disc's centre at

0.800\ \mathrm{m}

carries

M_2

❌ That is the numerator only.

M_1\frac{L_1}{2} + M_2 L_2 = 1.800\ \mathrm{kg\,m}

still has to be divided by the total mass

M_1 + M_2 = 3.00\ \mathrm{kg}

❌ Not quite. Each part contributes its own mass times the position of its own centre: the rod at

L_1/2

, the disc at

L_2

Show solution

The rod is uniform, so its own centre of mass is at its midpoint, $L_1/2 = 0.500\ \mathrm{m}$ from the pivot. The disc's centre of mass is its centre, $L_2 = 0.800\ \mathrm{m}$ from the pivot. Both lie on the line of the rod, so the combined centre lies on that line too:

(M_1 + M_2)\,L_{cm} = M_1\frac{L_1}{2} + M_2 L_2

L_{cm} = \frac{M_1\frac{L_1}{2} + M_2 L_2}{M_1 + M_2} = \frac{(2.00)(0.500) + (1.00)(0.800)}{2.00 + 1.00} = \frac{1.800}{3.00} = 0.600\ \mathrm{m}

Two limits check the formula:

$M_2 \to 0$ gives $L_{cm} \to L_1/2 = 0.500\ \mathrm{m}$ , the rod's own centre.
$M_1 \to 0$ gives $L_{cm} \to L_2 = 0.800\ \mathrm{m}$ , the centre of the disc.

For any masses in between, $L_{cm}$ lies between those two points, closer to the heavier part.

Problem 3 · From Release to the Bottom

Given: the same pendulum — $M_1 + M_2 = 3.00\ \mathrm{kg}$ , $L_{cm} = 0.600\ \mathrm{m}$ , $I_S = 1.327\ \mathrm{kg\,m^2}$ , $g = 9.8\ \mathrm{m/s^2}$ — released from rest at $\theta_i = 60\degree$ and swinging to the bottom, where $\theta = 0$ .

How much potential energy is given up?

What is the angular speed $\omega_f$ at the bottom?

✅ Correct! All

8.82\ \mathrm{J}

becomes

\tfrac{1}{2}I_S\omega_f^2

, so

\omega_f = \sqrt{2(8.82)/1.327} = 3.65\ \mathrm{rad/s}

❌ Check the height the centre of mass loses. It falls by

L_{cm}\big(1 - \cos\theta_i\big)

, not by

L_{cm}

and not by

L_{cm}\sin\theta_i

; with

\cos 60\degree = 0.5

that drop is

0.300\ \mathrm{m}

❌ Check the algebra of the last step. From

\Delta U = \tfrac{1}{2}I_S\omega_f^2

, multiply by two and divide by the full

I_S

before taking the positive root:

\omega_f = \sqrt{2\,\Delta U / I_S}

Show solution

Step 1 — the potential energy given up. Treat the potential energy as if the whole mass sat at the centre of mass. At release that point stands a height $L_{cm}\big(1 - \cos\theta_i\big)$ above the level it reaches at the bottom:

h = L_{cm}\big(1 - \cos 60\degree\big) = (0.600)(1 - 0.500) = 0.300\ \mathrm{m}

\Delta U = (M_1 + M_2)\, g\, h = (3.00)(9.8)(0.300) = 8.82\ \mathrm{J}

Step 2 — the kinetic energy gained. The pendulum starts from rest and the pivot does no work, so the whole $8.82\ \mathrm{J}$ appears as rotational kinetic energy about $S$ . Rod and disc are one rigid body turning with one angular speed:

\tfrac{1}{2} I_S \omega_f^2 = 8.82\ \mathrm{J}

\omega_f = \sqrt{\frac{2(8.82)}{1.327}} = \sqrt{13.30} = 3.65\ \mathrm{rad/s}

Why the kinetic energy cannot use $L_{cm}$ : a point particle of mass $3.00\ \mathrm{kg}$ at $0.600\ \mathrm{m}$ would have $I = ML_{cm}^2 = 1.08\ \mathrm{kg\,m^2}$ and would give $\omega_f = 4.04\ \mathrm{rad/s}$ . The real body has $I_S = 1.327\ \mathrm{kg\,m^2}$ , because its mass is spread out about the pivot rather than concentrated at one point.

Problem 4 · Swap the Disc for a Ring

Given: the disc is replaced by a thin ring of the same mass and radius, so $I_{cm} = M_2 R_2^2 = 0.040\ \mathrm{kg\,m^2}$ instead of $0.020\ \mathrm{kg\,m^2}$ . The masses, $L_1$ , $L_2$ and the release angle are unchanged. What happens to $L_{cm}$ and to $\omega_f$ ?

✅ Correct!

L_{cm}

knows only masses and positions, so it stays at

0.600\ \mathrm{m}

;

I_S

rises to

1.347\ \mathrm{kg\,m^2}

and

\omega_f

drops from

3.646

3.619\ \mathrm{rad/s}

at the same

\theta_i = 60\degree

❌ Check which side of the balance $I_S$ sits on. The energy released is unchanged, and

I_S

is in the denominator of

\omega_f^2 = 2\Delta U / I_S

— more inertia means a slower swing, not a faster one.

❌ $I_{cm}$ is one of the three terms of $I_S$ . Doubling it raises

I_S

0.020\ \mathrm{kg\,m^2}

, which changes the kinetic-energy side of the balance.

❌ The centre of mass does not move. A ring and a disc of the same mass and radius have their centres of mass in the same place; only the spread of mass about that centre differs.

❌ Not quite. Ask which of

L_{cm}

and

I_S

can even notice a change in

I_{cm}

Show solution

The potential-energy side. $L_{cm} = \dfrac{M_1\frac{L_1}{2} + M_2L_2}{M_1+M_2}$ contains only masses and the positions of the two centres. A ring has its centre of mass at its centre, exactly like the disc, so

L_{cm} = 0.600\ \mathrm{m} \quad \text{(unchanged)}, \qquad \Delta U = (3.00)(9.8)(0.600)\big(1 - \cos 60\degree\big) = 8.82\ \mathrm{J} \quad \text{(unchanged)}

The kinetic-energy side. $I_S$ does notice, because $I_{cm}$ is one of its three terms:

I_S = 0.667 + 0.040 + 0.640 = 1.347\ \mathrm{kg\,m^2} \quad (\text{was } 1.327)

\omega_f = \sqrt{\frac{2(8.82)}{1.347}} = 3.619\ \mathrm{rad/s} \quad (\text{was } 3.646)

The same energy has to spin a body that resists a little more, so the pendulum passes the bottom about $0.8\%$ slower. This is the asymmetry of the whole problem in one move: the centre of mass fixes the energy released, but only the distribution of mass about the pivot fixes the speed it buys.

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