Physical pendulum — compound pendulum: Worked examples

This is the "throw everything at it" page for the compound pendulum. The parent note built the one master formula $T = 2 π \frac{I}{m g d}, L_{eq} = \frac{I}{m d} .$ Here we drill it through every kind of situation the topic can hand you — normal shapes, degenerate pivots, limits, a word problem, and an exam twist — so you never meet a case cold.

The scenario matrix

Before working anything, here is the full map of case-classes this topic can throw. Each worked example below is tagged with the cell(s) it covers.

#	Case class	What makes it tricky	Covered by
A	Standard shape, pivot off-CM	just apply the recipe	Ex 1 (rod), Ex 2 (disk)
B	Pivot AT the center of mass ( $d \to 0$ )	restoring torque dies, $T \to \infty$	Ex 3
C	Pivot at the minimum-period point ( $d = k$ )	derivative $= 0$ , special length $2 k$	Ex 4
D	Reversibility / center of oscillation	two different pivots, same $T$	Ex 5
E	Composite body (two masses combined)	must add $I$ and find combined CM	Ex 6
F	Large-amplitude limit	small-angle formula fails	Ex 7
G	Real-world word problem	translate words → $m, d, I$	Ex 8 (swinging sign)
H	Exam twist: solve for an unknown length	invert the formula	Ex 9

Every numeric answer below is machine-checked at the bottom.

Example 1 — Case A: uniform rod, pivot at one end

Worked example Rod swung from its end

A uniform rod, length $ℓ = 1.20 m$ , mass $m$ , swings about one end. Find $T$ (take $g = 9.81 m/s^{2}$ ).

Forecast: guess — is $T$ bigger or smaller than a simple pendulum of length $ℓ$ ? (Point mass at the tip would give $T = 2 π ℓ / g \approx 2.20$ s.) Write your guess down.

Step 1 — moment of inertia about the pivot. $I = \frac{1}{3} m ℓ^{2} .$ Why this step? The pivot is at the end, and the standard Moment of inertia of a rod about its end is $\frac{1}{3} m ℓ^{2}$ — the mass is spread all the way along the rod, so far bits count heavily.

Step 2 — distance to the CM. $d = \frac{ℓ}{2} = 0.60 m .$ Why this step? A uniform rod's CM is at its geometric middle; gravity acts there.

Step 3 — plug in. $T = 2 π \frac{\frac{1}{3} m ℓ ^{2}}{m g ( ℓ /2 )} = 2 π \frac{2 ℓ}{3 g} = 2 π \frac{2 ( 1.20 )}{3 ( 9.81 )} \approx 1.79 s .$ Why this step? The recipe. Notice $m$ cancels — a heavier rod of the same length swings at the same rate.

Verify: $L_{eq} = \frac{2 ℓ}{3} = 0.80 m < ℓ$ , so the rod swings like a shorter simple pendulum → faster than the point-mass-at-tip guess ( $1.79 < 2.20$ s). Consistent with the mass-near-pivot-speeds-things-up rule. Units: $m / (m/s^{2}) = s$ . ✓

Example 2 — Case A: uniform disk, pivot at the rim

Worked example Disk hung from its edge

A uniform disk, radius $R = 0.15 m$ , mass $m$ , hangs from a point on its rim. Find $T$ and $L_{eq}$ . ( $g = 9.81$ .)

Forecast: the pivot is a distance $R$ from the CM. Do you expect $L_{eq}$ to be more or less than $R$ ? (Recall $L_{eq} = k^{2} / d + d > d$ always.)

The figure shows the disk (lavender), its CM marked at the center, and the coral square marking the rim pivot $O$ . The mint arrow is the shift vector from CM to pivot — its length is exactly $R$ , which is both the distance $d$ AND the amount added to $I$ via the parallel-axis theorem. The butter arrow at the CM is gravity $m g$ pulling straight down.

Step 1 — $I$ about the CM, then shift to the pivot. $I_{c m} = \frac{1}{2} m R^{2}, I = I_{c m} + m d^{2} = \frac{1}{2} m R^{2} + m R^{2} = \frac{3}{2} m R^{2} .$ Why this step? The Parallel axis theorem shifts the axis from the CM to the rim, adding $m d^{2} = m R^{2}$ . Look at the mint arrow in the figure: the shift distance IS $R$ .

Step 2 — $d = R$ . Why this step? The CM of a uniform disk is its center; the rim pivot is exactly one radius away.

Step 3 — plug in. $T = 2 π \frac{\frac{3}{2} m R ^{2}}{m g R} = 2 π \frac{3 R}{2 g} = 2 π \frac{3 ( 0.15 )}{2 ( 9.81 )} \approx 0.95 s .$ Why this step? Substitute $I = \frac{3}{2} m R^{2}$ and $d = R$ into the master formula; one power of $R$ and the mass $m$ cancel, leaving a clean $3 R /2 g$ that depends only on the radius and $g$ .

Verify: $L_{eq} = I / (m d) = \frac{3}{2} R = 0.225 m > R = 0.15 m$ . ✓ Bigger than $d$ , exactly as forecast. Units check as before.

Example 3 — Case B: pivot AT the center of mass (degenerate)

Worked example What happens if you nail it through the middle?

Take the same rod as Ex 1 but pivot it exactly at its CM, then at a tiny distance $d = 0.01 m$ from the CM. What is $T$ ?

Forecast: with the pivot at the CM, will it swing fast, slow, or not at all?

Step 1 — pivot exactly at CM ( $d = 0$ ). $τ = - m g d sin θ = 0 for all θ .$ Why this step? Recall $τ$ is the restoring torque (defined at the top). Gravity now acts through the pivot — zero lever arm ( $d = 0$ ), so $τ = 0$ . The rod is in neutral equilibrium; released at any angle it just stays there. There is no oscillation, so $T$ is undefined (formally $T \to \infty$ ).

Step 2 — pivot a whisker away, $d = 0.01 m$ . Use $T = 2 π (k^{2} + d^{2}) / (g d)$ with the rod's radius of gyration about the CM $k = ℓ / 12$ (Radius of gyration). For $ℓ = 1.20$ : $k^{2} = ℓ^{2} /12 = 0.12 m^{2}$ . $T = 2 π \frac{0.12 + ( 0.01 ) ^{2}}{9.81 ( 0.01 )} = 2 π \frac{0.1201}{0.0981} \approx 6.95 s .$ Why this step? As $d \to 0$ the numerator $\to k^{2}$ (finite) but the denominator $\to 0$ , so $T$ blows up — a huge, sluggish swing.

Verify: compare to the end-pivot rod ( $T = 1.79$ s). Moving the pivot toward the CM made the period much larger (6.95 s), confirming $T \to \infty$ as $d \to 0$ . This is the counter-intuitive Forecast-then-Verify point from the parent note. ✓

Example 4 — Case C: the minimum-period pivot ( $d = k$ )

Worked example Where does the rod swing fastest?

For the $ℓ = 1.20 m$ rod, find the pivot distance $d$ giving the shortest period, and that $T_{m i n}$ .

Forecast: somewhere between the end ( $d = 0.6$ ) and the CM ( $d = 0$ ) sits the fastest spot. Guess where.

The figure plots period $T$ (vertical) against pivot distance $d$ (horizontal) for this rod. It is a U-shaped valley: $T$ shoots up on the left as $d \to 0$ (pivot near CM, restoring torque dies) and rises again on the right as $d$ grows. The coral dot marks the bottom of the valley at $d = k \approx 0.346$ m, and the mint square marks the end pivot ( $d = 0.60$ , $T \approx 1.79$ s) sitting higher up the right wall. The whole point of the example is to find that coral dot.

Step 1 — write the thing to minimise. $T$ is smallest exactly when $L_{eq} = k^{2} / d + d$ is smallest (since $T = 2 π L_{eq} / g$ grows with $L_{eq}$ ). So we minimise $f (d) = \frac{k ^{2}}{d} + d$ . Why this step? Instead of wrestling the whole square root, we minimise the simpler quantity inside it — the square root is increasing, so the minimum sits at the same $d$ .

Step 2 — take the derivative and understand each piece. $\frac{df}{dd} = \frac{d}{dd} (\frac{k ^{2}}{d}) + \frac{d}{dd} (d) = - \frac{k ^{2}}{d ^{2}} + 1.$ Why this step? We hunt the valley bottom, where the slope $df / dd$ is zero. The two derivatives: (i) $d = d^{1}$ differentiates to $1$ (a straight line of slope $1$ ); (ii) $k^{2} / d = k^{2} d^{- 1}$ differentiates by the power rule $d^{- 1} \to - 1 \cdot d^{- 2}$ , giving $- k^{2} / d^{2}$ . Intuition: making $d$ a little bigger shrinks the $k^{2} / d$ term (dividing by more), so its slope is negative — hence the minus sign.

Step 3 — set the slope to zero and solve. $- \frac{k ^{2}}{d ^{2}} + 1 = 0 \Rightarrow d^{2} = k^{2} \Rightarrow d = k = \frac{ℓ}{12} \approx \frac{1.20}{12} = 0.3464 m .$ Why this step? Zero slope = bottom of the valley. The two competing effects (torque dying vs. inertia growing) exactly balance when $d = k$ .

Step 4 — the minimum period. $T_{m i n} = 2 π \frac{2 k}{g} = 2 π \frac{2 ( 0.3464 )}{9.81} \approx 1.67 s .$ Why this step? At $d = k$ , $L_{eq} = k^{2} / k + k = 2 k$ , so $T = 2 π 2 k / g$ .

Verify: $1.67 s < 1.79 s$ (end pivot) — yes, faster. And $d = 0.346 < 0.6$ , matching the coral dot's position to the left of the mint square in the figure. ✓

Example 5 — Case D: reversibility (center of oscillation)

Worked example Two pivots, one period

Rod as before ( $ℓ = 1.20$ ), pivoted at the end ( $d_{1} = 0.60$ , $L_{eq} = 0.80$ ). Show that pivoting instead at the center of oscillation gives the same $T$ . This underlies Kater's pendulum.

Forecast: the center of oscillation lies $L_{eq} = 0.80$ m from the end pivot. How far is that point from the CM?

Step 1 — locate the center of oscillation. It sits $L_{eq} = 0.80 m$ from the end pivot, i.e. $0.80 - 0.60 = 0.20 m$ past the CM (on the far side). So the new pivot distance is $d_{2} = 0.20 m$ . Why this step? The center of oscillation is by definition a distance $L_{eq}$ from the pivot along the CM line.

Step 2 — period about the new pivot $d_{2} = 0.20$ . With $k^{2} = 0.12$ : $T_{2} = 2 π \frac{k ^{2} + d _{2}^{2}}{g d _{2}} = 2 π \frac{0.12 + 0.04}{9.81 ( 0.20 )} = 2 π \frac{0.16}{1.962} \approx 1.79 s .$

Step 3 — compare to Ex 1. Ex 1 gave $T_{1} = 1.79 s$ from the end pivot. Why this step? $T_{1} = T_{2}$ is the reversibility theorem: the two "conjugate" pivots swap roles and keep the same period.

Verify: $L_{eq}$ from the new pivot $= k^{2} / d_{2} + d_{2} = 0.12/0.20 + 0.20 = 0.80 m$ — same effective length, hence same $T$ . Symmetric, as promised. ✓

Example 6 — Case E: composite body (rod + bob)

Worked example Rod with a lump on the end

A uniform rod, $ℓ = 1.00 m$ , mass $M = 0.50 kg$ , pivoted at the top end, carries a small dense bob of mass $m_{b} = 0.50 kg$ fixed at the bottom tip. Find $T$ . ( $g = 9.81$ .)

Forecast: the bob drags the CM downward toward the tip. Will $T$ rise or fall versus the bare rod?

Step 1 — combined center of mass (from the pivot). $d = \frac{M ( ℓ /2 ) + m _{b} ℓ}{M + m _{b}} = \frac{0.50 ( 0.50 ) + 0.50 ( 1.00 )}{1.00} = 0.75 m .$ Why this step? CM of a composite is the mass-weighted average of the parts' positions. The heavy tip pulls the CM down from $0.50$ to $0.75$ m.

Step 2 — total moment of inertia about the pivot. $I = rod, end \frac{1}{3} M ℓ^{2} + point bob at tip m_{b} ℓ^{2} = \frac{1}{3} (0.50) (1.00)^{2} + (0.50) (1.00)^{2} = 0.1667 + 0.50 = 0.6667 kg\cdotm^{2} .$ Why this step? Moments of inertia add about a common axis. A point mass at distance $ℓ$ contributes $m_{b} ℓ^{2}$ .

Step 3 — plug in with total mass. $T = 2 π \frac{I}{( M + m _{b} ) g d} = 2 π \frac{0.6667}{( 1.00 ) ( 9.81 ) ( 0.75 )} \approx 1.89 s .$ Why this step? In the master formula the " $m$ " is the total swinging mass $M + m_{b}$ , because gravity acts on the whole body through the combined CM at distance $d$ . Use the combined $I$ and combined $d$ we just found, not the parts' separate values.

Verify: bare rod (Ex 1 style, $ℓ = 1.0$ ) gives $T = 2 π 2 (1.0) / (3 \cdot 9.81) \approx 1.64$ s. The bob raised $T$ to $1.89$ s — heavier far end means bigger $I$ growing faster than $d$ , so slower. Physically reasonable. ✓

Example 7 — Case F: large-amplitude limit (formula breaks)

Worked example When the small-angle formula lies

The end-pivoted rod (Ex 1, $T_{0} = 1.79 s$ for tiny swings) is released from $θ_{0} = 9 0^{\circ}$ . Estimate the true period and see how far off the simple formula is.

Forecast: big swing — does the real period go up or down compared to $1.79$ s?

Step 1 — why the simple formula fails. We used $sin θ \approx θ$ , valid only for small $θ$ . At $9 0^{\circ}$ that approximation understates the slow-down near the turning points, where the restoring torque $m g d sin θ$ is weaker than the linear $m g d θ$ would suggest. Why this step? Honesty about the domain of validity — the parent note's third mistake.

Step 2 — convert the angle to radians, then apply the correction. The amplitude-correction series requires the amplitude $θ_{0}$ in radians, because it comes from a Taylor expansion of $sin θ$ where $θ$ must be in radians. Convert first: $θ_{0} = 9 0^{\circ} = 90 \times \frac{π}{180} rad = \frac{π}{2} \approx 1.5708 rad .$ The leading amplitude correction is $T \approx T_{0} (1 + \frac{1}{16} θ_{0}^{2} + \dots), θ_{0} in radians,$ $T \approx 1.79 (1 + \frac{( 1.5708 ) ^{2}}{16}) = 1.79 (1 + 0.1542) \approx 2.07 s .$ Why this step? The $\frac{1}{16} θ_{0}^{2}$ term is the first term of expanding the exact elliptic-integral period; it must use radians, so forgetting to convert $9 0^{\circ}$ (using $9 0^{2}$ instead of $1.570 8^{2}$ ) would give a nonsense answer. This is the classic unit trap.

Verify: $2.07 > 1.79 s$ — big swings are slower, up by about $15%$ . So quoting $1.79$ s here would be wrong by a noticeable margin. The correction term is positive and small-ish, as a first-order term should be. ✓

Example 8 — Case G: real-world word problem

Worked example The swinging shop sign

A rectangular shop sign, height $h = 0.80 m$ (vertical), hangs from a horizontal rod along its top edge, so it swings about that top edge. Treat it as a uniform flat plate, mass $m$ . How long does one full swing take? ( $g = 9.81$ .)

Forecast: it's like a plate pivoted at its top. Guess whether $L_{eq}$ is more or less than $h$ .

Step 1 — translate words into $m, d, I$ . Swing is about the top edge; only the vertical extent $h$ matters (the horizontal width lies along the pivot axis and doesn't affect the swinging moment). Model as a thin lamina swinging about its top edge = same as a rod of length $h$ about its end for this motion. Why this step? Rotation about the top edge only "feels" mass distributed in the vertical direction; the width is parallel to the axis.

Step 2 — the numbers. $I = \frac{1}{3} m h^{2}, d = \frac{h}{2} = 0.40 m .$ Why this step? Having reduced the plate to "a rod of length $h$ pivoted at its end" in Step 1, we reuse the rod-at-end results: $I = \frac{1}{3} m h^{2}$ and the CM sits at the mid-height $h /2$ .

Step 3 — period. $T = 2 π \frac{\frac{1}{3} m h ^{2}}{m g ( h /2 )} = 2 π \frac{2 h}{3 g} = 2 π \frac{2 ( 0.80 )}{3 ( 9.81 )} \approx 1.46 s .$ Why this step? Substitute $I$ and $d$ into the master formula; $m$ and one power of $h$ cancel, leaving the same $2 h /3 g$ shape as the rod — as it must, since we modelled it as a rod.

Verify: $L_{eq} = \frac{2 h}{3} = 0.533 m > d = 0.40$ m (always $L_{eq} > d$ ). A real sign swings noticeably slower than a $0.40$ m string-pendulum. Units of seconds. ✓

Example 9 — Case H: exam twist (invert for an unknown)

Worked example Find the hidden bob position

A light (massless) rod of length $ℓ = 1.00 m$ carries a point bob of mass $m$ at distance $x$ from the top pivot. Measured period $T = 1.60 s$ . Where is the bob? ( $g = 9.81$ .)

Forecast: a point mass on a light rod IS just a simple pendulum of length $x$ . So what should $x$ come out to?

Step 1 — set up $I$ and $d$ . $I = m x^{2}, d = x .$ Why this step? All the mass is the point bob at distance $x$ ; the rod is massless so contributes nothing.

Step 2 — the period reduces to the simple pendulum. $T = 2 π \frac{m x ^{2}}{m g x} = 2 π \frac{x}{g} .$ Why this step? $I / (m d) = x^{2} / x = x$ , so $L_{eq} = x$ — this special case degenerates back to the Simple pendulum. Good sanity anchor.

Step 3 — solve for $x$ . $x = \frac{g T ^{2}}{4 π ^{2}} = \frac{9.81 ( 1.60 ) ^{2}}{4 π ^{2}} = \frac{25.11}{39.48} \approx 0.636 m .$ Why this step? Square both sides of $T = 2 π x / g$ to get $T^{2} = 4 π^{2} x / g$ , then isolate $x$ . This is the "invert the formula" move the exam is testing.

Verify: $x = 0.636 m < ℓ = 1.00$ m, so the bob sits on the rod — physically valid. Plug back: $2 π 0.636/9.81 \approx 1.60$ s ✓, matching the given period. And it confirms the forecast that a point mass gives $L_{eq} = x$ . Units: metres. ✓

Case-coverage recap

Recall Which case is each example?

Ex 1 rod / Ex 2 disk ::: Case A (standard off-CM pivot) Ex 3 pivot at CM ::: Case B (degenerate, $T \to \infty$ ) Ex 4 fastest pivot ::: Case C ( $d = k$ , minimum period) Ex 5 two pivots same $T$ ::: Case D (reversibility, Kater's pendulum) Ex 6 rod plus bob ::: Case E (composite: add $I$ , weighted CM) Ex 7 $9 0^{\circ}$ release ::: Case F (large-amplitude, formula fails) Ex 8 shop sign ::: Case G (real-world translation) Ex 9 hidden bob ::: Case H (invert to solve for a length)