Simple vs Physical Pendulum: What Is the Difference?

A physical pendulum is any rigid body swinging about a fixed axis. A simple pendulum is the special case with all the mass at one distance from the pivot, so it acts as a point on a string. Put a point mass into the physical pendulum formula and it becomes the simple one exactly.

AP Physics: Unit 7 (topics 7.2 Frequency and Period of SHM, 7.5 Simple and Physical Pendulums, 5.4 Rotational Inertia). The physical pendulum belongs to AP Physics C: Mechanics only. Its Unit 7, Oscillations, is weighted at 10 to 15 percent of the exam over about 12 to 17 class periods, and Topic 7.5, Simple and Physical Pendulums, carries learning objective 7.5.A, describe the properties of a physical pendulum. EK 7.5.A.1 defines a physical pendulum as a rigid body that undergoes oscillation about a fixed axis. EK 7.5.A.2 gives the period for small amplitudes as derived from Newton's second law in rotational form, with the relevant equation T = 2 pi times the square root of I over m g d, and its sub-statements give the exact restoring torque as minus m g d sine theta at 7.5.A.2.i, the small-angle approximation at 7.5.A.2.ii, and the resulting second-order differential equation at 7.5.A.2.iii. EK 7.5.A.3 states that a simple pendulum is a special case of physical pendulums in which the hanging object can be modeled as a point mass at a distance from the pivot, with the relevant equation T = 2 pi times the square root of length over g. EK 7.5.A.4 adds the torsion pendulum, with the derived equation I alpha = minus k delta theta. Suggested skills for Topic 7.5 are 1.B, 2.A, 2.B, 3.A and 3.B. The phrase physical pendulum appears in no other CED of the four, and the physical pendulum period is printed only on the two Physics C booklets, while the simple pendulum period is printed on all four. The parallel axis theorem used throughout this page is EK 5.4.B.2, printed on the AP Physics 1 and C: Mechanics sheets as I prime equals I about the center of mass plus M d squared.

One is the special case of the other, and the CED says so

This is not two competing models of a swinging object. The simple pendulum is a physical pendulum in which all of the mass has been moved to a single distance from the pivot. AP Physics C: Mechanics states the relationship directly rather than leaving it to be noticed.

EK 7.5.A.1: a physical pendulum is a rigid body that undergoes oscillation about a fixed axis. That is the general object, and its period for small amplitudes is at EK 7.5.A.2:

T_{\text{phys}} = 2\pi \sqrt{\frac{I}{mgd}}

Here $I$ is the rotational inertia about the pivot axis, $m$ is the mass of the body, and $d$ is the distance from the pivot to the body's center of mass.

EK 7.5.A.3: a simple pendulum is a special case of physical pendulums in which the hanging object can be modeled as a point mass at a distance $\ell$ from the pivot point. The relevant equation attached to that statement is $T_p = 2\pi\sqrt{\ell/g}$ .

So the two formulas are one formula. The next section does the substitution that turns the first into the second, in two lines, and that collapse is the reason this page exists. Everything else here is about what you lose when the collapse is not available: which term goes missing, how big the error is, and which of the four AP courses ever asks.

Watch the collapse: put a point mass into the general formula

Take the physical pendulum period and impose the one condition EK 7.5.A.3 names, that the hanging object is a point mass at distance $\ell$ from the pivot.

A point mass at distance $\ell$ from the axis has rotational inertia $I = m\ell^2$ , which is EK 5.4.A.2 in AP Physics 1 and the single-object case of $I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$ on the sheet. And the center of mass of a single point mass is the point mass, so $d = \ell$ . Substitute both:

T_{\text{phys}} = 2\pi\sqrt{\frac{I}{mgd}} = 2\pi\sqrt{\frac{m\ell^2}{mg\ell}} = 2\pi\sqrt{\frac{\ell}{g}} = T_p

The mass cancels, one power of $\ell$ cancels, and what is left is the simple pendulum formula exactly. Not approximately, and not in a limit: it is an algebraic identity once $I = m\ell^2$ and $d = \ell$ are true together.

Both conditions have to hold. That is the part worth slowing down for, because each one fails in a different way.

$d = \ell$ says the center of mass is at the same distance as the mass you are counting. True for a point, false for a rod.
$I = m\ell^2$ says every gram is at distance $\ell$ . True for a point, false for a bob of real size and false for the string if the string has mass.

Running the substitution backwards gives the quantity that makes the comparison quantitative. Any physical pendulum swings with the period of a simple pendulum whose length is

\ell_{\text{eq}} = \frac{I}{md}

because substituting that $\ell_{\text{eq}}$ into $2\pi\sqrt{\ell/g}$ reproduces $2\pi\sqrt{I/(mgd)}$ . So the question is never whether a physical pendulum behaves like a simple one. It is which length to use. For a point mass, $\ell_{\text{eq}} = \ell$ . For a uniform rod pivoted at one end, $\ell_{\text{eq}} = 2L/3$ , which is neither the rod's length nor the distance to its center of mass. The CED does not give this quantity a name; it is worth carrying anyway, because it turns the whole comparison into one number.

Simple vs physical pendulum, side by side

Question you are asking	Simple pendulum	Physical pendulum
What is swinging	A point mass on a massless string	Any rigid body about a fixed axis
CED essential knowledge	7.5.A.3, and 7.2.A.1.ii	7.5.A.1 and 7.5.A.2
Period	$T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$	$T_{\text{phys}} = 2\pi\sqrt{\dfrac{I}{mgd}}$
What the length symbol means	Pivot to the point mass	Not applicable; you need $I$ and $d$ separately
Equivalent simple length	$\ell$ itself	$\ell_{\text{eq}} = I/(md)$
Rotational inertia needed	No	Yes, about the pivot axis
Depends on total mass	No	No, if the mass distribution is unchanged
Restoring influence	Gravitational force along the arc	Restoring torque about the pivot, EK 7.5.A.2.i
Exact restoring law	Proportional to $\sin\theta$	$\tau = -mgd\sin\theta$ , EK 7.5.A.2.i
Small-angle form	$\sin\theta \approx \theta$	$\tau = -mgd\theta = I\alpha$ , EK 7.5.A.2.ii
Courses that require it	All that teach oscillations	AP Physics C: Mechanics only
Printed on the sheet	All four booklets	Both Physics C booklets
Second law used	Translational	Rotational

Three rows deserve unpacking.

The mass row is the surprise. The simple pendulum's mass independence is famous. The physical pendulum's is not, and it is just as real: for a rigid body of fixed shape and uniform density, $I$ is proportional to the total mass, so the $m$ in the numerator cancels the $m$ in the denominator of $I/(mgd)$ . Double every gram of a swinging rod and its period does not change. What does change the period is moving mass around, because that changes $I$ and $d$ differently. Scaling the mass is free; redistributing it is not.

The rotational inertia row is what the extra work buys. A physical pendulum problem is a rotational problem. EK 7.5.A.2 says the period is derived from the application of Newton's second law in rotational form, so the chain runs torque, then rotational inertia, then angular acceleration. A simple pendulum problem never has to name an axis, because the point mass model throws that structure away.

The second-law row is why the pivot matters. Move the pivot on a physical pendulum and both $I$ and $d$ change, in different proportions, so the period changes. Worked example one moves the pivot of a rod and the period moves with it. On a simple pendulum, moving the pivot is just changing $\ell$ .

The case that separates them: a rod is not a bob on a string

A uniform rod of mass $0.60$ kg and length $0.90$ m hangs from a pivot at one end and swings through a small angle. What is its period?

There are three tempting answers and only one of them is right.

Model	Length used	Period
Simple pendulum, length equals the rod's length	$0.90$ m	$1.904$ s
Simple pendulum, length equals the distance to the center of mass	$0.45$ m	$1.346$ s
Physical pendulum, done properly	$\ell_{\text{eq}} = 0.60$ m	$1.555$ s

The two wrong answers bracket the right one, which is what makes this trap expensive: neither of them looks absurd. Using the full length is $22.5$ percent too slow. Using the distance to the center of mass is $13.4$ percent too fast. Both errors come from the same move, replacing a distributed rigid body with a point.

The ratios are clean enough to be worth remembering as a check. For a uniform rod pivoted at its end, $\ell_{\text{eq}} = 2L/3$ , so

\frac{T(\text{using } L)}{T_{\text{phys}}} = \sqrt{\frac{L}{2L/3}} = \sqrt{\frac{3}{2}} = 1.225, \qquad \frac{T(\text{using } L/2)}{T_{\text{phys}}} = \sqrt{\frac{L/2}{2L/3}} = \frac{\sqrt{3}}{2} = 0.866

The center of mass is the right place to put the gravitational force and the wrong place to put the mass. EK 2.6.A.1.iii says the gravitational force on a system can be considered to be exerted on the system's center of mass, and that is what puts $d$ into the restoring torque $\tau = -mgd\sin\theta$ . It does not license collapsing the body to a point for the purpose of computing $I$ , because rotational inertia depends on where every gram is, not on where the center of mass is. The two appearances of the body's geometry in $T_{\text{phys}} = 2\pi\sqrt{I/(mgd)}$ are doing different jobs, and only one of them is about the center of mass.

Worked example one runs the rod in full and then moves its pivot, which changes the period again without changing the rod.

How good is the point-mass model when the bob is real

Every laboratory simple pendulum is a physical pendulum, because bobs have size and strings have mass. The useful question is not whether the idealisation is exact but how much it costs, and the answer is: almost nothing, until the bob stops being small next to the string.

Take a uniform sphere of mass $m$ and radius $R$ on a string of negligible mass, with its center a distance $\ell$ from the pivot. The parallel axis theorem, EK 5.4.B.2 on both the AP Physics 1 and the C: Mechanics sheet as $I' = I_{\text{cm}} + Md^2$ , gives the rotational inertia about the pivot:

I = \tfrac{2}{5}mR^2 + m\ell^2

The center of mass is still at $d = \ell$ , so the period becomes $T = 2\pi\sqrt{\left(\tfrac{2}{5}R^2 + \ell^2\right)/(g\ell)}$ , and the equivalent length is $\ell_{\text{eq}} = \ell + \tfrac{2}{5}R^2/\ell$ . The correction term is the whole story: it grows as $R^2$ and shrinks as $1/\ell$ .

Bob radius $R$	String gives $\ell$	Simple model	Physical model	Error
$0.05$ m	$0.80$ m	$1.795$ s	$1.797$ s	$0.08$ percent
$0.20$ m	$0.40$ m	$1.269$ s	$1.331$ s	$4.9$ percent

At a ratio of $R/\ell = 1/16$ the correction is smaller than the reaction time of the student holding the stopwatch. At $R/\ell = 1/2$ it is bigger than any reasonable experimental uncertainty and a lab report that ignored it would be wrong.

So the models do not disagree about physics; they disagree about whether a term is worth carrying. Worked example two computes both cases and shows the term appearing and then dominating. This is also the shape of the answer an exam wants when it asks whether the simple pendulum model is appropriate: name the condition, $R \ll \ell$ , and say what happens when it fails.

The small-angle step is the same step in both, taken twice

Neither formula on this page is exact, and both fail for the same reason at the same place. AP Physics C: Mechanics sets it out at Topic 7.5 in three consecutive sub-statements, and the sequence is worth following because it is also the derivation.

EK 7.5.A.2.i. When displaced from equilibrium, the gravitational force exerted on a physical pendulum's center of mass provides a restoring torque. Derived equation: $\tau = -mgd\sin\theta$ . Note the sine. This line is exact.
EK 7.5.A.2.ii. For small amplitudes of motion, the small-angle approximation can be applied to the restoring torque. Derived equations: $\sin\theta \approx \theta$ and $\tau = -mgd\theta = I\alpha$ . This line is where the approximation enters, and it is the only place it enters.
EK 7.5.A.2.iii. The small-angle approximation and Newton's second law in rotational form yield a second-order differential equation that describes SHM: $\dfrac{d^2\theta}{dt^2} = -\omega^2\theta$ .

Reading the three in order tells you what the period formula is: the solution to step 3, valid to the extent that step 2 is valid. Compare $\tau = -mgd\theta = I\alpha$ with $\alpha = -(mgd/I)\theta$ and $\omega^2 = mgd/I$ , and $T = 2\pi/\omega$ gives $T_{\text{phys}} = 2\pi\sqrt{I/(mgd)}$ directly.

The simple pendulum inherits every word of this. Its statement in AP Physics 1 carries the restriction in the sentence itself: EK 7.2.A.1.ii gives the period of a simple pendulum displaced by a small angle. EK 7.1.A.2.iii says the motion of a pendulum with a small angular displacement can be modeled as simple harmonic motion because the restoring torque is proportional to the angular displacement, which is step 2 stated in words for the algebra-based course. See /glossary/small-angle-approximation for the approximation itself.

So the small angle is not a difference between the two pendulums. It is a condition they share, because they are the same system. The differences are entirely in how you get $I$ and $d$ . If you are looking for what separates simple from physical, the small angle is not it, and spring vs pendulum period is where the small-angle condition actually does distinguish one oscillator from another.

One case in Topic 7.5 sits outside this whole discussion. EK 7.5.A.4 describes a torsion pendulum, in which the restoring torque is proportional to the angular displacement of a rotating system, with the derived equation $I\alpha = -k\Delta\theta$ , and the CED's example is a horizontal disk suspended from a wire attached to its center of mass, undergoing rotational oscillations in the horizontal plane. There is no $g$ in it and no small-angle approximation, because the wire's restoring torque really is proportional to the angle rather than to its sine. A torsion pendulum is neither a simple nor a physical pendulum in the sense of this page.

Which courses ask, and where the confusion costs a mark

The scope difference is sharp, and worth knowing before you spend revision time on it.

AP Physics 1 requires the simple pendulum and never mentions a physical pendulum. The phrase does not appear in its Course and Exam Description.
AP Physics 2 mentions neither pendulum by name in its course framework, though its equation sheet prints the simple pendulum period.
AP Physics C: Mechanics is the only course with a topic devoted to this, Topic 7.5, Simple and Physical Pendulums, under learning objective 7.5.A, describe the properties of a physical pendulum. Suggested skills for the topic are 1.B, 2.A, 2.B, 3.A and 3.B.
AP Physics C: Electricity and Magnetism does not mention either, but its booklet carries the shared mechanics equation table, so both period formulas are printed on its sheet.

Notice that the AP Physics 1 treatment is not a simplification of the C: Mechanics one. It is the same content with the general case removed, which is why the C: Mechanics objective is phrased as describing the physical pendulum and lets the simple pendulum fall out of it at EK 7.5.A.3, rather than the other way round.

The errors this confusion produces, each a specific lost mark:

Using $T_p = 2\pi\sqrt{\ell/g}$ for a swinging rod, hoop or plate. The archetype. For a rod pivoted at one end it gives an answer $22.5$ percent too slow.
Using the distance to the center of mass as the simple pendulum length. The other half of the same error, and it is $13.4$ percent too fast for the same rod. The center of mass distance is $d$ in the physical pendulum formula, not a substitute for $\ell$ .
Taking $I$ about the center of mass instead of about the pivot. $T_{\text{phys}} = 2\pi\sqrt{I/(mgd)}$ needs the rotational inertia about the axis the body actually turns on. Use the parallel axis theorem to move a supplied center-of-mass value out to the pivot.
Cancelling the mass in $I/(mgd)$ without checking the geometry is fixed. The cancellation is legitimate for scaling a body uniformly, and illegitimate the moment you glue an extra mass onto one end, because then $I$ , $m$ and $d$ change by different factors.
Forgetting that a physical pendulum pivoted at its center of mass has no period at all. With $d = 0$ the restoring torque $\tau = -mgd\sin\theta$ vanishes, and the formula divides by zero. The body sits in equilibrium at every angle rather than oscillating.
Applying either formula at a large amplitude. Both rest on EK 7.5.A.2.ii, and neither survives a release angle where $\sin\theta$ has visibly parted company with $\theta$ .
Treating the torsion pendulum of EK 7.5.A.4 as a physical pendulum. Its restoring torque comes from a twisted wire, not from gravity, so $g$ does not appear and its period does not use $d$ at all.

For the CED framing, see Topic 7.5, Simple and Physical Pendulums and Topic 7.2 in C: Mechanics, or Topic 7.2 in AP Physics 1. The rotational inertia side is at Topic 5.4 and mass vs rotational inertia, and the full sheet is at the C: Mechanics formula page.

A swinging rod, three models, and only one right answer

A uniform rod of mass $0.60$ kg and length $0.90$ m is pivoted at one end and swings through a small angle in a vertical plane. Take the rod's rotational inertia about a perpendicular axis through one end as $I = \frac{1}{3}ML^2$ . Find its period as a physical pendulum, and find what a simple pendulum model would give using the full length and using the distance to the center of mass. Then move the pivot to a point $0.30$ m from the top end and find the new period. Use $g = 9.8\ \text{m/s}^2$ .

Rotational inertia about the end pivot. $I = \frac{1}{3}ML^2 = \frac{1}{3}(0.60)(0.90)^2 = \frac{1}{3}(0.60)(0.81) = 0.162\ \text{kg}\cdot\text{m}^2$ .
Distance from pivot to center of mass. The rod is uniform, so its center of mass is at the midpoint: $d = L/2 = 0.45\ \text{m}$ .
Period from EK 7.5.A.2. $T_{\text{phys}} = 2\pi\sqrt{\dfrac{I}{mgd}} = 2\pi\sqrt{\dfrac{0.162}{(0.60)(9.8)(0.45)}} = 2\pi\sqrt{\dfrac{0.162}{2.646}} = 2\pi\sqrt{0.061224} = 2\pi(0.24744) = 1.555\ \text{s}$ .
Equivalent simple length. $\ell_{\text{eq}} = \dfrac{I}{md} = \dfrac{0.162}{(0.60)(0.45)} = \dfrac{0.162}{0.27} = 0.600\ \text{m}$ , which is $\frac{2}{3}L$ as the symbolic result predicts. A simple pendulum $0.600$ m long has period $2\pi\sqrt{0.600/9.8} = 1.555\ \text{s}$ , matching.
Wrong model one, using the rod's full length. $2\pi\sqrt{0.90/9.8} = 2\pi\sqrt{0.091837} = 2\pi(0.30305) = 1.904\ \text{s}$ . That is $1.904/1.555 = 1.225$ times the correct value, or $22.5$ percent too slow, and $1.225 = \sqrt{3/2}$ exactly.
Wrong model two, using the center of mass distance. $2\pi\sqrt{0.45/9.8} = 2\pi\sqrt{0.045918} = 2\pi(0.21429) = 1.346\ \text{s}$ . That is $1.346/1.555 = 0.866$ times the correct value, or $13.4$ percent too fast, and $0.866 = \sqrt{3}/2$ exactly.
Move the pivot to $0.30$ m from the top. The center of mass is at $0.45$ m from the top, so now $d = 0.45 - 0.30 = 0.15\ \text{m}$ . Use the parallel axis theorem, EK 5.4.B.2, from the center-of-mass value $I_{\text{cm}} = \frac{1}{12}ML^2 = \frac{1}{12}(0.60)(0.81) = 0.0405\ \text{kg}\cdot\text{m}^2$ : $I' = I_{\text{cm}} + Md^2 = 0.0405 + (0.60)(0.15)^2 = 0.0405 + 0.0135 = 0.0540\ \text{kg}\cdot\text{m}^2$ .
New period. $T = 2\pi\sqrt{\dfrac{0.0540}{(0.60)(9.8)(0.15)}} = 2\pi\sqrt{\dfrac{0.0540}{0.882}} = 2\pi\sqrt{0.061224} = 1.555\ \text{s}$ . The same period as the end pivot, from a completely different $I$ and $d$ , because $\ell_{\text{eq}} = 0.0540/((0.60)(0.15)) = 0.600$ m again. Two different pivots on one rod can share a period, which no simple pendulum model would ever suggest.

The rod's period is $1.555$ s, matching a simple pendulum of length $\ell_{\text{eq}} = 0.600$ m, that is $\frac{2}{3}$ of the rod's length. Modelling it as a simple pendulum of length $0.90$ m gives $1.904$ s, $22.5$ percent too slow; using $0.45$ m gives $1.346$ s, $13.4$ percent too fast. Moving the pivot to $0.30$ m from the top gives $1.555$ s again, the same period from different values of $I$ and $d$ .

The collapse, checked numerically, then the term that breaks it

A bob of mass $0.25$ kg hangs on a string of negligible mass with its center $\ell = 0.80$ m below the pivot, and swings through a small angle. Find the period twice, once from the simple pendulum formula and once from the physical pendulum formula treating the bob as a point mass. Then repeat the physical pendulum calculation treating the bob as a uniform sphere of radius $R = 0.05$ m, and again for a fatter bob of radius $R = 0.20$ m at $\ell = 0.40$ m. Take the sphere's rotational inertia about its center as $\frac{2}{5}mR^2$ and use $g = 9.8\ \text{m/s}^2$ .

Simple pendulum formula. $T_p = 2\pi\sqrt{\ell/g} = 2\pi\sqrt{0.80/9.8} = 2\pi\sqrt{0.081633} = 2\pi(0.285714) = 1.7952\ \text{s}$ .
Physical pendulum formula with a point mass. $I = m\ell^2 = (0.25)(0.80)^2 = (0.25)(0.64) = 0.1600\ \text{kg}\cdot\text{m}^2$ , and $d = \ell = 0.80\ \text{m}$ . Then $T = 2\pi\sqrt{\dfrac{0.1600}{(0.25)(9.8)(0.80)}} = 2\pi\sqrt{\dfrac{0.1600}{1.960}} = 2\pi\sqrt{0.081633} = 1.7952\ \text{s}$ .
The two agree to every digit, which is what an identity looks like when you check it with numbers. Note where the agreement came from: $0.1600/1.960$ reduced to $0.80/9.8$ because the mass divided out and one power of the length divided out.
Small real bob, $R = 0.05$ m. Parallel axis theorem: $I = \frac{2}{5}mR^2 + m\ell^2 = \frac{2}{5}(0.25)(0.05)^2 + 0.1600 = (0.10)(0.0025) + 0.1600 = 0.00025 + 0.1600 = 0.16025\ \text{kg}\cdot\text{m}^2$ .
The center of mass has not moved, so $d = 0.80$ m still. $T = 2\pi\sqrt{\dfrac{0.16025}{1.960}} = 2\pi\sqrt{0.081760} = 2\pi(0.285937) = 1.7966\ \text{s}$ . Against $1.7952$ s, that is $0.08$ percent longer.
Fat bob, $R = 0.20$ m at $\ell = 0.40$ m. Now $m\ell^2 = (0.25)(0.16) = 0.0400$ and $\frac{2}{5}mR^2 = (0.10)(0.04) = 0.0040$ , so $I = 0.0440\ \text{kg}\cdot\text{m}^2$ and $d = 0.40$ m. $T = 2\pi\sqrt{\dfrac{0.0440}{(0.25)(9.8)(0.40)}} = 2\pi\sqrt{\dfrac{0.0440}{0.980}} = 2\pi\sqrt{0.044898} = 2\pi(0.211892) = 1.3314\ \text{s}$ .
Compare with the simple model there. $2\pi\sqrt{0.40/9.8} = 2\pi\sqrt{0.040816} = 2\pi(0.202031) = 1.2694\ \text{s}$ . The simple model is now $4.9$ percent fast, about sixty times the error of the first case.
Read the scaling. The correction to the equivalent length is $\frac{2}{5}R^2/\ell$ . Going from $R/\ell = 0.0625$ to $R/\ell = 0.50$ multiplied that ratio by $8$ , and the error grew by roughly $8^2$ , because the correction goes as $R^2$ .

For a point mass at $0.80$ m both formulas give $1.7952$ s, identically. Giving the bob a radius of $0.05$ m raises the period to $1.7966$ s, an error of $0.08$ percent in the simple model. A bob of radius $0.20$ m at $\ell = 0.40$ m swings at $1.3314$ s against the simple model's $1.2694$ s, an error of $4.9$ percent. The point-mass model is excellent while $R$ is small next to $\ell$ and fails as $R^2/\ell$ grows.

Using a physical pendulum to measure a rotational inertia

An irregular metal plate of mass $1.40$ kg is hung from a small hole and set swinging through a small angle. The pivot is $0.22$ m from the plate's center of mass, and the measured period is $1.35$ s. Find the plate's rotational inertia about the pivot, and about a parallel axis through its center of mass. Then find the equivalent simple pendulum length and compare it with the pivot distance. Use $g = 9.8\ \text{m/s}^2$ .

Rearrange the physical pendulum period for $I$ . From $T = 2\pi\sqrt{I/(mgd)}$ , square: $T^2 = 4\pi^2 I/(mgd)$ , so $I = \dfrac{mgd\,T^2}{4\pi^2}$ .
Numerator. $mgd = (1.40)(9.8)(0.22)$ . Take it in order: $(1.40)(9.8) = 13.72$ , then $(13.72)(0.22) = 3.0184\ \text{N}\cdot\text{m}$ . And $T^2 = (1.35)^2 = 1.8225\ \text{s}^2$ . The product is $(3.0184)(1.8225) = 5.5010$ .
Divide by $4\pi^2 = 39.478$ . $I = 5.5010/39.478 = 0.1393\ \text{kg}\cdot\text{m}^2$ about the pivot.
Move it to the center of mass with the parallel axis theorem. EK 5.4.B.2 reads $I' = I_{\text{cm}} + Md^2$ , so $I_{\text{cm}} = I' - Md^2 = 0.13934 - (1.40)(0.22)^2 = 0.13934 - (1.40)(0.0484) = 0.13934 - 0.06776 = 0.0716\ \text{kg}\cdot\text{m}^2$ .
Equivalent simple pendulum length. $\ell_{\text{eq}} = \dfrac{I}{md} = \dfrac{0.1393}{(1.40)(0.22)} = \dfrac{0.1393}{0.308} = 0.452\ \text{m}$ .
Compare. The pivot is $0.22$ m from the center of mass, and the plate swings like a simple pendulum more than twice that long. A simple pendulum of length $0.22$ m would have period $2\pi\sqrt{0.22/9.8} = 0.941\ \text{s}$ , against the measured $1.35$ s. Modelling the plate as a point at its center of mass would have been $30$ percent fast.
Why this measurement is worth making. Nothing here required knowing the plate's shape, and the plate is irregular, so no formula for $I$ was available. A stopwatch, a ruler and a balance produced a rotational inertia. This is the reverse of the usual exam direction, where $I$ is supplied and the period is asked for, and it is what the CED means at skill 3.A by creating experimental procedures appropriate for a given scientific question.

The plate's rotational inertia is $0.1393\ \text{kg}\cdot\text{m}^2$ about the pivot and $0.0716\ \text{kg}\cdot\text{m}^2$ about a parallel axis through its center of mass. Its equivalent simple pendulum length is $0.452$ m, more than twice the $0.22$ m pivot distance, so treating the plate as a point mass at its center of mass would have predicted $0.941$ s against the measured $1.35$ s.

Frequently asked questions

What is the difference between a simple pendulum and a physical pendulum?

A physical pendulum is any rigid body that oscillates about a fixed axis, and its period for small amplitudes is 2 pi times the square root of the rotational inertia about the pivot divided by the product of the mass, g, and the distance from the pivot to the center of mass. A simple pendulum is the special case in which the whole hanging object can be treated as a point mass at one distance from the pivot, and its period is 2 pi times the square root of that distance divided by g. The AP Physics C: Mechanics CED states the relationship at essential knowledge 7.5.A.3: a simple pendulum is a special case of physical pendulums in which the hanging object can be modeled as a point mass at a distance from the pivot point.

How does the physical pendulum formula reduce to the simple pendulum formula?

Put a point mass into it. A point mass at distance L from the axis has rotational inertia m times L squared, and the center of mass of a single point is that point, so the distance from the pivot to the center of mass is also L. Substituting both into 2 pi times the square root of I over m g d gives 2 pi times the square root of m L squared over m g L, and the mass cancels along with one power of L, leaving 2 pi times the square root of L over g. That is the simple pendulum period exactly, not approximately. Both conditions have to hold: every gram at the same distance, and the center of mass at that same distance.

Does the mass of a physical pendulum affect its period?

Not if you change the mass without changing the shape. Rotational inertia is proportional to the total mass for a body of fixed geometry and uniform density, so the mass in the numerator of I over m g d cancels the mass in the denominator, exactly as it does for a simple pendulum. Doubling every gram of a swinging rod leaves its period unchanged. What does change the period is redistributing mass, because moving mass changes the rotational inertia and the center of mass position by different factors. So scaling the mass is free and rearranging it is not.

Can you use the simple pendulum formula for a swinging rod?

No, and the error is large enough to cost the question. A uniform rod pivoted at one end swings like a simple pendulum whose length is two thirds of the rod's length, not the full length and not half of it. For a 0.90 m rod the correct period is 1.555 s; using the full 0.90 m gives 1.904 s, which is 22.5 percent too slow, and using the 0.45 m distance to the center of mass gives 1.346 s, which is 13.4 percent too fast. The right method is the physical pendulum formula with the rotational inertia taken about the pivot axis.

Is the physical pendulum on the AP Physics 1 exam?

No. The phrase physical pendulum does not appear in the AP Physics 1 Course and Exam Description, and its equation sheet does not print the physical pendulum period. Only AP Physics C: Mechanics requires it, at Topic 7.5, Simple and Physical Pendulums, under learning objective 7.5.A, describe the properties of a physical pendulum. Both Physics C booklets print the formula, because the Electricity and Magnetism booklet carries the same mechanics equation table as the Mechanics booklet. AP Physics 1 and AP Physics 2 print the simple pendulum period and stop there.

What is the equivalent length of a physical pendulum?

It is the rotational inertia about the pivot divided by the product of the mass and the pivot to center of mass distance. A simple pendulum of that length has exactly the same period as the physical pendulum, because substituting it into the simple pendulum formula reproduces the physical pendulum formula. For a uniform rod pivoted at one end it works out to two thirds of the rod's length. The AP CED does not give this quantity a name, but carrying it turns any comparison between the two models into a single number, and it makes clear that the question is never whether a rigid body behaves like a simple pendulum but which length to use.

Why does a physical pendulum pivoted at its center of mass not swing?

Because the restoring torque vanishes. Essential knowledge 7.5.A.2.i gives that torque as minus m g d times the sine of the angle, where d is the distance from the pivot to the center of mass. Set d to zero and the torque is zero at every angle, so there is nothing to return the body to any particular orientation and it simply stays where it is put. The period formula reflects this by dividing by zero. This is also why the position of the pivot relative to the center of mass, not just the size of the body, controls how fast a physical pendulum swings.