AP Physics C: Mechanics · Topic 7.5

Topic 7.5: Simple and Physical Pendulums

Unit 7: Oscillations10-15% of the multiple-choice section

A physical pendulum is a rigid body swinging about a fixed axis. Its period is two pi times the square root of the rotational inertia divided by mass times g times the distance to the centre of mass. A simple pendulum is the point-mass special case. No AP Physics 1 topic covers this.

AP Physics: Unit 7 (topics 7.5 Simple and Physical Pendulums). Topic 7.5 of the current AP Physics C: Mechanics course and exam description, inside Unit 7, which is weighted 10 to 15% of the multiple-choice section at about 12 to 17 class periods. This topic has NO AP Physics 1 counterpart: it is one of only four C: Mechanics topics sharing no title with an algebra-based topic, alongside 1.1, 1.5 and 2.9, and it is why C: Mechanics Unit 7 has five topics where AP Physics 1 Unit 7 has four. One learning objective, 7.5.A (describe the properties of a physical pendulum), and four numbered essential-knowledge statements. 7.5.A.1 defines a physical pendulum as a rigid body that undergoes oscillation about a fixed axis. 7.5.A.2 says that for small amplitudes its period is derived from Newton's second law in rotational form, with the Relevant equation T_phys = 2*pi*sqrt(I/(mgd)); 7.5.A.2.i gives the Derived equation tau = -mgd sin(theta) and says the gravitational force on the centre of mass provides the restoring torque; 7.5.A.2.ii gives the Derived equations sin(theta) approx theta and tau = -mgd theta = I alpha; 7.5.A.2.iii gives the second-order differential equation d2theta/dt2 = -omega^2 theta. 7.5.A.3 makes a simple pendulum a special case in which the hanging object can be modeled as a point mass at a distance l from the pivot, with the Relevant equation T_p = 2*pi*sqrt(l/g). 7.5.A.4 defines a torsion pendulum as a case of SHM where the restoring torque is proportional to the angular displacement of a rotating system, giving the horizontal disk on a wire as the example, with the Derived equation I alpha = -k delta-theta. Suggested skills 1.B, 2.A, 2.B, 3.A, 3.B: five, more than any other Unit 7 topic, and 1.B and 3.A appear nowhere else in the unit. Topic 7.5 prints no boundary statement; Unit 7 prints exactly one, under Topic 7.3. Equation-sheet facts verified against both Tables of Information rendered as images: the C: Mechanics sheet prints T_p, T_phys and T_s, the torque cross product, the parallel-axis theorem, I = integral r^2 dm, the point-mass sum and Newton's second law in rotational form; the AP Physics 1 sheet prints T_p and T_s but has no physical-pendulum entry and no integral form for rotational inertia. Not printed on the C sheet: the restoring torque, the small-angle approximation, the angular differential equation, I alpha = -k delta-theta, any torsion pendulum period (which appears nowhere in the CED), and any table of rotational inertias for named shapes. Neither of the CED's two Unit 7 sample multiple-choice questions aligns to 7.5.A. Sample instructional activity 5 sits on this topic: groups use a pendulum to determine the acceleration of gravity in the classroom, and the winning group is the one whose procedure includes the most error-reducing components, with timing multiple periods, linearizing data and very precisely finding the centre of mass of the bob given as the CED's examples. The minimum-period result for a rod pivoted at L/sqrt(12) from its centre is flagged on the page as a consequence of 7.5.A.2 rather than a CED statement.

What Topic 7.5 requires, and why it stands alone

Topic 7.5 is the fifth topic of Unit 7, and it is the reason AP Physics C: Mechanics has five topics in this unit where AP Physics 1 has four. There is no AP Physics 1 counterpart to this topic at all. It is one of only four topics in the whole of C: Mechanics that share no title with an algebra-based topic, alongside 1.1 Scalars and Vectors, 1.5 Motion in Two or Three Dimensions, and 2.9 Resistive Forces.

Learning objective 7.5.A: describe the properties of a physical pendulum.

Statement	What it says
7.5.A.1	A physical pendulum is a rigid body that undergoes oscillation about a fixed axis
7.5.A.2	For small amplitudes of motion, the period of a physical pendulum is derived from the application of Newton's second law in rotational form. Relevant equation: $T_{\text{phys}} = 2\pi\sqrt{\dfrac{I}{mgd}}$
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: $\tau = -mgd\sin\theta$
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 $\tau = -mgd\theta = I\alpha$
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: $\dfrac{d^2\theta}{dt^2} = -\omega^2\theta$
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. Relevant equation: $T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$
7.5.A.4	A torsion pendulum is a case of SHM where the restoring torque is proportional to the angular displacement of a rotating system. For example, a horizontal disk that is suspended from a wire attached to its center of mass may undergo rotational oscillations about the wire in the horizontal plane. Derived equation: $I\alpha = -k\Delta\theta$

The suggested skills are 1.B (create quantitative graphs with appropriate scales and units, including plotting data), 2.A (derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway), 2.B (calculate or estimate an unknown quantity with units from known quantities), 3.A (create experimental procedures that are appropriate for a given scientific question) and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim).

Five suggested skills, more than any other topic in Unit 7, which have four each. Two of them appear nowhere else in the unit: 1.B and 3.A. Those are the graph-with-real-data skill and the design-an-experiment skill, and together they say what this topic is for.

Topic 7.5 prints no boundary statement. Unit 7 prints exactly one across all five topics and it sits under Topic 7.3, where it limits how far the differential equation has to be taken.

Notice also the ordering across the unit. Topic 7.2 stated the simple-pendulum period, in essential knowledge 7.2.A.1.ii, without justifying it. This topic derives it, and derives it as a special case of something more general. The loop closes here.

The derivation, which the CED prints in pieces

Statements 7.5.A.2 through 7.5.A.2.iii are one derivation split into four numbered parts. Run end to end it is five lines, and skill 2.A means the exam can ask for exactly this.

Step 1, the geometry. A rigid body of mass $m$ hangs from a fixed axis. Its centre of mass sits a distance $d$ from that axis. Displace it by an angle $\theta$ from the vertical.

Step 2, the restoring torque. Statement 7.5.A.2.i: when displaced from equilibrium, the gravitational force exerted on the pendulum's centre of mass provides a restoring torque. The gravitational force $mg$ acts downward at the centre of mass, and the sheet's $\vec{\tau} = \vec{r} \times \vec{F}$ gives its magnitude as $mgd\sin\theta$ , with $d\sin\theta$ the perpendicular lever arm. The sign is negative because the torque acts to reduce $\theta$ :

\tau = -mgd\sin\theta

Step 3, the small-angle approximation. Statement 7.5.A.2.ii: for small amplitudes the approximation $\sin\theta \approx \theta$ can be applied, with $\theta$ in radians. Combine that with Newton's second law in rotational form, which the sheet prints as $\alpha_{\text{sys}} = \tau_{\text{net}}/I_{\text{sys}}$ :

\tau = -mgd\theta = I\alpha

Step 4, the differential equation. Statement 7.5.A.2.iii: since $\alpha = d^2\theta/dt^2$ , that rearranges to

\frac{d^2\theta}{dt^2} = -\frac{mgd}{I}\theta

which has the shape of $d^2\theta/dt^2 = -\omega^2\theta$ , the angular twin of Topic 7.3's Derived equation. So the motion is simple harmonic, and $\omega^2 = mgd/I$ .

Step 5, the period. Read the coefficient and convert with the printed $T = 2\pi/\omega$ :

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

which is 7.5.A.2's relevant equation, and it is printed on the equation sheet.

The important asymmetry: the destination is printed and the route is not. Steps 2, 3 and 4 are all labelled Derived equations by the framework and none of them appears on the sheet. What is printed is the torque cross product, Newton's second law in rotational form, the rotational inertia definitions, and the final period. That is the exact shape of a Mathematical Routines free-response question, which is worth 10 points at a suggested 20 to 25 minutes and lists skill 2.A among its skills.

One caution on symbols, because two distances get confused here. In $T_{\text{phys}}$ , the letter $d$ is the distance from the axis to the centre of mass, and the equation sheet's variable list defines $d$ as "distance". It is not the length of the body, not the distance to the far end, and not the radius of anything. In the parallel-axis theorem $I' = I_{\text{cm}} + Md^2$ , printed a few lines above on the same sheet, $d$ is the same distance, which is convenient but easy to take for granted.

Where the small-angle approximation lives, and what it costs

A pendulum is not simple harmonic motion. A pendulum at small amplitude is approximately simple harmonic motion. Being exact about which step introduces the approximation is worth points, because a question can ask you to name it.

The approximation enters at step 3 and nowhere else. Step 2's $\tau = -mgd\sin\theta$ is exact for a rigid body in a uniform gravitational field, at any angle. Replacing $\sin\theta$ with $\theta$ is where the physics is bent, and statement 7.5.A.2.ii is where the CED says so. Everything downstream inherits the approximation: the differential equation, the period, the claim that the period is independent of amplitude, and the energy expression.

Why this matters conceptually: with the exact torque, the restoring torque is not proportional to the angular displacement, so statement 7.1.A.2's condition for SHM fails and no single angular frequency exists. Only after the substitution is the torque proportional to the displacement.

How big is the error? $\sin\theta \approx \theta$ overstates the sine, and here is by how much:

Amplitude	$\theta$ in radians	$\sin\theta$	$\theta$ exceeds $\sin\theta$ by
5 degrees	0.0873	0.0872	0.13%
10 degrees	0.1745	0.1736	0.51%
15 degrees	0.2618	0.2588	1.15%
20 degrees	0.3491	0.3420	2.1%
30 degrees	0.5236	0.5000	4.7%

And the direction of the resulting error in the period is fixed, which is a claim you can defend without any correction formula. Since $\sin\theta < \theta$ for every $\theta > 0$ , the real restoring torque is always weaker than the linear model's. A weaker restoring torque means a slower return, so a real pendulum's period is longer than $T_{\text{phys}}$ gives, and the gap grows with amplitude. A real pendulum released from a large angle runs slow.

The CED gives no correction series and does not ask for one, so do not quote one. What it does ask for, through skill 3.B, is the qualitative claim with its justification, and the argument above is that justification.

Three consequences to carry:

Radians, always. $\sin\theta \approx \theta$ is false in degrees, and it is false by a factor of $180/\pi$ . Every $\theta$ in this topic is a radian measure, and a calculator in degree mode will produce numbers that look reasonable and are not.
The amplitude-independence of 7.3.A.5 is conditional here. It is a statement about SHM, and a pendulum is SHM only to this approximation.
A lab that swings its pendulum too far biases $g$ low. A measured period longer than the model's, fed through $g = 4\pi^2\ell/T^2$ , returns a value of $g$ that is too small.

Rotational inertia is the whole difference

Compare the two printed periods and one symbol is the difference:

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

A simple pendulum's period is set by a length. A physical pendulum's is set by how the mass is distributed, through the rotational inertia $I$ about the pivot. Two bodies of the same mass whose centres of mass are the same distance from the pivot can have different periods, because they can have different $I$ .

This is why the topic depends on Unit 5 and Unit 6 in a way nothing in AP Physics 1's Unit 7 does. To use $T_{\text{phys}}$ you must first produce an $I$ about the pivot, and the sheet gives three routes to one:

Route	Printed	When to use it
$I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$	yes	discrete masses, or adding the parts of a compound body
$I = \int r^2\, dm$	yes, and absent from the AP Physics 1 sheet	a continuous body, especially one of varying density
$I' = I_{\text{cm}} + Md^2$	yes	a body whose value about its own centre you already have

No table of rotational inertias for named shapes appears anywhere in the Table of Information. There is no $\frac{1}{3}ML^2$ printed for a rod about its end and no $\frac{1}{2}MR^2$ printed for a disk. Exam questions that need one tend to state it in the stem, as the framework's own sample multiple-choice question 14 does for a uniform disk in Unit 6. The other route is to derive it, which is what $I = \int r^2\, dm$ is for. Topic 5.4 covers producing one, and Topic 6.1 covers what a rotating body does with it.

For a compound body, both of the first and third routes get used together: find each part's $I$ about its own centre of mass, shift each to the pivot with the parallel-axis theorem, then add. Worked example 1 does exactly that for a rod carrying a disk, and shows how much the disk's own $\frac{1}{2}MR^2$ term is worth compared with its $Md^2$ term.

One cancellation worth knowing before you compute anything. The mass drops out of a physical pendulum's period whenever the body's shape is fixed. Both $I$ and $md$ are proportional to the total mass, so their ratio is not. A heavier rod of the same dimensions swings at exactly the same rate. That is the same reason the bob's mass does not matter for a simple pendulum, stated for an extended body, and it is why worked example 1's answer for the equivalent length contains no mass at all.

The simple pendulum as a special case, and the equivalent length

Statement 7.5.A.3 does something the algebra-based course never does: it makes the simple pendulum a derived special case rather than an independent fact. A simple pendulum is a physical pendulum in which the hanging object can be modeled as a point mass at a distance $\ell$ from the pivot point.

Check that it collapses correctly. A point mass $m$ at distance $\ell$ has $I = m\ell^2$ by the printed sum, and its centre of mass is the point itself, so $d = \ell$ . Substitute:

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

The general formula reduces to the printed simple-pendulum result, which is 7.5.A.3's relevant equation and also 7.2.A.1.ii. Two masses cancel and one power of $\ell$ survives.

Running that backwards gives the single most useful idea in this topic. Define the equivalent length of a physical pendulum as

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

Then $T_{\text{phys}} = 2\pi\sqrt{\ell_{\text{eq}}/g}$ exactly, so any physical pendulum keeps time like a simple pendulum of that length. It is a single number that summarises the whole body, and it turns every physical-pendulum question into a simple-pendulum question once you have it.

The equivalent length is not the distance to the centre of mass, and it is not the length of the body. For a uniform rod pivoted at one end, $I = \frac{1}{3}mL^2$ and $d = L/2$ , so

\ell_{\text{eq}} = \frac{mL^2/3}{mL/2} = \frac{2L}{3}

Two thirds of the rod's length, which is more than $d = L/2$ and less than $L$ . That inequality is general: $\ell_{\text{eq}} > d$ always, because the parallel-axis theorem gives $I = I_{\text{cm}} + md^2$ and therefore $\ell_{\text{eq}} = d + I_{\text{cm}}/(md)$ , which exceeds $d$ by a positive amount for any body with size. So a physical pendulum always swings more slowly than a point mass placed at its centre of mass, and more quickly than a point mass placed at its far end if the far end is beyond the equivalent length.

Using $T_p = 2\pi\sqrt{\ell/g}$ for an extended body, with $\ell$ taken as some convenient length on the diagram, is the most available error in Topic 7.5 and it is what 7.5.A.3's careful wording is guarding against. The permission is to model the hanging object as a point mass, and a rod, a hoop or a plate cannot be.

The minimum period, and where a rod swings fastest

This section is a consequence of 7.5.A.2 rather than a statement in the CED, and it is flagged as such. It is worth knowing because it is counterintuitive, it makes a good experimental-design question, and it explains data that otherwise looks like an error.

Take a uniform rod of length $L$ and drill a pivot hole a distance $h$ from its centre. The parallel-axis theorem gives $I = \frac{1}{12}mL^2 + mh^2$ , and the centre of mass is at $d = h$ , so

\ell_{\text{eq}} = \frac{I}{mh} = \frac{L^2/12 + h^2}{h} = \frac{L^2}{12h} + h

Now watch both ends of the range. As $h$ shrinks toward zero, the first term blows up and the period grows without limit, which makes sense: pivot exactly at the centre of mass and there is no restoring torque at all, so it never swings back. As $h$ grows toward $L/2$ , the second term grows and the period grows again.

Somewhere in between, the period is shortest. The two terms are equal when $h^2 = L^2/12$ , that is $h = L/\sqrt{12} = 0.2887L$ , and there $\ell_{\text{eq}} = 2h$ , its smallest value. For a 1.000 m rod that is a pivot 0.289 m from the centre, an equivalent length of 0.577 m, and a period of

T_{\min} = 2\pi\sqrt{\frac{0.5774}{9.8}} = 1.525\ \mathrm{s}

Three things follow that are worth having:

Moving the pivot closer to the centre of mass makes a physical pendulum slower, not faster. That is the opposite of the simple-pendulum intuition that shorter means quicker, and it is the standard trap here. The distance $d$ appears in the denominator of $T_{\text{phys}}$ .
Near the minimum, the period barely changes. A rod pivoted at 0.20 m, 0.30 m or 0.40 m from its centre gives periods that differ by only about 3%, even though the pivot moved by a factor of two. Data that looks flat over that range is correct, not sloppy.
Two different pivot positions give the same period. Setting $\ell_{\text{eq}}$ equal for two values of $h$ gives $h_1 h_2 = L^2/12$ , so pivots at 0.10 m and at 0.833 m from the centre would match, if the rod were long enough to have both.

Worked example 2 uses the same family of pendulums to extract $g$ , and the flatness above is exactly why a single measurement is a poor way to do it and a graph is a good one.

The torsion pendulum, where g never appears

Statement 7.5.A.4 closes the unit with a case that belongs to no other AP physics course. Quoted whole:

"A torsion pendulum is a case of SHM where the restoring torque is proportional to the angular displacement of a rotating system. For example, a horizontal disk that is suspended from a wire attached to its center of mass may undergo rotational oscillations about the wire in the horizontal plane."

Its Derived equation is

I\alpha = -k\Delta\theta

Read the statement's first sentence as a definition and it is 7.1.A.2 restated for rotation: the restoring quantity is proportional to the displacement, so the motion is simple harmonic exactly, with no approximation anywhere. There is no $\sin\theta$ to linearise, because the wire's restoring torque really is proportional to the twist. In that sense a torsion pendulum is a cleaner oscillator than the gravitational kind.

The period follows from the coefficient in the usual way, and it is the one period in this unit the CED never prints:

\frac{d^2\theta}{dt^2} = -\frac{k}{I}\theta \quad \Longrightarrow \quad \omega = \sqrt{\frac{k}{I}} \quad \Longrightarrow \quad T = 2\pi\sqrt{\frac{I}{k}}

There is nothing to quote and nothing on the sheet, so reading $\omega^2 = k/I$ off the coefficient is the only route. That makes this the clearest demonstration in the unit of why the method matters more than the formulas.

Now look at what is missing from that period. No $g$ . No length. No mass except through $I$ . The consequences:

A torsion pendulum keeps the same period on the Moon, in orbit, or anywhere else. A physical pendulum's period goes as $1/\sqrt{g}$ and lengthens wherever the gravitational field is weaker.
It therefore measures inertia, not weight. That is one answer to the unit's own essential question about how an astronaut can be "weighed" in space, with the scare quotes the CED itself uses: an oscillator of this kind determines a rotational inertia or a mass, and does it in free fall where a scale reads zero. Worked example 3 carries that out as a two-stage calibration.
A spring oscillator does the same job in translation, since $T_s = 2\pi\sqrt{m/k}$ has no $g$ in it either. The two are the same idea in linear and rotational form.

A notation warning, because it is the sharpest in the unit. The symbol $k$ in 7.5.A.4 is a torsion constant with units of newton metres per radian, a torque per unit angle. The same letter on the same equation sheet is defined in the variable list as "spring constant", in newtons per metre. They are different quantities with different dimensions, and the only defence is to check units before substituting. The unit check for the period is worth doing once: $\mathrm{kg \cdot m^2}$ divided by $\mathrm{N \cdot m/rad}$ is $\mathrm{s^2}$ , since a newton is $\mathrm{kg \cdot m/s^2}$ and the radian is dimensionless.

Skill 3.A, and the lab this topic is built for

Skill 3.A, create experimental procedures that are appropriate for a given scientific question, is listed for Topic 7.5 and for no other topic in Unit 7. Skill 1.B, create quantitative graphs with appropriate scales and units including plotting data, is the same. Between them they point at the Experimental Design and Analysis free-response question, which is worth 10 points at a suggested 25 to 30 minutes and lists skills 1.B, 2.B, 2.D and 3.A.

The CED's fifth sample instructional activity is a Topic 7.5 desktop experiment, and it names the criteria it wants: divide students into groups, have each group use a pendulum to determine the acceleration of gravity in the classroom, and let the winning group be the one whose procedure includes the most components for reducing error, with three examples given in the CED itself, timing multiple periods, linearizing the data, and very precisely finding the centre of mass of the bob.

Those three are worth taking apart, because each maps to a specific sentence you can write on an exam.

Timing multiple periods. Reaction-time uncertainty is a fixed number of seconds per timing, so timing 25 cycles and dividing cuts its effect on $T$ by 25. Because $g$ goes as $1/T^2$ , a fractional error in $T$ doubles into $g$ , which makes this the highest-value step available.

Linearizing the data. Square the printed formula so the relationship becomes a straight line, then take a slope. For a simple pendulum, $T^2 = (4\pi^2/g)\ell$ , so $T^2$ against $\ell$ has slope $4\pi^2/g$ . For a physical pendulum the algebra is richer and the graph is better, because it yields two independent routes to $g$ from one line. Worked example 2 does that version. Linearization is a slope, and a slope uses every data point, which is why it beats averaging single-point calculations.

Finding the centre of mass. Both printed formulas depend on a distance to the centre of mass: $\ell$ for a simple pendulum, $d$ for a physical one. An error there propagates directly into $g$ . For a compound body the centre of mass has to be computed rather than eyeballed, using the sheet's $\vec{x}_{\text{cm}} = \sum m_i\vec{x}_i / \sum m_i$ or its integral form.

A fourth component the CED does not list but which the small-angle section above justifies: keep the amplitude small and say so, because a large swing lengthens the measured period and biases $g$ low. Naming the assumption is itself a procedure component.

Because the accepted value of $g$ is itself worth care on this exam, compare a lab result against the value the is g 9.8 or 10 guide explains rather than against a remembered one. The Table of Information for this course prints $g = 9.8\ \mathrm{m/s^2}$ .

What the sheet prints, how 7.5 is tested, and the sibling page that does not exist

Checked against the Table of Information in the AP Physics C: Mechanics course and exam description and the corresponding appendix of the AP Physics 1 course and exam description, both rendered as images.

Relation	C: Mechanics sheet	AP Physics 1 sheet
$T_p = 2\pi\sqrt{\ell/g}$	yes	yes
$T_{\text{phys}} = 2\pi\sqrt{I/(mgd)}$	yes	no
$T_s = 2\pi\sqrt{m/k}$	yes	yes
$\vec{\tau} = \vec{r} \times \vec{F}$	yes	no, it prints $\tau = r_\perp F = rF\sin\theta$
$I' = I_{\text{cm}} + Md^2$	yes	yes
$I = \int r^2\, dm$	yes	no
$\alpha_{\text{sys}} = \tau_{\text{net}}/I_{\text{sys}}$	yes	yes
$\tau = -mgd\sin\theta$	no, it is 7.5.A.2.i	not in that course
$\sin\theta \approx \theta$	no, it is 7.5.A.2.ii	not in that course
$d^2\theta/dt^2 = -\omega^2\theta$	no, it is 7.5.A.2.iii	not in that course
$I\alpha = -k\Delta\theta$	no, it is 7.5.A.4	not in that course
Torsion pendulum period	no, nowhere in the CED	not in that course
Table of rotational inertias for named shapes	no	no

So both pendulum periods are printed here and only the simple one is printed for the algebra-based course, which follows from that course not having this topic. Everything between the two printed periods, the restoring torque, the approximation and the differential equation, is yours to produce.

On assessment: Unit 7 is weighted 10 to 15% of the multiple-choice section, at about 12 to 17 class periods. The exam is 3 hours, with 42 multiple-choice questions in 85 minutes and 4 free-response questions in 95 minutes, each half the score, and a four-function, scientific, or graphing calculator is allowed on both sections. On the multiple-choice section skill 2.A carries 25 to 30%, 2.B carries 20 to 25% and 3.B carries 15 to 25%; skills 1.B and 3.A are marked not applicable there, since Science Practice 1 is not assessed on that section and 3.A is a free-response skill. On the free-response section Practice 1 carries 20 to 35%, Practice 2 carries 40 to 45% and Practice 3 carries 30 to 35%. Required course content can be assessed with any skill.

Neither of the CED's two Unit 7 sample multiple-choice questions aligns to 7.5.A: question 5 aligns to 7.2.A and question 12 to 7.3.A. None of the four sample free-response questions aligns to Unit 7. The Progress Check for Unit 7 runs about 18 multiple-choice questions and 4 free-response questions, one of each type.

There is no AP Physics 1 page to send you to from here, because AP Physics 1 has no Topic 7.5. Its Unit 7 has four topics and stops before pendulums as rigid bodies, and its equation sheet has no physical-pendulum entry. If you arrived here from the algebra-based course, the page that matches your framework is the AP Physics 1 Unit 7 hub, and the parts of this page that apply to you are the simple pendulum and the small-angle approximation. The rotational inertia, the physical pendulum and the torsion pendulum are not on your exam.

AP Physics C: Mechanics is a calculus-based, college-level course, equivalent to a first course in an introductory college sequence in calculus-based physics. Its stated prerequisite is that students have taken or are concurrently taking calculus.

Where to go next: Topic 7.2 states the simple-pendulum period this topic derives, Topic 7.4 works out a physical pendulum's energy, Topic 5.4 is where the rotational inertia comes from and Topic 5.6 is the law the derivation rests on. The Unit 7 hub lists all five topics, and the torque guide owns the lever-arm routine.

A compound pendulum: a rod carrying a disk

A uniform rod of mass $M_r = 0.30$ kg and length $L = 0.50$ m is pivoted at one end. A uniform disk of mass $M_d = 0.50$ kg and radius $R = 0.080$ m is rigidly attached with its centre at the rod's far end, in the plane of the swing. The rod's rotational inertia about its end is $\frac{1}{3}M_r L^2$ and the disk's about its own centre is $\frac{1}{2}M_d R^2$ . Take $g = 9.8\ \mathrm{m/s^2}$ . Find (a) the rotational inertia about the pivot, (b) the distance to the centre of mass, (c) the period, (d) the equivalent simple-pendulum length, and (e) how wrong two tempting shortcuts would be.

Declare the convention: distances are measured from the pivot along the rod, and the small-angle model of 7.5.A.2 is assumed, so the amplitude is a few degrees.
(a) Build $I$ about the pivot part by part, using the printed sum $I_{\text{tot}} = \sum I_i$ with the printed parallel-axis theorem for the disk.
Rod: $I_r = \frac{1}{3}M_r L^2 = \frac{1}{3}(0.30)(0.50)^2 = \frac{1}{3}(0.30)(0.2500) = 0.02500\ \mathrm{kg \cdot m^2}$ .
Disk, about its own centre: $\frac{1}{2}M_d R^2 = \frac{1}{2}(0.50)(0.080)^2 = \frac{1}{2}(0.50)(6.400 \times 10^{-3}) = 1.600 \times 10^{-3}\ \mathrm{kg \cdot m^2}$ . Its centre is $L = 0.50$ m from the pivot, so $I' = I_{\text{cm}} + M_d L^2 = 1.600 \times 10^{-3} + (0.50)(0.2500) = 0.001600 + 0.1250 = 0.1266\ \mathrm{kg \cdot m^2}$ .
Total: $I = 0.02500 + 0.1266 = 0.1516\ \mathrm{kg \cdot m^2}$ .
(b) The total mass is $M = 0.30 + 0.50 = 0.80$ kg. Using the printed centre-of-mass sum with the rod's centre at $L/2$ : $d = \dfrac{(0.30)(0.250) + (0.50)(0.500)}{0.80} = \dfrac{0.07500 + 0.2500}{0.80} = \dfrac{0.3250}{0.80} = 0.40625\ \mathrm{m}$ .
(c) $Mgd = (0.80)(9.8)(0.40625) = 3.185\ \mathrm{N \cdot m}$ , so the printed $T_{\text{phys}} = 2\pi\sqrt{I/(Mgd)} = 2\pi\sqrt{0.1516/3.185} = 2\pi\sqrt{0.0475981} = 2\pi(0.218170) = 1.3708 \approx 1.37\ \mathrm{s}$ .
(d) $\ell_{\text{eq}} = \dfrac{I}{Md} = \dfrac{0.1516}{(0.80)(0.40625)} = \dfrac{0.1516}{0.3250} = 0.46646 \approx 0.466\ \mathrm{m}$ .
Check that it reproduces the period: $2\pi\sqrt{0.46646/9.8} = 2\pi\sqrt{0.0475982} = 1.3708\ \mathrm{s}$ , matching part (c) to every digit. Note $\ell_{\text{eq}} = 0.466$ m sits between $d = 0.406$ m and $L = 0.500$ m, as the general inequality $\ell_{\text{eq}} > d$ requires.
(e) Shortcut one, treating the whole thing as a simple pendulum of length 0.50 m because that is where the disk is: $2\pi\sqrt{0.500/9.8} = 2\pi\sqrt{0.0510204} = 1.4192 \approx 1.42\ \mathrm{s}$ . That is 3.5% too long. The rod's mass is spread nearer the pivot than the disk, which pulls the equivalent length in.
Shortcut two, keeping the compound treatment but modelling the disk as a point mass and dropping its $\frac{1}{2}M_d R^2$ : $I = 0.02500 + 0.1250 = 0.1500\ \mathrm{kg \cdot m^2}$ , giving $2\pi\sqrt{0.1500/3.185} = 1.3635 \approx 1.36\ \mathrm{s}$ , which is 0.53% too short.
The comparison between the two shortcuts is the lesson. The disk's own rotational inertia contributes $1.600 \times 10^{-3}$ against its $Md^2$ contribution of $0.1250$ , a ratio of about 1.3%, because the radius is small compared with the distance to the pivot. The first shortcut costs about six times as much as the second. Decide which terms matter by comparing them, not by habit.
Finally, note that no mass survives in the equivalent length if the shape is held fixed: doubling both $M_r$ and $M_d$ doubles $I$ and doubles $Md$ , so $\ell_{\text{eq}}$ and the period are unchanged. Changing the ratio of the two masses does change them.

(a) $I = 0.1516\ \mathrm{kg \cdot m^2}$ , of which the disk contributes 0.1266. (b) $d = 0.406$ m. (c) $T = 1.37$ s. (d) $\ell_{\text{eq}} = 0.466$ m, which reproduces the period exactly and lies between $d$ and $L$ . (e) Treating it as a simple pendulum of length 0.50 m gives 1.42 s, 3.5% high; dropping the disk's own rotational inertia gives 1.36 s, 0.53% low.

Determining g from a physical pendulum, with two routes from one graph

A uniform rod of length $L = 1.000$ m has pivot holes drilled at several distances $h$ from its centre. A student times 25 oscillations at each hole and records the period. Take the rod's rotational inertia about its centre as $\frac{1}{12}mL^2$ . The data are $h = 0.100$ m with $T = 1.9390$ s, $h = 0.200$ m with $T = 1.5761$ s, $h = 0.300$ m with $T = 1.5260$ s, and $h = 0.400$ m with $T = 1.5653$ s. Find (a) how to linearize, (b) $g$ from the slope, (c) $g$ from the intercept, and (d) the rod's length from the graph alone, with no value of $g$ .

(a) Start from the printed $T_{\text{phys}} = 2\pi\sqrt{I/(mgd)}$ with $d = h$ and, by the printed parallel-axis theorem, $I = \frac{1}{12}mL^2 + mh^2$ . The mass cancels: $T = 2\pi\sqrt{\dfrac{L^2/12 + h^2}{gh}}$ .
Square it and multiply through by $h$ to clear the denominator: $T^2 h = \dfrac{4\pi^2}{g}\left(\dfrac{L^2}{12} + h^2\right)$ .
That is a straight line if you plot $y = T^2 h$ against $x = h^2$ , with slope $\dfrac{4\pi^2}{g}$ and vertical intercept $\dfrac{4\pi^2 L^2}{12g}$ . Plotting $T$ against $h$ instead gives a curve with a minimum, which no slope can be read from.
$h$ (m) $T$ (s) $x = h^2$ (m $^2$ ) $y = T^2h$ (m $\cdot$ s $^2$ )
0.100 1.9390 0.0100 0.37597
0.200 1.5761 0.0400 0.49682
0.300 1.5260 0.0900 0.69860
0.400 1.5653 0.1600 0.98007
(b) Take the slope between the first and last points, which is what a student does by hand with a ruled line: slope $= \dfrac{0.98007 - 0.37597}{0.1600 - 0.0100} = \dfrac{0.60410}{0.1500} = 4.0273\ \mathrm{s^2/m}$ .
Then $g = \dfrac{4\pi^2}{\text{slope}} = \dfrac{39.478}{4.0273} = 9.8027 \approx 9.80\ \mathrm{m/s^2}$ . A least-squares line through all four points gives a slope of $4.0277\ \mathrm{s^2/m}$ and $g = 9.802\ \mathrm{m/s^2}$ , so the two-point estimate is not costing anything here, but a best fit over every point is what the Experimental Design and Analysis question asks for.
(c) The intercept from the same line: $b = y_1 - (\text{slope})x_1 = 0.37597 - (4.0273)(0.0100) = 0.37597 - 0.040273 = 0.33570\ \mathrm{m \cdot s^2}$ .
Since $b = \dfrac{4\pi^2 L^2}{12g}$ and $L = 1.000$ m is known, $g = \dfrac{4\pi^2 L^2}{12 b} = \dfrac{39.478}{12(0.33570)} = \dfrac{39.478}{4.0284} = 9.8000 \approx 9.80\ \mathrm{m/s^2}$ .
Two independent numbers from one graph, agreeing to three significant figures, and both matching the 9.8 m/s $^2$ printed on this course's Table of Information. That agreement is itself evidence the model fits, and saying so is a skill 3.C justification.
(d) Now the elegant part. Divide the intercept by the slope: $\dfrac{b}{\text{slope}} = \dfrac{4\pi^2 L^2/(12g)}{4\pi^2/g} = \dfrac{L^2}{12}$ . The $g$ and the $4\pi^2$ both cancel.
$\dfrac{0.33570}{4.0273} = 0.083357\ \mathrm{m^2}$ , so $L = \sqrt{12(0.083357)} = \sqrt{1.00028} = 1.0001 \approx 1.000\ \mathrm{m}$ , recovering the rod's length from timings alone with no gravitational constant involved.
One thing the data itself shows: the periods at $h = 0.200$ , 0.300 and 0.400 m differ by only a few percent, and the shortest is at $h = 0.300$ m. That is the minimum-period effect, whose predicted location is $h = L/\sqrt{12} = 0.2887$ m with $T_{\min} = 1.525$ s. The measured minimum sits right there, and a student who timed only two nearby holes would conclude, wrongly, that the pivot position does not matter.

$h$ (m)	$T$ (s)	$x = h^2$ (m $^2$ )	$y = T^2h$ (m $\cdot$ s $^2$ )
0.100	1.9390	0.0100	0.37597
0.200	1.5761	0.0400	0.49682
0.300	1.5260	0.0900	0.69860
0.400	1.5653	0.1600	0.98007

(a) Plot $T^2h$ against $h^2$ : the slope is $4\pi^2/g$ and the intercept is $4\pi^2L^2/(12g)$ . (b) Slope $4.027\ \mathrm{s^2/m}$ , giving $g = 9.80\ \mathrm{m/s^2}$ . (c) Intercept $0.3357\ \mathrm{m \cdot s^2}$ , giving $g = 9.80\ \mathrm{m/s^2}$ independently. (d) The intercept divided by the slope is $L^2/12$ with $g$ cancelling, returning $L = 1.000$ m. The data also shows the predicted minimum period of 1.525 s near $h = 0.289$ m.

A torsion pendulum used as an inertia balance

A uniform disk of mass 0.60 kg and radius 0.100 m is hung horizontally from a vertical wire attached at its centre and set into rotational oscillation in the horizontal plane. Its rotational inertia about the wire is $\frac{1}{2}MR^2$ . It completes one oscillation every 2.40 s. An irregular object of unknown shape is then hung from the same wire and oscillates with a period of 3.70 s. Find (a) the wire's torsion constant, (b) the unknown object's rotational inertia, (c) a route to (b) that never needs the torsion constant, and (d) what changes if the whole apparatus is taken into orbit.

Declare the convention: $\Delta\theta$ is the angular displacement from the untwisted position, positive in the direction of the initial twist, so the wire's restoring torque is negative there.
Set up the model from 7.5.A.4, whose Derived equation is $I\alpha = -k\Delta\theta$ . Since $\alpha = d^2\theta/dt^2$ , that is $\dfrac{d^2\theta}{dt^2} = -\dfrac{k}{I}\theta$ , which has the shape of 7.5.A.2.iii, so $\omega^2 = k/I$ and $T = 2\pi\sqrt{I/k}$ . None of that period is printed on the equation sheet; it comes from reading the coefficient.
(a) The known disk: $I_1 = \frac{1}{2}MR^2 = \frac{1}{2}(0.60)(0.100)^2 = \frac{1}{2}(0.60)(0.0100) = 3.000 \times 10^{-3}\ \mathrm{kg \cdot m^2}$ .
Its angular frequency is $\omega_1 = 2\pi/T_1 = 2\pi/2.40 = 2.6180\ \mathrm{rad/s}$ , so $\omega_1^2 = 6.8539\ \mathrm{rad^2/s^2}$ .
$k = I_1\omega_1^2 = (3.000 \times 10^{-3})(6.8539) = 0.020562 \approx 0.0206\ \mathrm{N \cdot m/rad}$ .
Check the units, since this $k$ is a torsion constant and not the spring constant the sheet's variable list names: $\mathrm{kg \cdot m^2}$ times $\mathrm{s^{-2}}$ is $\mathrm{kg \cdot m^2/s^2}$ , which is $\mathrm{N \cdot m}$ , and per radian since the radian is dimensionless. Correct.
(b) The unknown object on the same wire: $\omega_2 = 2\pi/3.70 = 1.6982\ \mathrm{rad/s}$ , so $\omega_2^2 = 2.8838\ \mathrm{rad^2/s^2}$ and $I_2 = \dfrac{k}{\omega_2^2} = \dfrac{0.020562}{2.8838} = 7.1302 \times 10^{-3}\ \mathrm{kg \cdot m^2}$ .
(c) The shorter route: because the wire is the same, $k$ is common to both runs, so $I \propto T^2$ . Then $I_2 = I_1\left(\dfrac{T_2}{T_1}\right)^2 = (3.000 \times 10^{-3})\left(\dfrac{3.70}{2.40}\right)^2 = (3.000 \times 10^{-3})(1.54167)^2 = (3.000 \times 10^{-3})(2.37674) = 7.1302 \times 10^{-3}\ \mathrm{kg \cdot m^2}$ .
Identical to part (b) to every digit, and it never evaluated $k$ . That is the honest way to run this measurement: the wire's property cancels, so an uncertainty in $k$ cannot contaminate the answer. Calibrate with a body whose rotational inertia you can compute, then compare periods.
Note also that the unknown's shape was never needed. That is the point of an inertia balance: it returns $I$ for an object you cannot integrate over.
(d) Nothing changes. The period $T = 2\pi\sqrt{I/k}$ contains no gravitational field strength, no length and no mass except through $I$ , so the apparatus reads the same in orbit as on a bench. Compare a physical pendulum, whose period goes as $1/\sqrt{g}$ and which does not oscillate at all in free fall, since the restoring torque $-mgd\sin\theta$ vanishes with the apparent $g$ .
That contrast is the CED's own essential question about weighing an astronaut in space, answered concretely. A torsion or spring oscillator measures inertia; a pendulum measures inertia and the gravitational field together, and a scale measures neither in free fall.

(a) $k = 0.0206\ \mathrm{N \cdot m/rad}$ , a torque per unit angle, not a spring constant. (b) $I_2 = 7.13 \times 10^{-3}\ \mathrm{kg \cdot m^2}$ . (c) The ratio route gives the same value from $I_2 = I_1(T_2/T_1)^2$ , with the wire's constant cancelling entirely. (d) Nothing changes in orbit, because no $g$ appears in the torsion period, unlike a pendulum, which does not swing at all in free fall.

Frequently asked questions

What is a physical pendulum in AP Physics C Mechanics?

Essential knowledge 7.5.A.1 defines it as a rigid body that undergoes oscillation about a fixed axis. Unlike a simple pendulum, whose mass is treated as concentrated at a point, a physical pendulum's mass is spread out, so how it is distributed matters. Statement 7.5.A.2 gives its small-amplitude period as two pi times the square root of the rotational inertia about the pivot divided by the product of mass, gravitational field strength, and the distance from the pivot to the centre of mass. That formula is printed on the AP Physics C: Mechanics equation sheet. AP Physics 1 has no equivalent topic and no such entry on its sheet.

What is the formula for the period of a physical pendulum?

Two pi times the square root of the rotational inertia divided by the product of the mass, the gravitational field strength and the distance from the axis to the centre of mass. Every symbol needs care: the rotational inertia is about the pivot, not about the centre of mass, and the distance is to the centre of mass, not to the far end of the body or to any other convenient point on the diagram. The formula is printed on the AP Physics C: Mechanics equation sheet. The three steps that produce it, the restoring torque, the small-angle approximation and the resulting differential equation, are not printed and are the derivation an exam question asks for.

How is a physical pendulum different from a simple pendulum?

A simple pendulum's period depends only on a length and the gravitational field strength. A physical pendulum's depends on how its mass is distributed, through the rotational inertia about the pivot. Essential knowledge 7.5.A.3 makes the simple pendulum a special case of the physical one: a point mass at a distance from the pivot has a rotational inertia equal to the mass times that distance squared, and substituting reduces the general formula to the familiar one. The general result is often written using an equivalent length, the rotational inertia divided by mass times the centre-of-mass distance, which is always larger than the centre-of-mass distance for any body with size.

Why is a pendulum only approximately simple harmonic motion?

Because the restoring torque is proportional to the sine of the angular displacement rather than to the displacement itself. Essential knowledge 7.5.A.2.i gives that torque as minus the mass times the gravitational field strength times the centre-of-mass distance times the sine of the angle, which is exact. Statement 7.5.A.2.ii then applies the small-angle approximation, replacing the sine of the angle with the angle in radians, and only after that substitution is the torque proportional to the displacement, which is what simple harmonic motion requires. Since the sine of an angle is always smaller than the angle, a real pendulum's restoring torque is weaker than the model's, so its period is slightly longer and grows with amplitude.

What is a torsion pendulum and what is its period?

Essential knowledge 7.5.A.4 defines it as a case of simple harmonic motion where the restoring torque is proportional to the angular displacement of a rotating system, and gives as its example a horizontal disk suspended from a wire attached to its centre of mass, undergoing rotational oscillations about the wire in the horizontal plane. Its derived equation sets the rotational inertia times the angular acceleration equal to minus a torsion constant times the angular displacement. The period is two pi times the square root of the rotational inertia divided by the torsion constant, and that period is printed nowhere in the course description, so you obtain it by reading the coefficient off the differential equation. It contains no gravitational field strength at all.

Does the mass affect a physical pendulum's period?

Not if the shape is held fixed. The rotational inertia about the pivot and the product of mass and centre-of-mass distance are both proportional to the total mass, so the mass cancels out of their ratio and therefore out of the period. A heavier rod of identical dimensions swings at exactly the same rate as a lighter one. What does change the period is changing how the mass is distributed: moving mass further from the pivot, or attaching a second body, changes the rotational inertia and the centre-of-mass distance by different factors. The same cancellation is why a simple pendulum's bob mass does not matter.

Where should you pivot a rod so it swings fastest?

At a distance from its centre equal to its length divided by the square root of twelve, which is about 0.289 of the length. This follows from the printed physical-pendulum period with the parallel-axis theorem, and it is a consequence of the framework rather than a statement in it. Pivot closer to the centre and the restoring torque weakens faster than the rotational inertia falls, so the period grows without limit as the pivot approaches the centre of mass, where the pendulum will not swing at all. Pivot further out and the rotational inertia grows faster, so the period grows again. Near the minimum the period is very insensitive to the pivot position, which is worth knowing when data looks suspiciously flat.