A physical pendulum is any rigid body that swings about a fixed pivot and, for small angular displacements, shows simple harmonic motion because gravity on its center of mass produces a restoring torque. The restoring torque is τ = −mgd sinθ (d = distance from pivot to COM). Using Newton’s 2nd law for rotation and the small-angle approximation (sinθ ≈ θ) you get Iα = −mgdθ, which leads to SHM with period T_phys = 2π√(I/(mgd)). A simple pendulum is just a special case of a physical pendulum where all the mass is treated as a point mass at distance ℓ from the pivot; its period becomes T = 2π√(ℓ/g). On the AP exam you should be able to derive T_phys from τ = Iα and apply the small-angle approximation (CED 7.5.A.2–7.5.A.3). For a clear CED-aligned summary and practice, check the Topic 7.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA) and try problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

For small oscillations the restoring torque is set by gravity while the resistance to angular acceleration is set by the pendulum’s inertia—and both scale with mass, so mass cancels. Start from the CED result for a physical pendulum: τ = −mgd sinθ ≈ −mgd θ and Iα = τ, so d²θ/dt² = −(mgd/I) θ → ω² = mgd/I. Thus T = 2π √(I/(mgd)) (Topic 7.5.A.2). For a simple pendulum (point mass at distance ℓ): I = mℓ² and d = ℓ, so T = 2π √(ℓ/g). The m drops out, so period depends on length ℓ and g, not mass. Physically: heavier mass gives a bigger gravitational pull (bigger restoring torque) but also more inertia (harder to accelerate) in the same proportion, so net oscillation speed is unchanged. For AP exam practice, you should be able to derive T from rotational Newton’s 2nd law using the small-angle approximation (see the Topic 7.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA) and more practice problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics)).

Think of the small-angle approximation as a math shortcut that turns a pendulum’s nonlinear motion into simple harmonic motion (SHM). For a physical pendulum the restoring torque is τ = −mgd sinθ. If θ is small (usually θ ≲ 10° or θ ≲ 0.17 rad, where the error is ≲1%), you can replace sinθ with θ (in radians): sinθ ≈ θ. That makes τ ≈ −mgd θ, so Iα = −mgd θ → (d²θ/dt²) = −(mgd/I) θ. That’s the standard SHM form d²θ/dt² = −ω²θ, with ω² = mgd/I, and period T = 2π√(I/(mgd)). For a simple pendulum (point mass at distance ℓ) this reduces to T = 2π√(ℓ/g). On the AP exam you’ll be expected to (a) show τ = −mgd sinθ, (b) apply sinθ ≈ θ for small amplitudes, and (c) derive the SHM differential equation and T (CED 7.5.A.2–3). For a quick review see the Topic 7.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA) and try practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Simple pendulum: idealized point mass m on a massless string of length ℓ that swings about a pivot. Restoring torque comes from gravity acting at the point mass; for small angles the motion is SHM with period Tp = 2π√(ℓ/g). It’s a special case of a physical pendulum (CED 7.5.A.3). Physical pendulum: any rigid body that oscillates about a pivot. Gravity on the center of mass produces a restoring torque τ = −mgd sinθ (d = distance from pivot to CM). For small θ use sinθ ≈ θ and apply Iα = τ to get SHM and Tphys = 2π√(I/(mgd)) (CED 7.5.A.1–A.2). Torsion pendulum: a rotating object suspended by a wire (or rod) where the restoring torque is elastic, τ = −kΔθ, not gravity. Equation Iα = −kΔθ gives SHM with angular frequency ω = √(k/I) and T = 2π√(I/k) (CED 7.5.A.4). For AP prep, know the derivations, small-angle approximation, and how I, d, and k affect T. Review the Topic 7.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Good question—you use torque because the object rotates about a pivot, so rotation equations (not linear force equations) give the restoring “push.” For a physical pendulum the weight mg acts at the center of mass a distance d from the pivot; that force produces a torque τ = −mgd sinθ about the pivot (CED 7.5.A.2.i). Newton’s second law for rotation is τ = Iα, so −mgd sinθ = I(d²θ/dt²). For small θ we use sinθ ≈ θ to get I(d²θ/dt²) = −mgd θ, which is the SHM equation d²θ/dt² = −(mgd/I) θ. That gives ω² = mgd/I and T = 2π sqrt(I/(mgd)) (CED 7.5.A.2.ii–iii). So force matters (it creates torque), but torque and moment of inertia are the right quantities to describe rotational oscillation and the period on the AP exam. For a quick refresher, see the Topic 7.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA). For practice, try rotational SHM problems on Fiveable’s practice page (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Start with Newton’s 2nd law for rotation about the pivot: τ = Iα. The weight mg acting at the center of mass a distance d from the pivot produces a restoring torque τ = −mgd sinθ (CED 7.5.A.2.i). For small oscillations use the small-angle approximation sinθ ≈ θ (7.5.A.2.ii), so −mgd θ = I (d^2θ/dt^2). Rearrange to the standard SHM form: d^2θ/dt^2 + (mgd / I) θ = 0. Compare with d^2θ/dt^2 + ω^2 θ = 0, so ω^2 = mgd / I and ω = sqrt(mgd / I). The period follows: T = 2π / ω = 2π sqrt( I / (mgd) ), which is the CED result (7.5.A.2). For a simple pendulum treat the mass as a point mass at distance ℓ so I = mℓ^2 and d = ℓ, giving T = 2π sqrt(ℓ/g) (7.5.A.3). For more on physical pendulums and exam-style derivations see the Topic 7.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA). For extra practice, try problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Short answer: a physical pendulum’s period includes I because the object’s mass is spread out, and how that mass is distributed about the pivot affects how hard it is to rotate. Quick explanation: start with Newton’s second law for rotation (CED 7.5.A.2): τ = Iα. The restoring torque from gravity is τ = −mgd sinθ (d = distance from pivot to center of mass). For small θ, sinθ ≈ θ, so −mgd θ = I d²θ/dt². That yields ω² = mgd / I and T = 2π √(I / (mgd)). For a simple pendulum the bob is treated as a point mass at distance ℓ, so its moment of inertia about the pivot is I = mℓ². Plugging that in gives T = 2π √(ℓ / g)—the I is still there algebraically but reduces to mℓ², canceling the m and leaving ℓ/g. So the difference is physical: simple pendulum = point mass (I = mℓ²); physical pendulum = extended rigid body so you must use its actual moment of inertia. For the AP exam you may be asked to derive T from τ = Iα and the small-angle approximation (see CED 7.5.A.2). For a concise study guide and practice problems, check the Topic 7.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA) and the Unit 7 overview (https://library.fiveable.me/ap-physics-c-mechanics/unit-7).

Small-angle approximation means you replace sinθ with θ (only when θ is in radians). It’s a linearization: sinθ ≈ θ for small θ so the restoring torque τ = −mgd sinθ becomes τ ≈ −mgd θ, which gives Iα = −mgd θ and leads to the SHM equation d²θ/dt² = −(mgd/I)θ and the small-amplitude period T = 2π√(I/(mgd)) (CED 7.5.A.2). When can you use it? Practically when θ is small enough that the error is negligible—commonly θ ≲ 0.1–0.2 rad (≈6°–12°). At ~0.1 rad error is <<1%; by ~0.2 rad error is a few percent. For larger amplitudes the motion is nonlinear: period depends on amplitude and sinθ ≠ θ, so the simple-pendulum period T = 2π√(ℓ/g) and the physical-pendulum formula from the CED no longer accurately apply. For AP prep, remember the CED expects you to apply the small-angle approximation to derive the SHM differential equation and period (see the Topic 7.5 study guide: https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA). For extra practice, try problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Use the physical-pendulum formula from rotational Newton’s law and the small-angle approximation. If a rigid body of mass m is pivoted a distance d from its center of mass and has moment of inertia I about the pivot, then for small angular displacements T_phys = 2π sqrt( I / (m g d) ). How to use it: - Compute I about the pivot. If you know I_cm (about the center of mass), use the parallel-axis theorem: I_pivot = I_cm + m d^2. - d is the distance from pivot to the center of mass. - The small-angle approximation (sin θ ≈ θ) is required for SHM and the period formula to hold. Special case: a simple pendulum (point mass at distance ℓ) is T = 2π sqrt(ℓ/g), which you get by setting I = mℓ^2 and d = ℓ in the physical formula. This is exactly the CED result for Topic 7.5 (see the study guide: https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA). For more practice, check the unit page (https://library.fiveable.me/ap-physics-c-mechanics/unit-7) and AP-style problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Good question—sin θ ≈ θ is the small-angle approximation and it’s what lets a pendulum behave like SHM. Two ways to see it: 1) Math (short): For θ in radians, the Taylor series is sin θ = θ − θ^3/6 + … For |θ| ≲ 0.1 rad (~6°) the θ^3 term is tiny, so sin θ ≈ θ. 2) Intuition: For small angles the arc length s on a circle ≈ radius × angle (s ≈ rθ), and the vertical/ restoring displacement is proportional to that arc. So the component of gravity that gives the torque (mgd sin θ) is nearly proportional to θ when θ is small. Important: θ must be in radians for the approximation to be valid. In the CED derivation (Topic 7.5) you replace sin θ by θ to get τ = −mgd θ = Iα, which leads to d²θ/dt² = −ω²θ and the period T = 2π√(I/(mgd)). For more review and practice on pendulums, see the Topic 7.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA) and try problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Move the pivot and you change both d (distance from pivot to center of mass) and the moment of inertia about the pivot, so the period changes. For small oscillations a physical pendulum has T = 2π sqrt( I / (m g d) ), with I = Icm + m d^2 (parallel-axis theorem). So T(d) = 2π sqrt( (Icm + m d^2) / (m g d) ). What that means in plain terms: - If the mass is basically a point (Icm ≈ 0), T → 2π sqrt(d/g) (the simple-pendulum formula). - For an extended body, moving the pivot changes both numerator and denominator, so the dependence isn’t linear. - There’s a particular pivot distance d = sqrt(Icm / m) that minimizes T. Pivoting much closer or much farther than that increases the period. On the AP exam you should be able to derive T(d) from τ = Iα and the small-angle approximation (sinθ ≈ θ) and use the parallel-axis theorem (CED 7.5.A.2–3). For a quick review see the Topic 7.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA). For more practice problems, check the unit practice set (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Gravity creates the restoring torque that makes a pendulum oscillate. For a physical pendulum, the weight mg acts at the center of mass a distance d from the pivot; small angular displacements give τ = −mgd sinθ ≈ −mgd θ. Plugging into rotational Newton (Iα = τ) gives I d²θ/dt² = −mgd θ, so d²θ/dt² = −(mgd/I) θ. That constant (mgd/I) is ω², so T = 2π/ω = 2π sqrt(I/(mgd)). For a simple pendulum the mass is a point at distance ℓ so I = mℓ² and d = ℓ, producing T = 2π sqrt(ℓ/g). Notice g sits in the denominator because larger g → bigger restoring torque → larger ω → shorter period. This chain (τ = −mgd sinθ → small-angle → SHM → T formula) is exactly what the CED asks you to know (7.5.A.2). For a quick study hit, see the Topic 7.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA) and try practice problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

A torsion pendulum is a rotating SHM system where the restoring torque comes from twisting (torsion) in a rod or wire rather than gravity. If a rigid object (like a disk) is suspended by a thin wire and you twist it by an angle Δθ, the wire exerts a restoring torque proportional to that angle: τ = −k Δθ, where k is the torsion constant. Using rotational Newton’s second law, I α = −k θ, you get the SHM differential equation (d²θ/dt² = −(k/I) θ). The angular frequency is ω = √(k/I) and the period is T = 2π √(I/k). Compare to a physical pendulum (CED 7.5.A): there the restoring torque is mgd sinθ (gravity-based) and T = 2π √(I/(mgd)) for small angles; for a torsion pendulum the wire provides the torque instead. Torsion pendulums are useful for measuring moments of inertia or small torques (e.g., Cavendish-style experiments). For the AP CED coverage of physical and torsion pendulums see the Topic 7.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA). For more unit review and lots of practice, check the Unit 7 overview (https://library.fiveable.me/ap-physics-c-mechanics/unit-7) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Simple pendulums are easier, but real objects aren’t point masses—they have size and mass distribution. A physical pendulum lets you model any rigid body swinging about a pivot; its period comes from Newton’s 2nd law in rotational form, T_phys = 2π√(I / (m g d)), where I is the object's moment of inertia and d is the COM distance to the pivot. That extra I and d matter when the mass isn’t concentrated at one point (use the parallel-axis theorem when needed). On the AP exam you may be asked to derive that SHM differential equation using τ = Iα and the small-angle sinθ ≈ θ approximation (skills 2.A and 3.B in the CED). So learn physical pendulums to handle real setups, do symbolic derivations, and connect torque, COM, and I. Want a focused review and examples? Check the Topic 7.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

You were supposed to measure the period of oscillation and the geometry/mass distribution of your pendulum so you could connect the measured period to the CED formulas and (if asked) determine g or the object’s moment of inertia. Specifically: - For a simple pendulum: measure the length ℓ and the period T (time, e.g., for 10 oscillations, divide by 10). Use T = 2π√(ℓ/g) so T² vs ℓ is linear (slope = 4π²/g). - For a physical pendulum: measure the mass m, distance d from pivot to center of mass, and T. Use Tphys = 2π√(I/(m g d)). If the body is a known shape, use its I about the CM and the parallel-axis theorem to solve for I or g. - Keep angles small (< ~10°), repeat trials, time many oscillations to reduce random error, and plot T² vs the relevant geometric quantity for linear analysis. This lab practices the AP LAB skill (3.A) and the LAB-style FRQ experimental analysis. For a quick review, see the Topic 7.5 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/5-simple-and-physical-pendulums/study-guide/m0lcXe33VLYhg8EA) and more practice problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).