Simple and Physical Pendulums: Study Guide (Physics C: Mechanics 7.5)

How this fits with Physics 1. Physics C: Mechanics and Physics 1 are separate courses. Physics 1 meets only the simple pendulum formula, in its Topic 7.2 study guide. This guide is the calculus-based treatment: it derives the motion of any rigid pendulum from Newton’s second law in rotational form, then shows the simple pendulum and the torsion pendulum as cases of the same idea.

What a physical pendulum is

A physical pendulum is a rigid body that swings about a fixed axis that does not pass through its centre of mass. A sign swinging from a hook through its top edge, a leg swinging from the hip, a ruler hanging from a nail through one of its holes: all are physical pendulums.

Three quantities describe it:

• m, its total mass.
• d, the distance from the pivot axis to the centre of mass.
• I, its rotational inertia about the pivot axis. Usually you find I_cm first and then use the parallel axis theorem from Topic 5.4: I = I_cm + md².

We measure the angular displacement θ from the equilibrium position, where the centre of mass hangs directly below the pivot. Choose a sign convention, for example “counterclockwise positive”, and keep θ in radians.

The restoring torque from gravity

Take torques about the pivot. The pivot force passes through the axis, so it gives no torque. Gravity acts at the centre of mass, a distance d from the pivot. When the body is turned through θ, the lever arm of the weight is d sin θ, so

τ = −mgd sin θ

The minus sign says the torque always acts back towards equilibrium: θ positive gives a negative torque, and θ negative gives a positive torque. That is what makes it a restoring torque.

The small-angle approximation

The torque depends on sin θ, not on θ. That is not quite the “restoring quantity proportional to displacement” that SHM needs. But for small angles in radians, sin θ ≈ θ. The table shows how good this is:

θ (degrees)	θ (rad)	sin θ	θ larger than sin θ by	True period longer than the small-angle value by
5	0.0873	0.0872	0.13%	0.05%
10	0.1745	0.1736	0.51%	0.19%
15	0.2618	0.2588	1.15%	0.43%
20	0.3491	0.3420	2.06%	0.77%
30	0.5236	0.5000	4.72%	1.74%

The last column is background only: it comes from an exact calculation that is beyond this course. What you need is the trend. At larger amplitudes the true torque, mgd sin θ, is weaker than the linear model mgdθ, so the pendulum returns more slowly and the true period is a little longer than the formula predicts. Below about 15°, the error is under half a percent.

With sin θ ≈ θ, the torque becomes

τ ≈ −mgdθ

From τ = Iα to SHM

Newton’s second law in rotational form about the fixed pivot is τ_net = Iα, with α = d²θ/dt². Put in the small-angle torque:

1. I (d²θ/dt²) = −mgdθ
2. d²θ/dt² = −(mgd/I) θ

Compare this with the SHM equation from Topic 7.3, d²x/dt² = −ω²x. It has the same form, with θ in place of x. So the pendulum performs angular simple harmonic motion with

ω² = mgd/I, so ω = √(mgd/I) and T = 2π/ω = 2π√(I/(mgd))

The solution is θ(t) = θ_max cos(ωt + φ). You need to know this solution and recognise the equation as SHM; the course does not ask you to prove that the solution works.

Notice three things:

• I is about the pivot. Using I_cm instead is the commonest error.
• The amplitude θ_max is not in T (within the small-angle approximation).
• ω is a constant, not the angular velocity. The angular velocity is dθ/dt = −ωθ_max sin(ωt + φ). Its largest size is ωθ_max, at the bottom of the swing.

The simple pendulum: a special case

A simple pendulum is a small, dense bob on a light string of length l (pivot to the centre of the bob). Model the bob as a point mass. Then:

• d = l, because all the mass is at distance l;
• I = ml², because a point mass at distance l has rotational inertia ml².

Substitute into the physical pendulum result:

T = 2π√(ml²/(mgl)) = 2π√(l/g)

The mass cancels because both I and the torque are proportional to m, and all the mass sits at the same distance. For a real body the mass is spread out, so I and md are not simply related, and the shape matters.

A useful idea: any physical pendulum has the same period as a simple pendulum of length l_eq = I/(md). This is always longer than d.

The torsion pendulum

A torsion pendulum is a body hung from a wire attached at its centre of mass, for example a horizontal disk on a vertical wire. Twist the disk through θ and the wire twists back with a torque proportional to the angle:

τ = −κθ

where κ (kappa) is the torsion constant of the wire, in N·m/rad. Then Iα = −κθ gives

d²θ/dt² = −(κ/I) θ, so ω = √(κ/I) and T = 2π√(I/κ)

Two differences from a gravity pendulum:

• No small-angle step. If the wire obeys τ = −κθ, the motion is SHM at any amplitude for which that holds.
• No g. The disk turns in a horizontal plane, so its weight gives no restoring torque. A torsion pendulum keeps the same period on the Moon or in orbit.

Worked example 1: a plank hung from a hole

Question. A uniform plank of mass 1.2 kg and length 0.90 m hangs from a nail through a hole 0.15 m from one end. It swings with small amplitude. Take counterclockwise as positive. Find (a) d, (b) I about the nail, (c) the period, and (d) the length of a simple pendulum with the same period. (I_cm of a uniform rod = ML²/12.)

1. (a) The centre of mass is at the middle, 0.45 m from the end. d = 0.45 − 0.15 = 0.30 m.
2. (b) I_cm = (1.2)(0.90)²/12 = 0.081 kg·m². Parallel axis: I = 0.081 + (1.2)(0.30)² = 0.081 + 0.108 = 0.189 kg·m².
3. (c) mgd = (1.2)(9.8)(0.30) = 3.53 N·m. ω = √(3.53 ÷ 0.189) = 4.32 rad/s. T = 2π/ω = 1.45 s.
4. (d) l_eq = I/(md) = 0.189 ÷ (1.2 × 0.30) = 0.525 m. Check: 2π√(0.525/9.8) = 1.45 s.

Check the mass. Doubling m doubles I and mgd, so T is unchanged. The period depends only on how the mass is spread: T = 2π√((I/m)/(gd)).

Check the model. Treating the plank as a simple pendulum of length d = 0.30 m gives 1.10 s, which is far too short. A plank is not a point mass.

Worked example 2: measuring rotational inertia with a torsion pendulum

Question. A uniform disk of mass 2.0 kg and radius 0.15 m hangs from a wire through its centre. It twists back and forth with period 2.40 s. A small machine part is fixed on the disk, centred on the wire, and the period becomes 3.10 s. Find (a) the torsion constant κ and (b) the rotational inertia of the part about the wire. (I of a uniform disk about its axis = ½MR².)

1. (a) I_disk = ½(2.0)(0.15)² = 0.0225 kg·m². From T = 2π√(I/κ), κ = 4π²I/T² = 4π²(0.0225) ÷ (2.40)² = 0.154 N·m/rad.
2. (b) With the part: I_total = κT²/(4π²) = 0.154 × (3.10)² ÷ 4π² = 0.0375 kg·m².
3. I_part = 0.0375 − 0.0225 = 0.0150 kg·m².

A shortcut. κ is the same in both runs, so I ∝ T². I_total = 0.0225 × (3.10/2.40)² = 0.0225 × 1.668 = 0.0375 kg·m². This avoids rounding κ.

Interpretation. No value of g was used. The same method works in orbit, which is why oscillation is a good way to measure inertia where weighing is impossible.

Designing a pendulum experiment

A typical task asks you to find g, or a rotational inertia, from timing data. Good procedures share these features:

• Time many oscillations, say 20, and divide. A reaction-time error of about 0.2 s then becomes about 0.01 s per period.
• Start timing as the body passes equilibrium, where it moves fastest and the instant is sharpest.
• Keep the amplitude small (under about 10°), so the small-angle model holds.
• Measure d to the centre of mass, found for example by balancing the body on an edge.
• Vary one quantity and linearise. For a uniform bar with several holes, write I = m(k² + d²), where k² = I_cm/m. Then T²d = (4π²/g)d² + (4π²/g)k². A graph of T²d against d² is a straight line with slope 4π²/g and intercept (4π²/g)k². You can find g from the slope and k² from intercept ÷ slope. Practice Question 6 uses this.
• Repeat and look for outliers before drawing the best-fit line.

Common misconceptions

• Using I_cm in T = 2π√(I/mgd). The body turns about the pivot, so I must be about the pivot (Worked example 1).
• Treating every pendulum as simple. T = 2π√(l/g) needs all the mass at one distance. A plank or hoop needs the physical pendulum formula.
• Measuring d to the far end, or to the bottom of the bob. d runs from the pivot to the centre of mass.
• Using degrees in sin θ ≈ θ. The approximation is only true in radians.
• Confusing ω with dθ/dt. ω = √(mgd/I) is fixed; the angular speed varies and peaks at ωθ_max.
• “The period never depends on mass.” Scaling all the mass changes nothing, but moving or adding mass in one place changes I and d differently (Practice Question 7).
• “A torsion pendulum needs gravity.” Its restoring torque comes from the wire, so g is absent.
• “Large swings have the same period.” Only approximately. Beyond small angles the period grows slightly.

Where this leads

This is the last topic of the course. It pulls together torque, rotational inertia and Newton’s second law in rotational form (Units 5 and 6) with the SHM ideas of Topics 7.1 to 7.4. Look back at Topic 7.4, Energy of Simple Harmonic Oscillators: the same pendulum can be analysed with energy, using ½I(dθ/dt)² for kinetic energy and mgd(1 − cos θ) for gravitational potential energy. Now try the practice questions, then use the revision notes and the checklist. When you are done, return to the course roadmap to plan your full-course revision.