Physical Pendulum Calculator
Physical Pendulum: T = 2π√(I/mgd), and Where the Simple Formula Fails
A real swinging object is not a bob on a string. Its mass is spread out, so its period depends on the moment of inertia about the pivot and on how far the centre of mass sits from that pivot — not on its length. Pick a shape and this returns the period, the length of the simple pendulum that would match it, and what the amplitude costs you.
Period of a real swinging body, and its equivalent length
A uniform rod of mass 1.6 kg and length 0.45 m hangs from a pivot at one end and swings with an amplitude of 8°. Taking g as 9.81 m/s², find its period.
The general period, the equivalent length, and the simple case inside it
- I
- moment of inertia about the pivot, not about the centre of mass. For anything not pivoted at its own centre this means applying the parallel-axis theorem first: I = Icm + md². Using Icm here is the commonest error on the page, and it always makes the period come out too short
- d
- the distance from the pivot to the centre of mass. Not the length of the object, not half of it unless the object happens to be uniform and pivoted at one end. It is what supplies the restoring torque: a body pivoted exactly at its centre of mass has d = 0, no restoring torque, an infinite period, and simply does not swing
- Leq = I/md
- the equivalent simple-pendulum length, sometimes called the length of the equivalent simple pendulum or the distance to the centre of oscillation. A simple pendulum of this length has exactly this period. It is the single most useful number on the page, because it reduces any swinging body to a string-and-bob you already understand
- what has vanished
- the mass. I is proportional to m for any uniform body, and m appears in the denominator too, so it cancels exactly. A steel rod and a wooden rod of the same length keep the same time. What has not vanished is the shape: the distribution of mass is in I and does not cancel
- the amplitude term
- the first term of an infinite series, T = T₀(1 + θ₀²/16 + 11θ₀⁴/3072 + …), with θ₀ in radians. At 8° it adds about 0.12 per cent; at 30° about 1.7 per cent; at 60° the first term alone gives 6.9 per cent and is itself no longer accurate
- why boards set this
- because it is the first place the simple pendulum formula visibly fails, and because it needs the parallel-axis theorem, the centre of mass and simple harmonic motion together. Compare it with the simple pendulum page, which is the d = L, I = mL² corner of this one
Worked example
A uniform rod of mass 1.6 kg and length 0.45 m hangs from a pivot at one end and swings with an amplitude of 8°. Taking g as 9.81 m/s², find its period.
Moment of inertia about the pivot. For a uniform rod about a perpendicular axis at one end, I = mL²/3 = (1.6)(0.45)²/3 = (1.6)(0.2025)/3 = 0.1080 kg m². Note it is about the pivot, so mL²/3 and not mL²/12.
Pivot to centre of mass. The rod is uniform, so its centre of mass is at the midpoint: d = L/2 = 0.225 m. This is the number people replace with L, and doing so makes the period come out a factor of √2 too small.
Substitute. T = 2π√(I/mgd) = 2π√(0.1080 / (1.6 × 9.81 × 0.225)). The denominator is 1.6 × 9.81 × 0.225 = 3.5316, so the ratio is 0.030581 and its square root is 0.174874.
Period. T = 2π × 0.174874 = 1.0988 s. Each half swing takes 0.549 s, and the frequency is 0.910 Hz.
The equivalent length, which is the real insight. Leq = I/md = 0.1080/(1.6 × 0.225) = 0.300 m, which is exactly 2L/3. Check it against the simple pendulum: 2π√(0.300/9.81) = 1.0988 s, the same number. So a 0.45 m rod swings like a 0.30 m string.
And the trap, in one line. If you had used the simple formula with the rod’s own length, 2π√(0.45/9.81) = 1.346 s — 22.5 per cent too slow. The rod is faster than a string of its own length because its mass is on average much closer to the pivot than the far end is.
Amplitude. 8° is 0.1396 rad, so the correction factor is 1 + 0.1396²/16 = 1.00122 and the true period is about 1.1001 s. That is 0.12 per cent, comfortably inside normal laboratory timing error at this amplitude.
Equivalent length for each shape, and how it compares with the object
| Body, pivoted as shown | I about the pivot | d | L_eq = I/md | Against its own size |
|---|---|---|---|---|
| Uniform rod, at one end | mL²/3 | L/2 | 2L/3 | Swings like 67% of its length |
| Uniform disc, on its rim | 3mR²/2 | R | 3R/2 | Swings like 75% of its diameter |
| Thin hoop, on its rim | 2mR² | R | 2R | Swings like its own diameter, exactly |
| Point mass on a light rod | mL² | L | L | The simple pendulum: L_eq is the length |
| Solid sphere, hung by a point on its surface | 2mR²/5 + mR² | R | 7R/5 | Not offered above; shown because the method is the same |
What the amplitude costs, for the default rod
| Amplitude θ₀ | In radians | θ₀²/16 | T (s) | Error if ignored |
|---|---|---|---|---|
| 2° | 0.0349 | 0.000076 | 1.0989 | 0.008% |
| 5° | 0.0873 | 0.000476 | 1.0993 | 0.048% |
| 8° — the default | 0.1396 | 0.001218 | 1.1001 | 0.122% |
| 15° | 0.2618 | 0.004283 | 1.1035 | 0.428% |
| 30° | 0.5236 | 0.017133 | 1.1176 | 1.713% |
| 45° | 0.7854 | 0.038553 | 1.1401 | 3.855% |
Why length is the wrong variable, and what the centre of oscillation is
The simple pendulum formula is a special case, and school papers meet the general one as soon as the bob stops being a point. T = 2π√(L/g) assumes all the mass sits at distance L from the pivot. A swinging rod, door, bicycle wheel or human leg has mass spread along its length, most of it closer to the pivot than the far end is, and it therefore swings faster than a string of the same length. The general result, T = 2π√(I/mgd), replaces the single length with two quantities: I, which says how the mass is spread, and d, which says where its centre is.
Where the formula comes from. Displace the body by θ. Its weight mg acts at the centre of mass, a distance d from the pivot, so the restoring torque is −mgd sinθ. Newton’s second law for rotation gives Iα = −mgd sinθ. That is not simple harmonic motion, because of the sine. Replace sinθ with θ, valid for small angles, and it becomes α = −(mgd/I)θ, which is simple harmonic motion with ω² = mgd/I and therefore T = 2π√(I/mgd). Every step of that is the simple pendulum derivation with I in place of mL² and d in place of L, which is why the two formulas look so alike.
The equivalent length is the number to take away. Leq = I/md is the length of the simple pendulum with the same period, and the point on the body at that distance from the pivot is called the centre of oscillation. For a uniform rod from one end it is 2L/3, which is two-thirds of the way down — not the centre of mass at L/2, and not the tip. Hit a bat or a racquet at its centre of oscillation and the handle feels no jolt, which is why the same point is also called the sweet spot. That is a real and checkable consequence of a formula that otherwise looks like bookkeeping.
The mass genuinely does cancel. For any uniform body I is proportional to m, so m appears in the numerator and the denominator of I/mgd and disappears. A steel rod and a balsa rod of identical dimensions have identical periods. What does not cancel is the shape: a rod and a hoop of the same outer size have different I and different d and swing at quite different rates. If a question gives you a mass for a physical pendulum period, it is either a distraction or it is wanted for something else, such as the restoring torque or the energy.
The special case worth knowing: d = 0. Pivot a body exactly at its centre of mass and the restoring torque vanishes. It does not oscillate at all; it stays wherever you put it. The formula agrees — with d = 0 the period is infinite — and this is the physics behind a balanced wheel, a compass needle on a jewelled bearing, and the fact that a stick balanced on its centre of mass has no preferred orientation. As d gets small the period gets very long and very sensitive, which is why real seconds pendulums were built with the bob well away from the pivot.
Where this goes wrong in exams. Four reliable mistakes. One: using Icm instead of I about the pivot — always add md² first. Two: putting the object’s length in for d instead of the distance to the centre of mass; for a rod that is a factor of two, which becomes √2 in the period. Three: forgetting that the amplitude correction wants radians, so θ₀²/16 with 8 in it rather than 0.1396 gives a nonsensical four per cent. Four: quoting T as the time for a half swing. T is one complete oscillation, there and back, and timing twenty of them and dividing by twenty is the standard laboratory defence against both that and reaction-time error.
Frequently asked questions
Why does a metre rule swing faster than a one-metre string?
Because most of the rule’s mass is nearer the pivot than the far end is, so there is less inertia to overcome for the same restoring torque. Quantitatively, Leq = 2L/3, so a 1.00 m rule keeps time like a 0.667 m string: 1.638 s against 2.006 s with g = 9.81. The ratio is √(2/3) = 0.816 for any uniform rod, whatever it is made of.
Do I use I about the centre of mass or about the pivot?
About the pivot, always. The body is rotating about the pivot, so that is the axis the equation of motion is written about. If you only know Icm, add md² using the parallel-axis theorem first. Using Icm alone always gives a period that is too short, and for a rod at one end it is too short by a factor of two.
What exactly is the centre of oscillation?
The point on the body at distance Leq = I/md from the pivot. A point mass placed there on a light rod would swing with exactly the period the whole body has. It has a second property that makes it famous: an impulse delivered there produces no reaction at the pivot, which is why a cricket bat or tennis racquet struck at that point does not sting the hands. Pivot and centre of oscillation are also interchangeable — hang the body from the centre of oscillation and the old pivot becomes the new one, with the same period.
Does the amplitude really not matter?
It matters, but weakly, and only through the square. At 5° the period is 0.05 per cent longer than the small-angle value, at 15° about 0.4 per cent, at 30° about 1.7 per cent. Below about 10° this is smaller than you can time by hand, which is why lab instructions specify a small swing. It is a systematic error, so averaging more swings will not remove it.
Which value of g should I put in?
Whichever your own data sheet prints, so your working matches the mark scheme. Standard gravity is 9.806 65 m/s² exactly — a defined conventional value, not a measurement — while Cambridge 9702 prints 9.81 and many NCERT examples use 9.8. Between 9.8 and 9.81 a period moves by 0.05 per cent, far less than the amplitude effect, so the choice almost never changes an answer to three significant figures.
Can I use this to measure g?
Yes, and it is the classic use: Kater’s reversible pendulum measures g to a few parts per million by exploiting the pivot and centre of oscillation being interchangeable. Adjust the two knife edges until the period is identical about both, and then Leq is simply the distance between them — a length you can measure very accurately — so g = 4π²Leq/T² without having to know I, m or d at all.
Related calculators
References
- NCERT, Physics: Textbook for Class XI, Part I, Chapter 7 “Systems of Particles and Rotational Motion”. Read 11 November 2026. Table 7.1 gives eight bodies and no more: thin circular ring about the perpendicular axis through its centre (MR²) and about a diameter (MR²/2); thin rod perpendicular to the rod at its midpoint (ML²/12); circular disc about the perpendicular axis through its centre (MR²/2) and about a diameter (MR²/4); hollow cylinder about its own axis (MR²); solid cylinder about its own axis (MR²/2); solid sphere about a diameter (2MR²/5). The theorem of perpendicular axes is §7.10 and the theorem of parallel axes is §7.10.1, stated there without proof. The thin spherical shell is not in Table 7.1, which is why this page derives 2MR²/3 for you instead of quoting it.
- PHYSICS (042) Curriculum, Class XI–XII, Central Board of Secondary Education, session 2025–26. Unit V, Motion of System of Particles and Rigid Body, reads in part: “Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications” and “Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation)”. Unit III, Laws of Motion, reads in part: “Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road)”. The parenthesis “no derivation” is why this page gives you the standard results rather than the integrals behind them. Quoted to establish what is in scope, never to claim any endorsement by the board.
- NIST, Fundamental Physical Constants — Complete Listing, 2022 CODATA adjustment, Standard Reference Database 121. The line this batch leans on is “standard acceleration of gravity 9.806 65 (exact) m s−2”. The word exact is the whole point: gn is a conventional value fixed by the 3rd CGPM in 1901, not a measurement of anywhere, so it carries no uncertainty and it is not the g where you are standing.
physics.nist.gov/cuu/Constants/ - Cambridge International AS & A Level Physics 9702, syllabus for examination in 2025, 2026 and 2027. Its data sheet gives the acceleration of free fall as 9.81 m s−2. Worth knowing what this syllabus leaves out, because it explains why a question from one board can look unanswerable to a student taught on another: topic 3.2 asks only for “a qualitative understanding of frictional forces and viscous/drag forces including air resistance” and states that “no treatment of the coefficients of friction and viscosity is required”. Quoted by outcome number so the difference is checkable rather than asserted.