The physical pendulum calculator above handles the real case: a rigid body of arbitrary shape swinging about a pivot that is not at its centre of mass. Unlike the idealised textbook pendulum, a physical pendulum has its mass spread out, so its period depends on how that mass is distributed as well as on where the pivot is. The governing relation is T = 2π√(I ÷ mgd), where I is the moment of inertia about the pivot axis and d is the distance from the pivot to the centre of mass.
Arb Digital publishes free physics calculators that state the assumption behind each number rather than presenting one figure and moving on. This page is deliberately separate from the simple pendulum calculator, which assumes a point mass on a massless string and cannot represent a swinging rod, a hanging sign or a compound balance at all. If your pendulum's supporting structure has meaningful mass, that page will not match your measurement and this one will.
What This Physical Pendulum Calculator Does
Restoring torque on a displaced rigid body is mgd sin θ. For small angles sin θ approaches θ, giving a torque proportional to displacement, which is the definition of simple harmonic motion. Equating that to I multiplied by angular acceleration yields the period relation above. Everything on this page follows from those two lines.
The presets carry the standard moments of inertia already shifted to the pivot using the parallel axis theorem. A uniform rod pivoted at one end has I = mL²/3 with the centre of mass at L/2. A uniform disc pivoted at its rim has I = 3mR²/2 with d = R. A thin ring pivoted at its rim has I = 2mR². A solid sphere pivoted at its surface has I = 7mR²/5.
The headline figure is the small-angle period. The grid adds the frequency, the equivalent simple pendulum length, the radius of gyration and the period corrected for the amplitude you entered. That last figure is where a real measurement will land, because real pendulums are never released at zero amplitude.
How to Use It
- Pick the preset that matches your body, or go custom. The presets cover the four shapes that appear most often in teaching and in simple hardware. Anything else needs a moment of inertia from a solid model or a measurement.
- Take the moment of inertia about the pivot, not the centre of mass. This is the single most common error. If you have the centre-of-mass value Icm, the parallel axis theorem gives I = Icm + md².
- Measure d to the centre of mass, not to the far end. For a uniform rod pivoted at one end that is half the length; for a non-uniform body it has to be found by balancing it.
- Adjust g if precision matters. The default is the standard value, but local gravity varies by about half a per cent between the equator and the poles, which shifts the period by a quarter of a per cent.
- Read the amplitude-corrected period before comparing with a stopwatch. At 15 degrees the correction is about 0.4 per cent, which over a hundred swings is a visible discrepancy.
The Formula and a Worked Example
The period is T = 2π√(I ÷ (mgd)). The default value of g is 9.806 65 m/s², the standard acceleration of gravity, which is an exactly defined quantity in the NIST reference on constants, units and uncertainty rather than a measured one.
Work the default through. A uniform rod of length 1 m and mass 2 kg, pivoted at one end, has I = 2 × 1² ÷ 3 = 0.6667 kg·m² and d = 0.5 m. The denominator is mgd = 2 × 9.80665 × 0.5 = 9.80665. The ratio is 0.6667 ÷ 9.80665 = 0.06798, its square root is 0.26073, and multiplying by 2π gives T = 1.6382 s. The frequency is 0.6104 Hz.
The equivalent simple pendulum length is Leq = I ÷ (md) = 0.6667 ÷ 1 = 0.6667 m. That is exactly two-thirds of the rod's length, and it is the length of massless string that would give a point mass the same period. Feed 0.6667 m into the simple pendulum relation T = 2π√(L/g) and you get 1.6382 s again, as you must.
The radius of gyration is k = √(I/m) = √0.3333 = 0.5774 m. It is the distance from the pivot at which a point mass equal to the body's mass would have the same moment of inertia. The amplitude correction uses the first term of the standard series expansion, T ≈ T₀(1 + θ²/16) with θ in radians; at 15° that is 1 + 0.2618²/16 = 1.00428, giving 1.6452 s.
Why the Mass Cancels, and When It Does Not
For any uniform body the moment of inertia is proportional to mass, so I = m × (something geometric). That mass then cancels against the m in the denominator, leaving a period that depends only on shape, size and gravity. A steel rod and a balsa rod of identical dimensions swing at identical rates. This is the rotational analogue of the fact that all objects fall at the same rate in vacuum, and it is why the presets on this page ignore the mass box entirely.
The cancellation fails as soon as the body is not uniform. Add a heavy bob to the end of a light rod and the mass distribution changes, so I and d both shift but not in the same proportion, and the period changes. This is precisely how a pendulum clock is regulated: moving the bob up or down changes both terms and the ratio between them.
It also fails when part of the system does not swing with the body, such as a heavy pivot bearing, or when the suspension itself flexes. And it fails for a body with distributed elasticity, which oscillates in bending modes as well as swinging as a rigid whole. The moment of inertia calculator handles the mass distribution term for shapes beyond the presets here, and the parallel axis theorem then shifts it to your pivot.
The Centre of Oscillation and Why It Matters
The equivalent simple pendulum length points to a specific place on the body called the centre of oscillation, and it has a physical consequence you can feel. If you strike the body at that point, the impulse produces no reaction force at the pivot. Strike anywhere else and the pivot receives a jolt.
Every racket, bat and hammer user has met this. The point on a cricket bat or tennis racket where a hit produces no sting in the hands is the centre of percussion, which coincides with the centre of oscillation for a body pivoted where the hands hold it. Move the impact away from it and the residual reaction at the handle is what stings.
There is a further symmetry worth knowing: pivot the body at its centre of oscillation instead and the original pivot becomes the new centre of oscillation, with an identical period. This reciprocity is the basis of Kater's reversible pendulum, an instrument used historically to measure the local acceleration due to gravity to high precision without ever needing to know the body's mass or its moment of inertia. Adjust the pendulum until it swings with the same period about both knife edges, and the distance between them is exactly the equivalent simple length, from which g follows directly.
Where the Small-Angle Assumption Breaks
The period formula on this page comes from replacing sin θ with θ, which is exact only in the limit of zero amplitude. The true period grows with amplitude, and the growth follows a series whose first correction term is θ²/16.
At 5 degrees the error is under 0.05 per cent and no ordinary measurement will see it. At 15 degrees it is about 0.4 per cent. At 30 degrees it reaches 1.7 per cent, at 60 degrees about 7 per cent, and at 90 degrees roughly 18 per cent. Beyond about 30 degrees the single-term correction reported in the grid itself becomes inadequate and the full elliptic integral is needed.
Two other departures matter in practice. Air resistance and pivot friction damp the swing, which slowly reduces amplitude and therefore slowly reduces the period as the motion decays — a pendulum released at a large angle speeds up as it winds down. And a pivot with meaningful friction or clearance introduces its own errors, which is why precision pendulums use knife edges or flexure suspensions. The physical pendulum treatment behind all of this is set out in Georgia State University's HyperPhysics page on the physical pendulum, which derives the rotational form directly from torque and angular acceleration.
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Browse All Free Tools Talk to Arb DigitalCommon Mistakes to Avoid
- Using the centre-of-mass moment of inertia — the formula needs the value about the pivot, so add md² to the centre-of-mass figure before entering it.
- Measuring d to the end of the body — it is the distance to the centre of mass, which for a uniform rod is half the length, not the whole length.
- Treating a rod as a simple pendulum of its full length — a uniform rod pivoted at one end swings like a simple pendulum of two-thirds its length, and the difference in period is over 20 per cent.
- Comparing a large-amplitude measurement to the small-angle period — at 30 degrees the true period is already 1.7 per cent longer, which accumulates visibly over many swings.
- Expecting mass to change the answer — for any uniform body it cancels completely, and only a non-uniform mass distribution alters the period.
Related Free Tools From Arb Digital
For the idealised point-mass case, use the simple pendulum calculator, which quantifies the large-angle error for that geometry. Get the mass distribution term for shapes beyond the presets here from the moment of inertia calculator, and the general oscillator relations from the simple harmonic motion calculator. The angular velocity calculator and the torque calculator cover the rotational quantities the derivation is built on, while the kinetic energy calculator and the force calculator handle the energy and force sides of the same motion. Everything Arb Digital publishes is on the free online tools hub.
Frequently Asked Questions
A simple pendulum is an idealisation: a point mass on a massless string, whose period depends only on the string length and gravity. A physical pendulum is any real rigid body swinging about a pivot, and its period depends on how its mass is distributed through the moment of inertia as well as on the distance from pivot to centre of mass.
Not for a uniform body. The moment of inertia is proportional to mass, so the mass cancels against the same term in the denominator and the period depends only on shape, size and gravity. A steel rod and a wooden rod of identical dimensions swing at identical rates. Mass only matters when the distribution is non-uniform, such as a heavy bob on a light rod.
The one about the pivot axis, not about the centre of mass. If you have the centre-of-mass value, apply the parallel axis theorem and add the mass multiplied by the square of the pivot-to-centre-of-mass distance. Entering the centre-of-mass value directly gives a period that is far too short.
Because the equivalent simple length is the moment of inertia divided by the mass times the pivot distance. For a rod pivoted at one end that is one-third of the mass times length squared, divided by mass times half the length, which reduces to two-thirds of the length. Treating the rod as a full-length simple pendulum overestimates the period by more than 20 per cent.
It is the point on the body at the equivalent simple pendulum length from the pivot. An impulse delivered there produces no reaction force at the pivot, which is why it is also called the centre of percussion and why a bat or racket has a sweet spot that does not sting the hands. Pivot the body there instead and the original pivot becomes the new centre of oscillation with the same period.
The small-angle period is under 0.05 per cent out at 5 degrees and about 0.4 per cent out at 15 degrees. By 30 degrees the error is 1.7 per cent, by 60 degrees about 7 per cent, and by 90 degrees roughly 18 per cent. The single-term correction shown in the grid is adequate to about 30 degrees; beyond that the full elliptic integral is needed.
There is no restoring torque at all, because gravity acts through the pivot and its moment arm is zero. The body will stay in whatever orientation you leave it and will not oscillate, so the period is undefined rather than infinite in any useful sense. The calculator reports this case in words instead of returning a number.
This tool is provided for educational use. It applies the small-angle rigid-body relation with a first-order amplitude correction and models no air resistance, pivot friction, pivot clearance, suspension flexure or elastic deformation of the body. Precision gravimetry and instrument design require the full treatment and calibrated hardware.
