Physical Pendulum Calculator
A physical pendulum is any rigid body that swings about a fixed pivot under gravity. Unlike a simple pendulum (which concentrates mass at a point), a physical pendulum has a distributed mass with a known moment of inertia. Enter the moment of inertia, mass, pivot-to-centre-of-mass distance, initial angle, and gravitational acceleration below to get the oscillation period, frequency, angular frequency, equivalent simple-pendulum length, maximum potential energy, and maximum velocity at the bottom of the swing.
Formula
Worked example
A uniform thin rod of mass 2 kg and length 1.0 m pivots about one end. The moment of inertia about the pivot is I = (1/3)*m*L^2 = (1/3)*2*1^2 = 0.667 kg-m^2. The centre of mass is at d = 0.5 m from the pivot. With g = 9.81 m/s^2 and initial angle 15 deg: T = 2*pi*sqrt(0.667 / (2*9.81*0.5)) = 2*pi*sqrt(0.068) = 1.642 s. Frequency = 0.609 Hz. Energy = 2*9.81*0.5*(1-cos(15deg)) = 0.342 J.
What is a physical pendulum?
A simple (or mathematical) pendulum idealises the swinging mass as a point located at the end of a massless string. A physical pendulum is the real-world generalisation: any rigid body free to rotate about a fixed horizontal pivot under the influence of gravity. A swinging door, a metronome arm, a compound pendulum in a clock, and a gymnast on a high bar all behave as physical pendulums. The key difference from the simple case is that mass is distributed throughout the body, so the oscillation depends on how that mass is arranged relative to the pivot, captured by the moment of inertia I.
The formula and its terms
The period of small oscillations is T = 2*pi*sqrt(I / (m*g*d)), where I is the moment of inertia of the body about the pivot axis (in kg-m^2), m is the total mass (kg), g is gravitational acceleration (m/s^2), and d is the distance from the pivot to the centre of mass (m). The formula is exact in the small-angle limit (angles below about 15 degrees) and remains a good approximation up to roughly 23 degrees with less than 5 percent error. The equivalent simple-pendulum length L_eq = I / (m*d) lets you replace any physical pendulum with a point-mass pendulum of that length and get the same period. The energy at the maximum angle is E = m*g*d*(1 - cos(theta)), all of which converts to kinetic energy at the equilibrium position, giving maximum angular velocity omega_max = sqrt(2E / I) and maximum centre-of-mass speed v_max = omega_max * d.
Moment of inertia and the parallel-axis theorem
The moment of inertia I about the pivot is the key parameter that distinguishes a physical pendulum from a simple one. For standard shapes you can look up I_cm (the moment about the centre of mass) from tables, then apply the parallel-axis theorem: I_pivot = I_cm + m*d^2, where d is the distance from the centre of mass to the pivot. For example, a uniform thin rod of length L pivoting about one end has I_cm = (1/12)*m*L^2 and I_pivot = (1/12)*m*L^2 + m*(L/2)^2 = (1/3)*m*L^2. More complex shapes can be treated by summing contributions from their component parts.
Small-angle approximation and accuracy limits
The period formula T = 2*pi*sqrt(I/(mgd)) relies on the approximation sin(theta) = theta (in radians), which is accurate to 0.5 percent for angles up to about 10 degrees, and to 5 percent for angles up to about 23 degrees. At 45 degrees the error reaches around 18 percent, and the true period (given by an elliptic integral) is noticeably longer. For most pendulum clock and physics-lab applications the amplitude is kept small precisely to maintain constant period. If your initial angle is large, the calculator flags this and you should use the exact elliptic-integral solution or a numerical simulation for precision work.
Common moment of inertia values for uniform rigid bodies
| Shape | Axis | I about centre of mass |
|---|---|---|
| Thin rod, length L | Perpendicular to rod at CM | (1/12) m L^2 |
| Solid disk, radius R | Through centre, perpendicular to face | (1/2) m R^2 |
| Solid sphere, radius R | Any diameter | (2/5) m R^2 |
| Hollow sphere (thin), radius R | Any diameter | (2/3) m R^2 |
| Rectangular plate, w x h | Perpendicular to plate at CM | (1/12) m (w^2 + h^2) |
| Hollow cylinder, radii R1 and R2 | Cylinder axis | (1/2) m (R1^2 + R2^2) |
For a body of mass m. Use the parallel-axis theorem (I_pivot = I_cm + m*d^2) to shift from the centre of mass to any parallel pivot axis.
Frequently asked questions
How is a physical pendulum different from a simple pendulum?
A simple pendulum treats all the mass as concentrated at a single point at the end of a massless rod. A physical pendulum is a real rigid body with mass spread throughout. The period depends on the full moment of inertia about the pivot rather than just the length to the mass. A simple pendulum is actually a special case of a physical pendulum where the equivalent length L_eq equals the rod length.
What is the moment of inertia and how do I find it?
The moment of inertia I measures how mass is distributed relative to the rotation axis. Look up the formula for your shape in the reference table (or a physics textbook), calculate I_cm for that formula, then add m*d^2 using the parallel-axis theorem to shift from the centre of mass to your pivot. For irregular bodies, I can be determined experimentally by measuring the oscillation period T and rearranging: I = (T / (2*pi))^2 * m * g * d.
What does the equivalent simple-pendulum length mean?
L_eq = I / (m*d) is the length of a simple (point-mass) pendulum that would oscillate with exactly the same period. It is useful for quick comparisons and for visualising the effective geometry of a complex shape. For a uniform rod pivoting about one end, L_eq = (2/3)*L, which is two-thirds of the rod length.
How do I set gravitational acceleration for another planet?
Change the gravitational acceleration field. Standard values: Moon 1.62 m/s^2, Mars 3.72 m/s^2, Earth 9.81 m/s^2, Jupiter 24.79 m/s^2. Because the period scales as 1/sqrt(g), a pendulum on the Moon swings about 2.46 times slower than the same pendulum on Earth.
At what angle does the small-angle approximation break down?
The approximation is within 0.5 percent error up to about 10 degrees, within 5 percent up to about 23 degrees, and within 18 percent up to 45 degrees. For precise work above 20 degrees, use the exact solution involving a complete elliptic integral of the first kind, or a numerical ODE solver.