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Rigid-Body Pendulum Period & Pivot Inertia

Find a physical pendulum’s small-angle period from pivot-axis inertia, mass, gravity and center-of-mass offset.

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How to use this calculator

  1. Enter the known values in the units shown. Results update as you type.
  2. Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
  3. Use the worked example to check the method. Reset restores the starting fields.

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Inputs and results stay in this browser tab. Bookify does not upload or store the values you enter.

Formula and method

For pivot-axis inertia 2 kg·m², mass 2 kg, offset 1 m and gravity 1 m/s², the equivalent simple-pendulum length is 1 m and the period is 2π seconds.

Equivalent length L = I/(md); period T = 2π√(L/g); frequency f = 1/T

Worked example

Enter these known values and leave the other values blank.

Moment of inertia about pivot axis I
2 kg·m²
Pendulum mass m
2 kg
Pivot-to-center-of-mass distance d
1 m
Gravitational acceleration
1 m/s²
Equivalent simple-pendulum length L
1 m
Small-angle oscillation period
6.283 sec
Oscillations per second
0.15915 Hz

Assumptions and limitations

  • Use the moment of inertia about the actual pivot axis, not about the center of mass. The parallel-axis theorem requires I ≥ md².
  • The body is rigid, the pivot is fixed, and oscillations are small about a stable hanging equilibrium. Damping and driving forces are excluded.
  • All quantities are positive. The center of mass must be below the pivot at equilibrium; a zero offset provides no gravitational restoring torque in this model.
  • Starting gravity is the rounded 9.81 m/s² or 32.185 ft/s². Change it for the intended location. The equivalent length is not generally the physical length of the body.

Common questions

Which moment of inertia belongs here?

Use inertia about the pivot axis. If you have center-of-mass inertia, add mass times the squared perpendicular distance between the parallel axes.

Does this work for a large swing?

The displayed period uses the small-angle approximation. Large-amplitude oscillations have a different period.

References

Bookify checks pivot-axis inertia against the parallel-axis lower bound I ≥ md². Rotational inertia in pound-force foot seconds squared uses exact international pound-force and foot definitions. Pound-mass square feet use the exact international pound and square international foot, replacing a rounded inertia factor. Feet per second squared use exactly 0.3048 meters per second squared, replacing a rounded reciprocal factor.

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