Moment of inertia and the parallel-axis theorem: rod, disc, sphere, cylinder calculator

Choose a body and an axis, enter the mass and the sizes: the calculator finds the moment of inertia about the central axis and about a parallel axis at the distance d (the parallel-axis theorem), shows the solution Given, SI, Solution, Answer and the graph I(d). There are also modes for a system of bodies (a flywheel with rods and loads) and for a flat plate of simple figures with cut-outs.

What is the moment of inertia

The moment of inertia I measures the inertia of a body rotating about an axis: I = Σ mᵢ rᵢ² over all particles, where rᵢ is the distance to the axis. In rotation it plays the role of mass: M = I ε, the angular momentum is L = I ω and the rotational energy is E = I ω² / 2. The unit is kg·m². The radius of gyration k = √(I / m) is the distance at which the whole mass would be a point.

Moments of inertia of homogeneous bodies (about central axes)

Body and axis	I_c
Thin rod, axis through the centre ⟂ rod	m l² / 12
Disc, axis of symmetry	m R² / 2
Disc, axis along a diameter	m R² / 4
Thin ring (hoop), axis of symmetry	m R²
Ring, axis along a diameter	m R² / 2
Hollow cylinder, axis of symmetry	m (R₁² + R₂²) / 2
Solid cylinder, axis of symmetry	m R² / 2
Cylinder, axis through the centre ⟂ axis	m (3R² + L²) / 12
Solid sphere	2 m R² / 5
Thin spherical shell	2 m R² / 3
Rectangular plate, axis ⟂ through the centre	m (a² + b²) / 12

The parallel-axis (Steiner) theorem

The moment of inertia about an axis parallel to the central one at a distance d is I = I_c + m d². The central moment is the smallest among parallel axes. A rod of length l about its end: I = m l²/12 + m (l/2)² = m l²/3. A disc about an axis on its rim: 3 m R²/2. A sphere about a tangent: 7 m R²/5. For a system of bodies the moments add: I = Σ (I_c,i + mᵢ dᵢ²).

A flat plate of any shape

For a homogeneous thin plate of simple figures (holes are negative areas) the centre of gravity and the second moments of area come from the same calculation as in the centre of gravity calculator. The moment of inertia about the axis perpendicular to the plate through the centre of gravity is J_C = σ (I_x + I_y), where σ = m / A is the surface density. About another parallel axis use the parallel-axis theorem.

Rotational dynamics

A hoop of 2 kg and radius 0.5 m has I = 0.5 kg·m²; a torque of 3 N·m gives ε = M/I = 6 rad/s². A body rolling down without slipping accelerates the less the larger I/(m R²) is: sphere 2/5, cylinder 1/2, hoop 1.

FAQ

How to find the moment of inertia of a rod?

About the axis through the centre perpendicular to the rod I = m l²/12, about the axis through the end m l²/3 (the parallel-axis theorem with d = l/2). For a rod of 3 kg and 2 m they are 1 and 4 kg·m².

What does the parallel-axis theorem say and when does it apply?

I = I_c + m d², where I_c is the moment about the axis through the centre of mass and d the distance to a parallel axis. It applies only to axes parallel to the central one. A non-parallel axis needs the inertia tensor.

What are the moments of inertia of a sphere, a disc, a ring and a cylinder?

Solid sphere 2 m R²/5, thin shell 2 m R²/3, disc and solid cylinder m R²/2, ring and thin tube m R², all about the axis through the centre. Choose the body and the calculator puts in the formula and shows the solution.

How to find the moment of inertia of a system of bodies or of a composite plate?

Add the moments of the parts about one common axis, each by the parallel-axis theorem. The mode System of bodies takes rows with the kind of the body, its mass, sizes and the distance to the axis; the mode Flat plate works with rectangles and circles, also cut out, and uses the same centre-of-gravity calculation as the theoretical mechanics page.

How are the moment of inertia, the torque and the angular acceleration related?

M = I ε, like F = m a. Enter a torque M and the calculator gives the angular acceleration; enter an angular velocity ω to get the angular momentum L = I ω and the rotational energy I ω²/2.