Theorems of Moment of Inertia

Computing the moment of inertia (MOI) from scratch about an arbitrary axis can be tedious. Two clean theorems — parallel axis and perpendicular axis — let us derive MOI about new axes from known results.

Concept

Parallel Axis Theorem. Let $I_{\text{CM}}$ be the MOI of a body of mass $M$ about an axis through its centre of mass. Then the MOI about any axis parallel to this, at distance $d$, is

$I = I_{\text{CM}} + M d^2$

Applies to all bodies.

Perpendicular Axis Theorem. For a planar (flat) lamina lying in the $xy$-plane, the MOI about the $z$-axis (perpendicular to the plane) equals the sum of MOIs about two perpendicular axes in the plane:

$I_z = I_x + I_y$

Applies only to plane laminae.

Derivation

Parallel Axis Theorem. Place the body's centre of mass at the origin. Let the original axis be the $z$-axis through CM. Let the new parallel axis pass through point $(a, b, 0)$. For a particle at $(x, y, z)$, distance from CM axis: $r^2 = x^2 + y^2$. Distance from new axis: $r'^2 = (x-a)^2 + (y-b)^2 = x^2 + y^2 - 2ax - 2by + a^2 + b^2$.

So

$I_{\text{new}} = \sum m_i r_i'^2 = \sum m_i (x_i^2 + y_i^2) - 2a \sum m_i x_i - 2b \sum m_i y_i + (a^2+b^2)\sum m_i$

Since the origin is at CM, $\sum m_i x_i = 0$ and $\sum m_i y_i = 0$:

$I_{\text{new}} = I_{\text{CM}} + M d^2$

where $d^2 = a^2 + b^2$.

Perpendicular Axis Theorem. Consider a thin lamina in the $xy$-plane. For an element at $(x, y)$:

• $I_x = \int y^2\,dm$ (axis along $x$, distance $|y|$).
• $I_y = \int x^2\,dm$ (axis along $y$, distance $|x|$).
• $I_z = \int (x^2 + y^2)\,dm$ (axis along $z$, distance $\sqrt{x^2+y^2}$).

Then

$I_z = I_x + I_y$

Note: The theorem only works for laminae because in 3D, $r^2$ for the $z$-axis is $x^2 + y^2$, not involving $z^2$ — fine for thin objects with $z \approx 0$.

Worked Example

(a) Find MOI of a thin ring (mass $M$, radius $R$) about a diameter.

(b) Find MOI of a thin ring about a tangent in its plane.

Solution:

(a) The ring is a planar lamina. MOI about its central perpendicular axis (the natural symmetry axis): $I_z = MR^2$. By symmetry, $I_x = I_y$ (two perpendicular diameters in the plane). By perpendicular axis theorem:

$I_z = I_x + I_y = 2I_{\text{diameter}}$

$I_{\text{diameter}} = \tfrac{1}{2}MR^2$

(b) A tangent line in the plane is parallel to a diameter, at distance $R$. By parallel axis theorem:

$I_{\text{tangent}} = I_{\text{diameter}} + MR^2 = \tfrac{1}{2}MR^2 + MR^2 = \tfrac{3}{2}MR^2$

Common Confusions

• Parallel axis theorem: $I_{\text{CM}}$ must be about the axis through the CM. You cannot apply it from an arbitrary parallel axis to another — only from the CM axis.
• Perpendicular axis theorem applies only to planar laminae. Don't use it on a solid sphere or a thick disc.
• $d$ in parallel axis is perpendicular distance between the two parallel axes, not arbitrary direction.
• The two perpendicular axes in the plane can be any two mutually perpendicular axes through the same point; not necessarily symmetry axes.

Key Takeaways

• Parallel axis: $I = I_{\text{CM}} + Md^2$.
• Perpendicular axis (laminae): $I_z = I_x + I_y$.
• Use to derive MOI about axes shifted parallel or perpendicular to known ones.
• For ring about diameter: $\tfrac{1}{2}MR^2$ via perp. axis.
• For ring about tangent: $\tfrac{3}{2}MR^2$ via parallel + perp. theorems.