Moment of inertia

The moment of inertia $I$ measures how difficult it is to make a body rotate about a given axis. It is the rotational analogue of mass: just as mass expresses a body’s resistance to changing its translational state of motion, the moment of inertia expresses its resistance to changing its state of rotation.

The crucial difference is that $I$ is not an intrinsic property of the body, but depends on how the mass is distributed relative to the chosen axis. The same object, rotated about different axes, has different moments of inertia. Mass located far from the axis counts much more, because to make it move at the same angular velocity it must be given a greater linear velocity.

Principle — Moment of inertia

For a system of point masses: $\ev{I = \sum_i m_i\,r_i^2}$ where $r_i$ is the distance of the $i$-th mass from the axis of rotation.

The factor $r_i^2$ (distance squared) explains why geometry matters so much: doubling the distance of a mass from the axis quadruples its contribution to $I$. Concentrating mass near the axis makes the body “agile” in rotation; moving it towards the edge makes it “sluggish”.

Notable moments of inertia (axis through the centre)

Body	Moment of inertia
Solid cylinder	$I = \tfrac{1}{2}mR^2$
Hollow cylinder (ring)	$I = mR^2$
Solid sphere	$I = \tfrac{2}{5}mR^2$
Hollow sphere	$I = \tfrac{2}{3}mR^2$
Rod (axis at centre)	$I = \tfrac{1}{12}mL^2$

For equal mass and radius, the ring has the largest moment of inertia (all the mass is at the maximum radius), the solid sphere the smallest (much mass close to the centre). This is exactly why, when let roll down the same slope, the solid sphere wins the race and the ring comes last.

The parallel axis theorem (Huygens–Steiner)

If the moment of inertia $I_\text{CM}$ about an axis through the centre of mass is known, the one about a parallel axis at distance $d$ is:

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

For example, for a uniform rod the moment about the centre is $\tfrac{1}{12}ML^2$; about one end (at distance $d = L/2$) it becomes

$I_\text{end} = \tfrac{1}{12}ML^2 + M\left(\tfrac{L}{2}\right)^2 = \tfrac{1}{3}ML^2$

a value four times larger, because now all the mass rotates, on average, farther from the axis.

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Topics: Dinamica rotazionale Concepts: Momento d’inerzia Objects: Corpo rigido

Related exercises: Problem — Hinged rod falling from horizontal · Problem — Ranking four rolling objects · Problem — Ranking the moments of inertia of three cylinders