Derive the expression for the moment of inertia of a uniform disc about an axis passing through the center and

III. Long Answer Questions · Q6

Q. Derive the expression for the moment of inertia of a uniform disc about an axis passing through the center and perpendicular to the plane.

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Concept understanding — Moment of Inertia

Moment of inertia ( $I$ ) is the rotational analogue of mass: just as mass measures a body's inherent resistance to a change in its state of linear motion (inertia), moment of inertia measures a body's resistance to a change in its state of rotational motion, for rotation about a specific axis. Its SI unit is kg m $^2$ , and its dimension is $[ML^2]$ .

For a single point mass $m_i$ at perpendicular distance $r_i$ from a fixed axis, the moment of inertia about that axis is $I=m_ir_i^2$ . For a rigid body made up of many such point masses, the moment of inertia about the axis is the sum over all of them:

I=\sum_i m_ir_i^2.

Key contrast with mass. Ordinary mass is (to excellent approximation, ignoring relativistic effects) a fixed, invariant property of a body. Moment of inertia is emphatically not invariant — it depends not only on how much mass a body has, but crucially on exactly how that mass is distributed relative to the chosen axis. The same rigid body has a different moment of inertia for every different axis of rotation one might choose, even axes lying entirely outside the physical body itself.

Computing $I$ for a continuous body. Treating an infinitesimally small mass element $dm$ , at perpendicular distance $r$ from the axis, as a point mass gives $dI=(dm)r^2$ ; integrating over the whole body gives the moment of inertia of any continuous, uniformly-distributed bulk object:

I=\int dI=\int r^2\,dm.

This integral, carried out with the mass element $dm$ expressed via the object's linear density $\lambda=M/\ell$ (for a rod), surface density $\sigma=M/(\pi R^2)$ (for a disc), or similar, yields the standard results for common shapes:

Uniform rod (mass $M$ , length $\ell$ ), axis through the center, perpendicular to the rod: $I=\dfrac{1}{12}M\ell^2$ ; axis through one end: $I=\dfrac{1}{3}M\ell^2$ .
Uniform ring (mass $M$ , radius $R$ ), axis through the center, perpendicular to the plane: $I=MR^2$ (every mass element is at the same distance $R$ from the axis, so the integral is trivial). …

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Step 1. Setting up the geometry.

Consider a solid disc of mass $M$ and radius $R$ , and find its moment of inertia about the axis through its center, perpendicular to the plane of the disc. Unlike a ring, a disc's mass is not concentrated at a single fixed distance from the axis — it is spread continuously from the center out to the rim. The disc can, however, be imagined as built up from a great many thin concentric rings of increasing radius, and the already-derived ring result can be reused for each one.

Step 2. An elemental ring.

Consider one such thin ring, of radius $r$ (where $0\le r\le R$ ), thickness $dr$ , and mass $dm$ . Being a ring itself, its own contribution to the moment of inertia is exactly the ring formula applied to this thin slice:

dI=(dm)\,r^2.

Step 3. Mass of the elemental ring.

Since the disc is uniform, its surface mass density (mass per unit area) is

\sigma=\frac{M}{\pi R^2}.

The area of the thin elemental ring is its circumference times its thickness, $2\pi r\,dr$ , so its mass is

dm=\sigma(2\pi r\,dr)=\frac{M}{\pi R^2}(2\pi r\,dr)=\frac{2M}{R^2}\,r\,dr.

Step 4. Substituting back.

dI=\frac{2M}{R^2}r^3\,dr.

Step 5. Integrating over every elemental ring.

Summing the contributions of every ring from the very center ( $r=0$ ) out to the rim ( $r=R$ ): …

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Q.Derive the expression for the moment of inertia of a uniform disc about an axis passing through the center and perpendicular to the plane.

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