Understanding the Moment of Inertia of a Disc

The moment of inertia of a disc is a fundamental concept in rotational dynamics, describing how mass is distributed with respect to a chosen axis of rotation. For JEE Main, understanding its formula, derivation, comparison with other shapes, and axis dependence is crucial for solving problems involving rotational motion and energy.

Definition and Physical Significance

The moment of inertia quantifies the resistance of a disc to changes in its rotational motion about a specified axis. It depends on the mass of the disc and the distribution of this mass relative to the rotation axis.

Formula and SI Unit for a Solid Disc

For a uniform solid disc of mass $M$ and radius $R$, the moment of inertia about the axis perpendicular to its plane and passing through its center is $I = \dfrac{1}{2}MR^2$. The SI unit of moment of inertia is kilogram meter squared (kg·m²).

Derivation of the Moment of Inertia of a Disc About Its Centre

The derivation uses calculus by dividing the disc into infinitesimal concentric rings of radius $r$ and thickness $dr$. The area of each ring is $2\pi r\, dr$.

If the surface mass density is $\sigma = \dfrac{M}{\pi R^2}$, then the mass of each ring is $dm = 2\pi r\, dr\, \sigma$.

The moment of inertia of the ring about the central axis is $dI = r^2\, dm = r^2 \cdot 2\pi r\, dr\, \sigma = 2\pi \sigma r^3\, dr$.

Integrate from $r=0$ to $r=R$:

$I = \int_{0}^{R} 2\pi \sigma r^3\, dr = 2\pi \sigma \int_{0}^{R} r^3\, dr = 2\pi \sigma \left[\dfrac{r^4}{4}\right]_0^R$

$I = 2\pi \sigma \dfrac{R^4}{4} = \dfrac{\pi \sigma R^4}{2}$

Substituting $\sigma = \dfrac{M}{\pi R^2}$, the equation becomes $I = \dfrac{1}{2}MR^2$.

A detailed understanding of this derivation enhances foundational knowledge of rotational dynamics. For further insight, review the page on Understanding Moment Of Inertia.

Moment of Inertia of a Disc About Different Axes

The moment of inertia of a disc varies with the choice of axis. For a central perpendicular axis, $I = \dfrac{1}{2}MR^2$. For any diameter in the plane, $I = \dfrac{1}{4}MR^2$, using the perpendicular axis theorem for planar bodies.

If the axis passes through the edge and is perpendicular to the plane, use the parallel axis theorem:

$I_{\text{edge}} = \dfrac{1}{2}MR^2 + MR^2 = \dfrac{3}{2}MR^2$

About a tangent in the plane, $I_{\text{tangent, in-plane}} = \dfrac{1}{4}MR^2 + MR^2 = \dfrac{5}{4}MR^2$.

These relations allow calculation of the moment of inertia for various axes essential in rotational motion problems. Connections with other planar shapes are discussed in Moment Of Inertia Of A Circle.

Moment of Inertia of a Disc with a Central Hole

For a disc with an inner hole of radius $r_1$ and an outer radius $r_2$ with total mass $M$, the mass of the hole is proportional to its area: $M_1 = M \dfrac{r_1^2}{r_2^2}$.

The moment of inertia about the central perpendicular axis is:

$I = \dfrac{1}{2}M r_2^2 - \dfrac{1}{2}M_1 r_1^2$

$I = \dfrac{1}{2}MR_2^2 - \dfrac{1}{2}M \dfrac{r_1^2}{r_2^2} r_1^2 = \dfrac{1}{2}M (r_2^2 - r_1^4 / r_2^2)$

This approach is required whenever a disc does not have uniform mass distribution across its entire area.

A comparative study with non-solid bodies such as rings and spheres can be explored in the article on Moment Of Inertia Of A Hollow Sphere.

Comparison with Other Rotational Bodies

The distribution of mass influences the moment of inertia. A thin ring of the same mass and radius as a disc has $I = MR^2$ about its central axis, which is greater than the disc’s moment of inertia since all mass is located at the maximum possible distance from the axis.

The moment of inertia of a hoop matches that of a ring. For in-depth understanding, refer to Moment Of Inertia Of An Ellipse for different geometries in rotational dynamics.

Application of Theorems to Calculate Moment of Inertia

Two fundamental theorems are often used: the parallel axis theorem and the perpendicular axis theorem. The parallel axis theorem states $I = I_{\text{center}} + Md^2$ where $d$ is the distance between axes. The perpendicular axis theorem for planar objects gives $I_{z} = I_{x} + I_{y}$ when $z$ is perpendicular to the $xy$-plane.

These theorems allow determination of a disc’s moment of inertia about axes that do not pass through its center or are not perpendicular to its plane. Further mathematical application is presented in Application Of Moment Of Inertia.

Examples: Numerical Calculation of Moment of Inertia

For a disc of mass $2\, \mathrm{kg}$ and radius $0.3\, \mathrm{m}$:

Central perpendicular axis: $I = \dfrac{1}{2} \times 2 \times (0.3)^2 = 0.09\, \mathrm{kg \cdot m^2}$

With a central hole of radius $0.1\, \mathrm{m}$: Mass of hole $= 2 \times \dfrac{(0.1)^2}{(0.3)^2} \approx 0.222\, \mathrm{kg}$

Moment of inertia of hole $= \dfrac{1}{2} \times 0.222 \times (0.1)^2 \approx 0.0011\, \mathrm{kg \cdot m^2}$

Net moment of inertia $= 0.09 - 0.0011 = 0.0889\, \mathrm{kg \cdot m^2}$

Physical Applications in Rotational Motion

The moment of inertia of a disc is crucial for analyzing rotational kinetic energy, $K = \dfrac{1}{2}I\omega^2$, where $\omega$ is angular velocity.

It is also used in calculating the angular acceleration of rolling discs, stability of rotating machinery, flywheel design, and evaluation of rolling systems. For more advanced structures, see the article on Moment Of Inertia Of A Cube.

Summary Table: Moments of Inertia for Different Shapes

Key Points and Common Mistakes

• Verify the specified rotation axis before calculation
• Distinguish between formulas for disc, ring, and hoop
• Apply the parallel and perpendicular axis theorems correctly
• Account for holes or non-uniform mass properly
• Use consistent SI units throughout calculations