Moment of Inertia of a Circle Explained with Formulas and Examples

The moment of inertia of a circle quantifies the resistance offered by a circular body to rotational motion about a specified axis. This property is central in rotational dynamics, strength of materials, and JEE Main Physics, especially for analyzing circular sections such as discs and rings.

Physical Meaning of Moment of Inertia for a Circle

The moment of inertia describes how mass is distributed relative to the axis of rotation. For a circle, this distribution depends on whether the circle is solid or hollow, and the position of the axis. Greater mass farther from the axis increases the moment of inertia.

Moment of Inertia of a Solid Circle: Core Formulas

For a solid circular disc of mass $M$ and radius $R$, the moment of inertia differs according to the orientation of the axis. The common cases relevant for JEE Main and board exams are summarized below.

The value $\dfrac{1}{2}MR^2$ is used when the axis passes perpendicularly through the centre of the disc. When the axis is along the diameter, the value becomes $\dfrac{1}{4}MR^2$.

Derivation of Moment of Inertia for a Solid Circle

The moment of inertia of a solid circle about an axis through its center and perpendicular to the plane can be derived by considering the disc as composed of thin concentric rings.

Consider a disc of mass $M$ and radius $R$. Select a differential ring of radius $r$ and thickness $dr$. The mass of the ring is $dm = \dfrac{M}{\pi R^2}\cdot 2\pi r dr$. The moment of inertia of this ring about the central axis is $dI = dm \cdot r^2$. Integrating from $r = 0$ to $R$ gives:

$I = \int_0^R r^2 \left(\dfrac{M}{\pi R^2}\cdot 2\pi r dr\right) = 2M \int_0^R \dfrac{r^3}{R^2} dr = 2M \left[\dfrac{r^4}{4R^2}\right]_0^R = 2M \cdot \dfrac{R^4}{4R^2} = \dfrac{1}{2}MR^2$

Thus, for a solid circle (disc), $I = \dfrac{1}{2}MR^2$ about the central perpendicular axis.

Moment of Inertia for Circular Rings and Hollow Circles

A thin ring of mass $M$ and radius $R$, with axis perpendicular to its plane through the center, has the moment of inertia:

$I = MR^2$

For a thick ring (annular disc) with inner radius $R_1$ and outer radius $R_2$, the moment of inertia about the center perpendicular axis is:

$I = \dfrac{M}{2}(R_1^2 + R_2^2)$

In problems involving circles with holes, subtract the moment of inertia of the missing part from that of the original disc.

Axis Orientation and Its Significance

The moment of inertia of a circular object varies significantly with the orientation of the axis. For planar bodies, the perpendicular axis theorem is commonly used: $I_z = I_x + I_y$, where $I_z$ is perpendicular to the plane and $I_x$, $I_y$ are in-plane axes.

Proper identification of the axis is essential in exam questions to avoid incorrect application of formulas. Reference to the Moment Of Inertia Overview gives more on the role of axis selection in rotational problems.

Moment of Inertia of a Circle in Terms of Diameter

If $d$ is the diameter of the circle, $R = \dfrac{d}{2}$. Thus, the moment of inertia for a solid circle about the diameter becomes $I = \dfrac{1}{4}M\left(\dfrac{d}{2}\right)^2 = \dfrac{Md^2}{16}$. For the perpendicular axis, $I = \dfrac{1}{2}M\left(\dfrac{d}{2}\right)^2 = \dfrac{Md^2}{8}$.

Polar Moment of Inertia for Circular Sections

The polar moment of inertia is used for torsional resistance calculations and is given for a circular area as $J = \dfrac{\pi R^4}{2}$. For mass-based moment of inertia (not area), use $I = \dfrac{1}{2}MR^2$ for the axis perpendicular to the plane.

A detailed comparison with related shapes is available in the Moment Of Inertia Of A Disc article, which highlights distinctions in rotational properties.

Solved Example: Solid Circle Moment of Inertia

Given: solid disc with mass $M = 0.8$ kg and radius $R = 0.2$ m. Calculate its moment of inertia about its central axis perpendicular to the plane.

Using $I = \dfrac{1}{2}MR^2$,

$I = \dfrac{1}{2} \times 0.8 \times (0.2)^2 = 0.016$ kg$\cdot$m$^2$

If the axis were along the diameter, use $I = \dfrac{1}{4}MR^2 = 0.008$ kg$\cdot$m$^2$.

Further solved examples involving tangent and non-standard axes can be found under advanced problems in Moment Of Inertia Of A Cube.

Key Points and Exam Practice for JEE

When solving moment of inertia problems involving circles, always check whether the object is solid, hollow, or annular, and identify the reference axis. Apply the correct formula as per the orientation and mass distribution.

• Determine if the object is a disc or a ring
• Identify axis: diameter, centroid, or perpendicular
• Use the perpendicular axis theorem for plane bodies
• Be careful with unit conversions (g $\to$ kg, cm $\to$ m)

Comparing circular and other standard shapes aids rapid problem-solving. Additional resources can be found at Moment Of Inertia Of A Hollow Sphere and Moment Of Inertia Of A Cone.

Moment of Inertia of a Circle: Related Shapes

The moment of inertia for a circle differs from other plane and solid figures due to unique mass or area distribution. Accurate knowledge is necessary for solving rotational motion and mechanical engineering problems in competitive exams.

Explore further comparisons with triangles at Moment Of Inertia Of A Triangle for diverse exam applications.