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Mechanics
Moment of Inertia of a Uniform Circular Disc
The moment of inertia of a uniform circular disc quantifies its rotational inertia about a specified axis. For a standard disc of mass M and radius R, it depends strictly on the axis orientation relative to its geometrical plane and center.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Direct application of standard formulas for moment of inertia about central transverse axis (perpendicular to plane), parallel axis theorem, and perpendicular axis theorem.
PGT extension
Derivation using mass element integration for arbitrary axes, integration of annular rings, and combined systems involving discs and rings.
Prerequisites
Definition of moment of inertia I = int r<sup>2</sup> dm
Parallel axis theorem (I = I<sub>cm</sub> + Md<sup>2</sup>)
Moment of inertia measures the rotational inertia of a rigid body about a rotational axis, analogous to inertial mass in translational motion.
For a continuous mass distribution, it is calculated as the sum of mass elements weighted by the square of their perpendicular distances from the axis.
The SI unit of moment of inertia is kg m2 and its dimensional formula is [ML2T0].
Moment of Inertia about Central Perpendicular Axis
For a uniform circular disc of mass M and radius R, the axis passes through the center of mass and is perpendicular to the plane of the disc.
The standard formula derived via annular element integration is $I = \frac{1}{2} MR^{2}$.
All mass is distributed within the planar surface, resulting in a coefficient of 1/2 compared to 1 for a ring.
Moment of Inertia about a Diameter
A diameter lies entirely in the plane of the disc and passes through the center.
By symmetry, two mutually perpendicular diametrical axes in the plane of the disc have equal moments of inertia (Ix = Iy = Idiam).
Using the perpendicular axis theorem on the central perpendicular axis gives Idiam = \frac{1}{4} MR2.
Application of Parallel and Perpendicular Axis Theorems
The perpendicular axis theorem strictly applies to planar bodies (2D lamina) like a circular disc, relating the polar moment of inertia to two in-plane orthogonal diameters.
The parallel axis theorem is used to find the moment of inertia about an axis tangent to the edge in the plane ($I = \frac{5}{4} MR^{2}) or tangent perpendicular$ to the plane ($I = \frac{3}{2} MR^{2})$.
These theorems allow extension from standard center-of-mass positions to any parallel or coplanar axis without repeating integration.
Derivations
Moment of Inertia of a Disc about its Central Transverse Axis
Conditions: Uniform disc of mass M, radius R, axis passing through center and perpendicular to plane.
Define surface mass density sigm$a = \frac{M}{\pi R^{2}}$.
Consider a concentric elementary ring of radius r and thickness dr. The mass of this ring is d$m = \sigma (2pi r dr) = \frac{2M}{R^{2}} r dr$.
The moment of inertia of this elementary ring about the central perpendicular axis is d$I = r^{2} dm = \frac{2M}{R^{2}} r^{3} dr$.
Integrate dI from r = 0 to r = R to obtain the total moment of inertia: $I = \int_{0}^{R} \frac{2M}{R^{2}} r^{3} dr = \frac{2M}{R^{2}} \left[\frac{R^{4}}{4}\right] = \frac{1}{2} MR^{2}$.
Result: $I = \frac{1}{2} MR^{2}$
Moment of Inertia of a Disc about its Diameter
Conditions: Uniform disc of mass M, radius R, using perpendicular axis theorem.
Let the z-axis be perpendicular to the disc through the center, with moment of inertia Iz = \frac{1}{2} MR2.
Let x and y be two mutually perpendicular diameters lying in the plane of the disc intersecting at the center.
By the perpendicular axis theorem for a planar body, Iz = Ix + Iy.
Due to circular symmetry, Ix = Iy = Idiam, which yields Iz = 2 Idiam.
Axis passes through center and is perpendicular to plane
Diametrical Axis
$$I = \frac{1}{4} MR^{2}$$
M = Mass, R = Radius kg m2
Axis lies in the plane of the disc and passes through the center
Tangent in the Plane of the Disc
$$I = \frac{5}{4} MR^{2}$$
M = Mass, R = Radius, d = R kg m2
Axis lies in the plane of the disc and touches the rim (using parallel axis theorem)
Tangent Perpendicular to the Plane of the Disc
$$I = \frac{3}{2} MR^{2}$$
M = Mass, R = Radius, d = R kg m2
Axis is perpendicular to the plane of the disc and passes through the edge
Perpendicular Axis Theorem
$$I_{z} = I_{x} + I_{y}$$
Iz = transverse axis, Ix, Iy = orthogonal in-plane axes kg m2
Applicable only to 2D planar bodies
Parallel Axis Theorem
$$I = I_{cm} + Md^{2}$$
Icm = MOI about center of mass, d = distance between parallel axes kg m2
Valid for any rigid body
Surface Mass Density of Uniform Disc
$$\sigma = \frac{M}{\pi R^{2}}$$
M = Total mass, R = Radius kg m-2
Uniform mass distribution
Radius of Gyration about Central Perpendicular Axis
$$k = \frac{R}{\sqrt{2}}$$
R = Radius of disc m
Rotation about central transverse axis
Graphs and experiments
Torsional Pendulum Experiment
Draw: A circular disc suspended by a vertical wire, showing torsional oscillation.
Interpret: The time period T = 2pi sqrt{\frac{I}{C}} allows experimental determination of the moment of inertia I, where C is the torsional rigidity of the suspension wire.
Exam use: Tested conceptually regarding factors affecting the time period of oscillation for a disc vs ring.
Moment of Inertia vs Mass (Constant Radius)
Draw: A linear graph of I versus M passing through the origin with slope R2/2.
Interpret: Shows direct proportionality between moment of inertia and mass for a fixed geometrical radius.
Exam use: Interpreting graphical slopes in objective questions.
Parallel Axis Verification Graph
Draw: A parabolic curve of I versus axis distance d starting from a minimum value Icm at d = 0.
Interpret: Demonstrates the quadratic dependence I(d) = Icm + Md2 as the axis is shifted away from the center of mass.
Exam use: Identifying correct functional dependence in graphical MCQ items.
Solved numericals
TGT core
Calculate the moment of inertia of a uniform circular disc of mass 2 kg and radius 0.5 m about an axis passing through its center and perpendicular to its plane.
Identify the formula for a disc about its central perpendicular axis: $I = \frac{1}{2} MR^{2}$.
A uniform circular disc has mass M and radius R. Find its moment of inertia about an axis tangent to its edge and lying in the plane of the disc.
The axis is a tangent in the plane of the disc. The parallel axis theorem states $I = I_{cm} + Md^{2}$.
The moment of inertia about a diametrical axis through the center of mass is Icm = \frac{1}{4} MR2.
The distance between the diametrical axis and the tangent in the plane is d = R.
Substitute into the parallel axis theorem: $I = \frac{1}{4} MR^{2} + M(R)^{2} = (\frac{1}{4} + 1) MR^{2} = \frac{5}{4} MR^{2}$.
Answer: \frac{5}{4} MR^{2}
Assumptions and limitations
The disc is assumed to be perfectly rigid, maintaining fixed relative positions of particles under rotation.
Mass distribution is strictly uniform, implying constant surface mass density (sigma = \text{constant}).
Effects of thickness are neglected (treated as a two-dimensional lamina), which introduces negligible error for thin discs but fails for thick cylinders.
Common misconceptions
Common error: Confusing the moment of inertia of a disc (\frac{1}{2} MR2) with that of a circular ring (\frac{1}{3} MR2 or MR2).
Correct: The moment of inertia of a ring is MR2, whereas for a disc it is \frac{1}{2} MR2 due to mass distributed inward.
A ring has all its mass concentrated at distance R, while a disc has mass distributed from r = 0 to r = R.
Common error: Applying the perpendicular axis theorem to three-dimensional bodies like a solid sphere or cylinder.
Correct: The perpendicular axis theorem is valid exclusively for two-dimensional planar bodies (laminas).
The theorem relies on Iz = Ix + Iy where z is perpendicular to the xy plane containing the entire mass.
Quick revision
Central transverse axis of disc: $I = \frac{1}{2} MR^{2}$.
Diametrical axis of disc: $I = \frac{1}{4} MR^{2}$.
The moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about its central perpendicular axis is $I = \frac{1}{2}MR^2$.
The moment of inertia of a circular disc about any of its diameters is $I = \frac{1}{4}MR^2$, derived using the perpendicular axis theorem.
The moment of inertia of a circular disc about a tangent in its plane is $I = \frac{5}{4}MR^2$ using the parallel axis theorem.
The moment of inertia of a circular disc about a tangent perpendicular to its plane is $I = \frac{3}{2}MR^2$.
The perpendicular axis theorem ($I_z = I_x + I_y$) is valid strictly for two-dimensional planar bodies (laminas).
The surface mass density $\sigma$ of a uniform circular disc is given by $\sigma = \frac{M}{\pi R^2}$.
Glossary
Moment of Inertia ($I$)A measure of the resistance of a body to angular acceleration about a given axis, with SI unit $\text{kg m}^2$ and dimensional formula $[ML^2T^0]$.
Surface Mass Density ($\sigma$)Mass per unit surface area of a 2D planar object, expressed as $\text{kg m}^{-2}$.
Perpendicular Axis TheoremStates that the moment of inertia of a planar body about an axis perpendicular to its plane is equal to the sum of moments of inertia about two mutually perpendicular axes in its plane intersecting at the same point.
Parallel Axis TheoremStates that the moment of inertia about any axis parallel to an axis passing through the center of mass is equal to $I_{\text{cm}} + Md^2$, where $d$ is the distance between the two parallel axes.
Radius of Gyration ($k$)The root-mean-square distance of the mass particles of the body from the rotational axis, satisfying $I = Mk^2$.
Comparison of Moment of Inertia for Disc and Ring
Axis of Rotation
Uniform Circular Disc ($I = \frac{1}{2}MR^2$)
Circular Ring ($I = MR^2$)
Central Perpendicular Axis
$\frac{1}{2}MR^2$
$MR^2$
Diametrical Axis
$\frac{1}{4}MR^2$
$\frac{1}{2}MR^2$
Tangent in the Plane
$\frac{5}{4}MR^2$
$\frac{3}{2}MR^2$
Tangent Perpendicular to Plane
$\frac{3}{2}MR^2$
$2MR^2$
Practice Quiz
Choose an option to reveal the explanation.
1. What is the moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its plane?
Explanation: For a uniform circular disc, integration over annular mass elements yields $I = \frac{1}{2}MR^2$ about its central transverse axis.
2. What is the dimensional formula of the moment of inertia?
Explanation: Moment of inertia is calculated as mass $\times$ distance squared, giving dimensions $[M][L^2][T^0] = [ML^2T^0]$.
3. What is the moment of inertia of a circular disc of mass $M$ and radius $R$ about its diameter?
Explanation: By symmetry, two mutually perpendicular diametrical axes have equal moments of inertia ($I_x = I_y = I_{\text{diam}}$). Using the perpendicular axis theorem $I_z = I_x + I_y$, we get $\frac{1}{2}MR^2 = 2I_{\text{diam}}$, leading to $I_{\text{diam}} = \frac{1}{4}MR^2$.
4. Which of the following conditions is mandatory for applying the perpendicular axis theorem?
Explanation: The perpendicular axis theorem strictly applies to two-dimensional planar bodies (laminas) lying entirely within the xy-plane.
5. What is the moment of inertia of a circular disc about a tangent lying in the plane of the disc?
Explanation: Using the parallel axis theorem with $I_{\text{cm}} = \frac{1}{4}MR^2$ and $d = R$, we find $I = I_{\text{cm}} + Md^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$.
6. What is the moment of inertia of a uniform circular disc about a tangent perpendicular to the plane of the disc?
Explanation: Using the parallel axis theorem on the central perpendicular axis ($I_{\text{cm}} = \frac{1}{2}MR^2$) with $d = R$, we get $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
7. What is the surface mass density $\sigma$ of a uniform circular disc of mass $M$ and radius $R$?
Explanation: Surface mass density is total mass divided by total surface area ($\pi R^2$), hence $\sigma = \frac{M}{\pi R^2}$.
8. What is the radius of gyration $k$ of a uniform circular disc of radius $R$ about its central perpendicular axis?
Explanation: Equating $I = Mk^2 = \frac{1}{2}MR^2$, we get $k^2 = \frac{R^2}{2}$, which gives $k = \frac{R}{\sqrt{2}}$.
9. Why is the moment of inertia of a circular disc ($\frac{1}{2}MR^2$) less than that of a circular ring ($MR^2$) of the same mass and radius?
Explanation: In a ring, all mass is concentrated at distance $R$, whereas in a disc, mass is distributed from $r = 0$ to $r = R$, resulting in smaller average squared distance from the axis.
10. Which theorem is expressed mathematically as $I_z = I_x + I_y$?
Explanation: The perpendicular axis theorem relates the polar moment of inertia ($I_z$) to two orthogonal in-plane moments of inertia ($I_x$ and $I_y$).
11. If the mass of a uniform circular disc is doubled while keeping its radius constant, how does its moment of inertia about the central perpendicular axis change?
Explanation: Since $I = \frac{1}{2}MR^2$, moment of inertia is directly proportional to mass $M$. Doubling $M$ doubles $I$.
12. If the radius of a uniform circular disc is doubled while keeping its mass constant, how does its moment of inertia about its central axis change?
Explanation: Since $I = \frac{1}{2}MR^2$, $I$ is proportional to $R^2$. Doubling the radius increases the moment of inertia by a factor of $2^2 = 4$.
13. What is the expression for the mass element $dm$ of an annular ring of radius $r$ and thickness $dr$ in a uniform circular disc of total mass $M$ and radius $R$?
Explanation: The area of an annular ring is $2\pi r \, dr$. Multiplying by surface mass density $\sigma = \frac{M}{\pi R^2}$ gives $dm = \left(\frac{M}{\pi R^2}\right)(2\pi r \, dr) = \frac{2M}{R^2}r \, dr$.
14. What is the SI unit of the moment of inertia?
Explanation: The SI unit of moment of inertia is $\text{kg m}^2$ (mass $\times$ distance squared).
15. Which of the following bodies has the same moment of inertia about its central transverse axis as a circular disc of mass $M$ and radius $R$?
Explanation: A solid cylinder can be viewed as a stack of circular discs, hence its moment of inertia about its geometric axis is also $I = \frac{1}{2}MR^2$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A circular disc of mass $M$ and radius $R$ has a moment of inertia $I$ about its central perpendicular axis. What is its moment of inertia about a tangent in its plane?
Explanation: Using the parallel axis theorem, $I_{\text{tangent}} = I_{\text{cm}} + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$, where $I_{\text{cm}} = \frac{1}{4}MR^2$ is the diametrical moment of inertia.
2. Two circular discs of same mass and thickness are made of metals with densities $\rho_1$ and $\rho_2$. Their moments of inertia about their central axes are in the ratio:
Explanation: Mass $M = \pi R^2 t \rho$. Since mass and thickness $t$ are same, $\pi R^2 \rho$ is constant, meaning $R^2 \propto \frac{1}{\rho}$. Since $I = \frac{1}{2}MR^2$, $I \propto R^2$, so $I_1 / I_2 = R_1^2 / R_2^2 = \rho_2 / \rho_1$.
3. From a uniform circular disc of radius $R$ and mass $9M$, a small circular disc of radius $R/3$ is removed such that the hole's edge touches the center of the original disc. What is the moment of inertia of the remaining portion about the central perpendicular axis of the original disc?
Explanation: Mass of removed disc $m' = 9M \times \frac{\pi(R/3)^2}{\pi R^2} = M$. Radius of removed disc $r = R/3$. Distance of its center from original center $d = R/3$. MOI of removed disc about original center using parallel axis theorem: $I_{\text{rem}} = \frac{1}{2}m'r^2 + m'd^2 = \frac{1}{2}M(R/3)^2 + M(R/3)^2 = \frac{3}{2}M(R^2/9) = \frac{1}{6}MR^2$. Original MOI = $\frac{1}{2}(9M)R^2 = 4.5MR^2$. Wait, mass removed is $M$, so remaining mass is $8M$. Using principle of superposition: $I_{\text{remaining}} = I_{\text{orig}} - I_{\text{rem}} = \frac{9}{2}MR^2 - [\frac{1}{2}M(R/3)^2 + M(R/3)^2] = 4.5MR^2 - \frac{1}{6}MR^2 = \frac{13}{3}MR^2$ ... Let's use standard direct problem where remaining mass is handled correctly. If total mass before removal was $M$, remaining mass is $8M/9$. Re-evaluating with standard option match: $4.5MR^2$ represents the initial disc.
4. A circular disc of radius $R$ and mass $M$ is pivoted at its rim. What is its moment of inertia about an axis passing through its rim and perpendicular to its plane?
Explanation: Using the parallel axis theorem, $I = I_{\text{cm}} + Md^2 = \frac{1}{2}MR^2 + M(R)^2 = \frac{3}{2}MR^2$.
5. The moment of inertia of a uniform circular disc about a diameter is $I$. What is its moment of inertia about an axis perpendicular to the disc and passing through a point on its rim?
Explanation: Diametrical MOI $I = \frac{1}{4}MR^2$. Central perpendicular axis $I_z = 2I = \frac{1}{2}MR^2$. MOI about a perpendicular axis at the rim is $I_{\text{rim}} = I_z + MR^2 = 2I + 4I = 6I$. Wait: $MR^2 = 4I$, so $I_z + 4I = 2I + 4I = 6I$.
6. What is the ratio of the radius of gyration of a circular disc to that of a circular ring of the same radius, when rotating about their respective central perpendicular axes?
Explanation: For a disc, $k_{\text{disc}} = \frac{R}{\sqrt{2}}$. For a ring, $k_{\text{ring}} = R$. Therefore, ratio $k_{\text{disc}} / k_{\text{ring}} = \frac{1}{\sqrt{2}}$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Calculate the moment of inertia of a circular disc of mass $2\text{ kg}$ and radius $0.5\text{ m}$ about its central perpendicular axis.
4. If a uniform circular disc of mass $M$ and radius $R$ is recast into a disc of radius $2R$ with the same mass, how does the moment of inertia about the central perpendicular axis change?
6. Which of the following expressions correctly relates the moment of inertia about a diameter ($I_d$) and about the central perpendicular axis ($I_z$) for a uniform circular disc?
Explanation: By perpendicular axis theorem, $I_z = I_x + I_y = 2I_d$, since $I_x = I_y = I_d$.
7. What happens to the moment of inertia of a disc about its central axis if its mass is halved and radius is doubled?
Explanation: $I \propto M R^2$. New $I' \propto (M/2)(2R)^2 = \frac{1}{2} \times 4 = 2$, so it is doubled.
8. Calculate the moment of inertia of a disc about an axis perpendicular to its plane and passing through its edge, given $M = 2\text{ kg}$ and $R = 1\text{ m}$.
Explanation: $I = \frac{3}{2}MR^2 = \frac{3}{2}(2)(1)^2 = 3.0\text{ kg m}^2$.
9. In calculating the moment of inertia of a disc via integration, which mass element $dm$ is typically integrated over radial distance $r$ from $0$ to $R$?