The moment of inertia is a quantitative measure of a rigid body's rotational inertia, representing its resistance to angular acceleration about a rotational axis. It depends on both the mass distribution relative to the axis of rotation and the chosen axis itself.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Calculation of moment of inertia for standard symmetrical bodies (ring, disc, rod, sphere, cylinder) using parallel and perpendicular axes theorems, and elementary problems on rotational kinetic energy.
PGT extension
General tensor formulation of inertia, principal moments of inertia, products of inertia, complex composite body integrations, and non-inertial frame rotational dynamics.
Prerequisites
Newton's laws of motion for rotational systems
Centre of mass definition and coordinate calculation
Calculus integration techniques for continuous mass distribution
Definition and Physical Significance
Mathematically defined as $I = \sum(m_i r_i^2)$ for discrete systems and $I = integral(r^2 dm)$ for continuous systems.
Acts as the rotational analogue of mass in linear translational motion.
Has SI units of kg m^2 and dimensional formula [ML^2].
Theorem of Perpendicular Axes
Applicable strictly to plane laminar (two-dimensional) bodies.
States that the moment of inertia about an axis perpendicular to the plane (I_z) equals the sum of moments of inertia about two mutually perpendicular axes in the plane meeting at the origin (I_x + I_y).
Expressed mathematically as $I_z = I_x + I_y$.
Theorem of Parallel Axes
Applicable to any rigid body of any shape (1D, 2D, or 3D).
States that the moment of inertia about any axis is equal to the moment of inertia about a parallel axis passing through the centre of mass plus the product of total mass and the square of the perpendicular distance between the axes.
Expressed mathematically as $I = I_cm + Md^2$.
Radius of Gyration
Defined as the root-mean-square distance of the constituent particles from the axis of rotation.
Expressed as $k = \sqrt(I / M)$, where M is total mass and I is moment of inertia.
Represents the geometric distribution of mass relative to the axis.
Derivations
Moment of Inertia of a Uniform Thin Rod
Conditions: Axis passing through its centre and perpendicular to its length, length L, mass M.
Consider a uniform thin rod of mass M and length L aligned along the x-axis from $x = -L/2$ to $x = L/2$.
Define linear mass density lambd$a = M/L$. Consider a small mass element d$m = \lambda dx = (M/L) dx at a distance x from the centre$.
Integrate the differential moment of inertia d$I = x^2 dm from -L/2$ to L/2.
Conditions: Axis passing through its centre and perpendicular to its plane, radius R, mass M.
Consider a thin circular ring of total mass M and radius R.
Since all mass elements of the ring are situated at the same perpendicular distance R from the axis passing through the centre perpendicular to the plane:
Write the definition of moment of inertia $I = integral r^2 dm$.
Substitute r = R to get $I = R^2 integral dm$.
Result: $I = MR^2$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
General Discrete System
$$I = \sum_{i=1}^{n} m_i r_i^2$$
$m_i = mass of particle i$, $r_i = perpendicular distance from axis$ kg m^2
Discrete point masses
General Continuous System
$$I = \int r^2 dm$$
dm = differential mass element, r = perpendicular distance kg m^2
Continuous mass distribution
Perpendicular Axes Theorem
$$I_z = I_x + I_y$$
I_x, $I_y = axes in plane; I_z = axis perpendicular$ to plane kg m^2
Planar (2D) bodies only
Parallel Axes Theorem
$$I = I_{cm} + M d^2$$
$I_cm = moment of inertia about centre of mass$, d = distance between parallel axes kg m^2
Any rigid body
Ring about Diametrical Axis
$$I = \frac{1}{2} MR^2$$
M = mass, R = radius kg m^2
Thin ring
Disc about Diametrical Axis
$$I = \frac{1}{4} MR^2$$
M = mass, R = radius kg m^2
Uniform circular disc
Solid Sphere about Diameter
$$I = \frac{2}{5} MR^2$$
M = mass, R = radius kg m^2
Uniform solid sphere
Hollow Sphere (Spherical Shell) about Diameter
$$I = \frac{2}{3} MR^2$$
M = mass, R = radius kg m^2
Thin spherical shell
Graphs and experiments
Torsional Pendulum Experiment
Draw: Graph of time period squared (T^2) versus moment of inertia (I) of added standard masses.
Interpret: A straight line passing through the origin or intercepting the axis confirming $T = 2pisqrt{I/C}$, allowing determination of torsional constant C.
Exam use: Directly tested in experimental physics viva and conceptual numericals on torsional oscillations.
Radius of Gyration Dependence on Axis Position
Draw: Plot of moment of inertia I versus squared distance x^2 from centre of mass using parallel axis theorem.
Interpret: Linear graph with slope equal to total mass M and intercept equal to I_cm.
Exam use: Used to evaluate graphical problems based on the parallel axes theorem.
Flywheel Rotational Dynamics
Draw: Plot of angular velocity versus time during acceleration and retardation phases.
Interpret: Slopes yield angular acceleration and deceleration, which help compute frictional torque and moment of inertia of the flywheel.
Exam use: Common laboratory method translated into conceptual MCQ items in PGT Physics.
Solved numericals
TGT core
Find the moment of inertia of a uniform circular disc of mass 2 kg and radius 0.5 m about an axis passing through its centre and perpendicular to its plane.
Identify the formula for the moment of inertia of a disc about its central transverse axis: $I = (1/2) M R^2$.
Calculate the moment of inertia of a thin uniform rod of mass 1.2 kg and length 1 m about an axis passing through one of its ends and perpendicular to its length.
Use the parallel axis theorem or direct integration for an axis at the end: $I = (1/3) M L^2$.
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.
First, find the moment of inertia of the disc about a diametrical axis in its plane: $I_diam = (1/4) M R^2$.
Apply the parallel axes theorem between the diametrical axis and the tangent axis in the plane, where the separation distance is d = R.
Calculate: $I_tangent = I_diam + M R^2 = (1/4) M R^2 + M R^2 = (5/4) M R^2$.
Answer: \frac{5}{4} MR^2
Assumptions and limitations
The body is treated as rigid, meaning distances between constituent particles remain invariant under applied forces and torques.
Common misconceptions
Common error: Moment of inertia is a constant intrinsic property of a body like mass.
Correct: Moment of inertia depends on the choice of the axis of rotation.
Mass is scalar and independent of position, whereas r in mr^2 is measured perpendicularly from a specific rotational axis.
Common error: Perpendicular axes theorem can be applied to any three-dimensional solid object.
Correct: The theorem is strictly valid only for two-dimensional planar laminar bodies.
For 3D bodies, $I_z = I_x + I_y does not hold$ because z-coordinates of mass elements are non-zero.
Quick revision
Moment of inertia ($I = \sum m_i r_i^2) measures rotational inertia$ and depends heavily on the axis of rotation.
General Discrete System: $I = \sum_{i=1}^{n} m_i r_i^2$
General Continuous System: $I = \int r^2 dm$
Must remember
Moment of inertia $I = \sum m_i r_i^2$ acts as the rotational analogue of mass in linear motion and has dimensions $[ML^2]$.
Theorem of Perpendicular Axes ($I_z = I_x + I_y$) is strictly applicable only to two-dimensional planar laminar bodies.
Theorem of Parallel Axes ($I = I_{cm} + Md^2$) is universally applicable to any rigid body of any shape.
Radius of gyration $k = \sqrt{I/M}$ represents the root-mean-square distance of mass distribution from the axis of rotation.
Glossary
Moment of Inertia ($I$)The measure of a body's resistance to angular acceleration about a given axis of rotation, defined by $\sum m_i r_i^2$ or $\int r^2 dm$.
Radius of Gyration ($k$)The root-mean-square distance of the constituent mass particles from the axis of rotation, given by $k = \sqrt{I/M}$.
Planar Lamiar BodyA two-dimensional flat object whose thickness is negligible, allowing the application of the perpendicular axes theorem.
Principal Moments of InertiaEigenvalues of the inertia tensor representing moments of inertia along principal axes where products of inertia vanish.
Perpendicular Axes vs. Parallel Axes Theorems
Feature
Perpendicular Axes Theorem
Parallel Axes Theorem
Applicability
Strictly 2D planar (laminar) bodies
Any rigid body (1D, 2D, or 3D)
Equation
$I_z = I_x + I_y$
$I = I_{cm} + Md^2$
Axis Configuration
Three mutually perpendicular axes meeting at a point in the plane
Two parallel axes, one passing through the centre of mass
Practice Quiz
Choose an option to reveal the explanation.
1. What are the SI units and dimensional formula of the moment of inertia?
Explanation: Moment of inertia is given by $I = \sum m_i r_i^2$, which has units of mass $\times$ distance squared ($\text{kg m}^2$) and dimensional formula $[ML^2]$.
2. Which of the following bodies satisfies the condition for applying the theorem of perpendicular axes?
Explanation: The theorem of perpendicular axes is strictly applicable only to two-dimensional planar (laminar) bodies. Among the choices, a thin circular ring is a planar lamina lying entirely in one plane.
3. The theorem of parallel axes is expressed mathematically as:
Explanation: The theorem of parallel axes states that $I = I_{cm} + Md^2$, where $I_{cm}$ is the moment of inertia about the centre of mass and $d$ is the distance between the parallel axes.
4. How is the radius of gyration $k$ related to the moment of inertia $I$ and total mass $M$?
Explanation: The radius of gyration is defined as $k = \sqrt{I / M}$, representing the root-mean-square distance of mass particles from the axis of rotation.
5. What is the moment of inertia of a thin uniform circular ring of mass $M$ and radius $R$ about its diametrical axis?
Explanation: For a thin ring about its diameter, due to symmetry about orthogonal diameters in the plane, $I_x = I_y = \frac{1}{2} I_z = \frac{1}{2} MR^2$.
6. What is the moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about a diametrical axis?
Explanation: For a circular disc, the moment of inertia about a transverse axis through its centre is $\frac{1}{2} MR^2$. By perpendicular axes theorem, about a diameter it is half of that, i.e., $\frac{1}{4} MR^2$.
7. What is the moment of inertia of a uniform solid sphere of mass $M$ and radius $R$ about its diameter?
Explanation: A uniform solid sphere has a moment of inertia about any diameter equal to $\frac{2}{5} MR^2$.
8. What is the moment of inertia of a thin spherical shell (hollow sphere) of mass $M$ and radius $R$ about its diameter?
Explanation: For a thin spherical shell (hollow sphere), the moment of inertia about its diameter is $\frac{2}{3} MR^2$.
9. Why is the moment of inertia not an intrinsic constant like mass?
Explanation: Moment of inertia depends on the distribution of mass relative to a specific axis of rotation ($I = \sum m_i r_i^2$), changing when the axis changes.
10. In PGT-level advanced mechanics, how is the general inertia of a rigid body represented for arbitrary rotations?
Explanation: For arbitrary three-dimensional rotations, the moment of inertia generalizes to a second-rank symmetric tensor whose components depend on mass distribution and coordinate axes.
11. What are the off-diagonal elements of the inertia tensor called?
Explanation: The off-diagonal terms of the inertia tensor matrix are known as the products of inertia (e.g., $I_{xy} = - \sum m_i x_i y_i$).
12. What condition defines the principal axes of a rigid body?
Explanation: Principal axes are specific orthogonal axes for which the products of inertia vanish, diagonalizing the inertia tensor.
13. If a planar body lies in the $xy$-plane, what does the perpendicular axes theorem state regarding $I_x$, $I_y$, and $I_z$?
Explanation: For a lamina in the $xy$-plane, the axis perpendicular to the plane is the $z$-axis, and the theorem states $I_z = I_x + I_y$.
14. What is the physical significance of the moment of inertia in rotational motion compared to mass in linear motion?
Explanation: Just as mass is the measure of inertia in linear motion resisting translational acceleration, moment of inertia is the measure of resistance to angular acceleration.
15. For a continuous mass distribution, how is the moment of inertia calculated?
Explanation: For continuous bodies, the discrete summation is replaced by integration over the mass elements: $I = \int r^2 dm$, where $r$ is the perpendicular distance from the axis.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. What is the moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its centre and perpendicular to its length?
Explanation: Integrating $\int x^2 dm$ from $-L/2$ to $L/2$ for a rod of linear mass density $\lambda = M/L$ yields $I = \frac{1}{12} ML^2$.
2. If the radius of gyration of a circular disc of mass $M$ and radius $R$ about an axis perpendicular to the disc and passing through its centre is $k$, what is the value of $k$?
Explanation: For a disc about its central transverse axis, $I = \frac{1}{2} MR^2$. Since $I = Mk^2$, we get $Mk^2 = \frac{1}{2} MR^2 \implies k = \frac{R}{\sqrt{2}}$.
3. Using the parallel axes theorem, find the moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through one of its ends and perpendicular to its length.
Explanation: Using $I = I_{cm} + Md^2$ where $I_{cm} = \frac{1}{12} ML^2$ and $d = L/2$, we get $I = \frac{1}{12}ML^2 + M(L/2)^2 = \left(\frac{1}{12} + \frac{1}{4}\right)ML^2 = \frac{1}{3} ML^2$.
4. Two rings of masses $M$ and $2M$ and radii $R$ and $2R$ respectively have moments of inertia $I_1$ and $I_2$ about their respective central transverse axes. What is the ratio $I_1 / I_2$?
Explanation: For a ring, $I = MR^2$. Thus $I_1 = M R^2$ and $I_2 = (2M)(2R)^2 = 8MR^2$. Therefore, $I_1 / I_2 = 1 / 8$.
5. What is the moment of inertia of a uniform circular disc about a tangent in its own plane?
Explanation: About a diameter, $I_{cm} = \frac{1}{4} MR^2$. Using parallel axes theorem for a tangent in its plane ($d = R$): $I = I_{cm} + Md^2 = \frac{1}{4} MR^2 + MR^2 = \frac{5}{4} MR^2$.
6. A uniform solid cylinder of mass $M$ and radius $R$ has a moment of inertia about its geometric axis equal to:
Explanation: A solid cylinder can be visualized as a stack of circular discs, so its moment of inertia about its central longitudinal axis is identical to that of a disc, i.e., $I = \frac{1}{2} MR^2$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. What is the moment of inertia of a thin circular ring of mass $M$ and radius $R$ about a tangent perpendicular to the plane of the ring?
Explanation: About the central transverse axis, $I_{cm} = MR^2$. Using parallel axes theorem with $d = R$: $I = MR^2 + M(R)^2 = 2 MR^2$.
2. If the moment of inertia of a body about an axis is changed by altering the axis of rotation, which of the following remains constant?
Explanation: Mass is an intrinsic scalar property of a body and remains constant regardless of the choice of axis, whereas moment of inertia and radius of gyration depend on the axis.
3. What is the radius of gyration of a uniform solid sphere of radius $R$ about its diameter?
Explanation: Equating $I = Mk^2$ to the solid sphere's moment of inertia $I = \frac{2}{5} MR^2$ gives $Mk^2 = \frac{2}{5} MR^2 \implies k = \sqrt{\frac{2}{5}} R$.
4. What is the moment of inertia of a thin uniform rectangular lamina of mass $M$, length $a$, and breadth $b$ about an axis passing through its centre and perpendicular to its plane?
5. In tensor mechanics, if the inertia tensor at a point is given by a $3 \times 3$ matrix, how many independent components does it possess due to symmetry?
Explanation: The inertia tensor is a symmetric tensor ($I_{ij} = I_{ji}$), meaning out of 9 components, only 6 are independent.
6. What is the moment of inertia of a uniform circular disc about a tangent perpendicular to its plane?
Explanation: For a disc, $I_{cm} = \frac{1}{2} MR^2$ about the central transverse axis. By parallel axes theorem with $d = R$: $I = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2$.
7. If four point masses $m$ are placed at the corners of a square of side $L$, what is the moment of inertia about an axis passing through the centre and perpendicular to the square's plane?
Explanation: Each mass is at a distance $r = \frac{L}{\sqrt{2}}$ from the centre. Thus $I = \sum m_i r_i^2 = 4 \cdot m \cdot \left(\frac{L}{\sqrt{2}}\right)^2 = 4m \left(\frac{L^2}{2}\right) = 2mL^2$.
8. Which theorem allows the calculation of moment of inertia of an arbitrary 3D rigid body about any parallel axis?
Explanation: The Theorem of Parallel Axes applies to rigid bodies of any shape (1D, 2D, or 3D) to find moments of inertia about parallel axes.
9. What is the radius of gyration of a thin spherical shell of radius $R$ about its diameter?
Explanation: For a spherical shell, $I = \frac{2}{3} MR^2$. Equating to $Mk^2$ gives $k = \sqrt{\frac{2}{3}} R$.
10. What happens to the moment of inertia of a rigid body if the axis of rotation is shifted closer to its centre of mass?
Explanation: According to the parallel axes theorem $I = I_{cm} + Md^2$, the moment of inertia is minimal when the axis passes through the centre of mass ($d=0$). Shifting the axis closer to the centre of mass decreases $d$ and hence decreases $I$.