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Mechanics
Moment of Inertia of a Thin Circular Ring
The moment of inertia of a thin circular ring of mass M and radius R about an axis perpendicular to its plane and passing through its centre is given by MR². It measures the rotational inertia of the ring against changes in its rotational motion.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Standard formulas for moment of inertia of a ring about standard axes (transverse axis through centre, diametrical axis, and tangent in the plane or perpendicular to the plane) using perpendicular and parallel axes theorems.
PGT extension
Integration-based derivations for ring and circular strip elements, application of parallel and perpendicular axes theorems to complex geometries involving rings, and combined rotational-translational motion involving rolling rings.
Prerequisites
Definition of moment of inertia (I = Σ m<sub>i</sub> r<sub>i</sub>² or ∫ r² dm)
Perpendicular axes theorem for planar bodies
Parallel axes theorem for any rigid body
Definition and Physical Significance
The moment of inertia (MI) of a rigid body depends on mass, mass distribution relative to the axis of rotation, and the orientation of the axis.
For a thin uniform ring, the entire mass M is concentrated at a constant perpendicular distance R from the central axis.
Its SI unit is kg·m² and its dimensional formula is [M L² T&sup0;].
MI about Transverse Axis through Centre
The axis passes through the centre of mass of the ring and is perpendicular to its plane.
Every differential element dm of the ring is at the same distance R from this axis.
Thus, the moment of inertia is strictly given by $I = MR^{2}$.
MI about a Diametrical Axis
A diameter lies in the plane of the ring and passes through its centre.
By symmetry, the moment of inertia about any two mutually perpendicular diameters in the plane are equal.
Applying the perpendicular axes theorem yields Idiameter = (1/2) MR².
MI about Tangential Axes
For a tangent in the plane of the ring, use the parallel axes theorem with the diametrical axis: $I = (1/2)MR^{2} + MR^{2} = (3/2)MR^{2}$.
For a tangent perpendicular to the plane of the ring, use the parallel axes theorem with the central transverse axis: $I = MR^{2} + MR^{2} = 2MR^{2}$.
Derivations
Moment of Inertia about Transverse Axis through Centre
Conditions: Thin uniform ring of mass M and radius R, axis perpendicular to the plane and passing through the centre.
Consider an infinitesimally small mass element dm of the ring.
The perpendicular distance of this element from the rotation axis is constantly equal to the radius R.
Integrate over the entire mass of the ring: $I = \int R^{2} dm$.
Since R² is constant, pull it outside the integral: $I = R^{2} \int dm$.
Substitute the total mass ∫ dm = M to get the final expression.
Result: $I = MR^{2}$
Moment of Inertia about a Diametrical Axis
Conditions: Thin uniform ring of mass M and radius R, axis lying along a diameter in the plane of the ring.
Choose two mutually perpendicular diameters lying in the plane of the ring as the x and y axes, and the transverse axis through the centre as the z axis.
By the perpendicular axes theorem, $I_z = I_x + I_y$.
Due to the circular symmetry of the ring, the moment of inertia about any diameter is equal, so $I_x = I_y = I_d$.
Substitute $I_z = MR^{2} into the equation: MR^{2} = 2I_d$.
Solve for the diametrical moment of inertia I_d.
Result: $I_d = \frac{1}{2} MR^{2}$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
MI about central transverse axis
$$I = MR^2$$
M = total mass of ring, R = radius kg·m²
Axis perpendicular to ring plane passing through centre
MI about a diameter
$$I = \frac{1}{2} MR^2$$
M = mass, R = radius kg·m²
Axis lies in the plane of the ring along a diameter
MI about tangent in the plane
$$I = \frac{3}{2} MR^2$$
M = mass, R = radius kg·m²
Tangent lies strictly in the plane of the ring
MI about tangent perpendicular to plane
$$I = 2MR^2$$
M = mass, R = radius kg·m²
Tangent is perpendicular to the plane of the ring
Perpendicular Axes Theorem
$$I_z = I_x + I_y$$
I_x, $I_y = MI about perpendicular axes in the plane; I_z = MI about axis perpendicular$ to plane kg·m²
Applicable only to planar (2D) bodies
Parallel Axes Theorem
$$I = I_{cm} + Md^2$$
$I_{cm} = MI about parallel axis through centre of mass$, d = distance between axes kg·m²
Applicable to any rigid body
Radius of Gyration for central axis
$$k = R$$
k = radius of gyration, R = radius of ring m
Axis is the central transverse axis
Radius of Gyration for diametrical axis
$$k = \frac{R}{\sqrt{2}}$$
k = radius of gyration, R = radius of ring m
Axis is along any diameter
Graphs and experiments
MI vs Square of Radius (R²)
Draw: A straight line passing through the origin plotting Moment of Inertia (I) on the y-axis against R² on the x-axis for a fixed mass M.
Interpret: The slope of the linear graph equals the mass M of the ring, verifying the direct proportionality $I = M(R^{2})$.
Exam use: Interpreting graphical representations of rotational inertia data in objective questions.
Torsional Pendulum Experiment
Draw: A ring suspended horizontally by a light wire, showing torsional oscillation.
Interpret: The time period $T = 2\pi\sqrt{I / C} allows experimental determination of the moment of inertia I$, where C is the torsional rigidity of the suspension wire.
Exam use: Questions linking experimental time period of suspension to moment of inertia.
Axis Orientation vs Moment of Inertia
Draw: A bar chart or polar plot comparing MI values for central transverse axis (MR²), diametrical axis (0.5MR²), in-plane tangent (1.5MR²), and perpendicular tangent (2MR²).
Interpret: Visually demonstrates that moment of inertia is not an inherent constant of a body but depends critically on the chosen axis of rotation.
Exam use: Conceptual ranking questions testing understanding of parallel and perpendicular axes theorems.
Solved numericals
TGT core
Calculate the moment of inertia of a ring of mass 2 kg and radius 0.5 m about an axis passing through its centre and perpendicular to its plane.
Identify given values: M = 2 kg, R = 0.5 m.
Use the formula for MI of a ring about its central transverse axis: $I = MR^{2}$.
A ring of mass M and radius R is recast into a ring of radius 2R keeping the mass constant. Find the ratio of their moments of inertia about their respective central transverse axes.
Write the initial moment of inertia: I1 = M R².
Write the new moment of inertia with radius 2R: I2 = M (2R)$^{2} = 4 M R^{2}$.
Take the ratio of I1 to I2: I1 / I2 = (M R²) / (4 M R²) = 1 / 4.
Answer: 1 : 4
Assumptions and limitations
The ring is assumed to be infinitely thin, meaning its cross-sectional radius is negligible compared to its outer radius R.
Common misconceptions
Common error: The moment of inertia of a ring depends on its angular velocity.
Correct: Moment of inertia is independent of angular velocity and angular acceleration.
MI is purely a geometric and mass distribution property of the rigid body for a chosen axis.
Common error: The formula $I = MR^{2} can be used$ for any axis passing through the centre.
Correct: $I = MR^{2}$ is valid only for an axis perpendicular to the plane of the ring passing through the centre.
For diametrical axes passing through the centre, the distance of mass elements varies from zero to R, giving $I = (1/2)MR^{2}$.
Quick revision
MI of ring about central transverse axis is MR², about diameter is 0.5MR², about in-plane tangent is 1.5MR², and about perpendicular tangent is 2MR².
MI about central transverse axis: $I = MR^2$
MI about a diameter: $I = \frac{1}{2} MR^2$
Must remember
The moment of inertia of a thin uniform ring of mass $M$ and radius $R$ about its central transverse axis is $I = MR^2$.
The moment of inertia of a thin ring about any of its diameters is $I = \frac{1}{2}MR^2$.
The moment of inertia of a ring about a tangent in its own plane is $I = \frac{3}{2}MR^2$.
The moment of inertia of a ring about a tangent perpendicular to its plane is $I = 2MR^2$.
Perpendicular axes theorem states $I_z = I_x + I_y$ and applies strictly to planar (2D) bodies.
Parallel axes theorem states $I = I_{cm} + Md^2$ and applies to any rigid body.
Glossary
Moment of Inertia ($I$)Measure of a body's resistance to angular acceleration about a given axis, quantified by $\sum m_i r_i^2$.
Radius of Gyration ($k$)Effective distance from the axis of rotation where the entire mass of the body can be assumed to be concentrated, given by $k = \sqrt{I/M}$.
Central Transverse AxisAn axis passing through the centre of mass of the ring and perpendicular to its plane.
Diametrical AxisAn axis lying in the plane of the ring and passing through its centre.
Moment of Inertia of Ring about Different Axes
Axis Description
Formula
Radius of Gyration ($k$)
Transverse axis through centre
$MR^2$
$R$
Diametrical axis
$\frac{1}{2}MR^2$
$R/\sqrt{2}$
Tangent in the plane
$\frac{3}{2}MR^2$
$\sqrt{3/2}R$
Tangent perpendicular to plane
$2MR^2$
$R\sqrt{2}$
Practice Quiz
Choose an option to reveal the explanation.
1. What is the moment of inertia of a thin circular ring of mass $M$ and radius $R$ about a transverse axis passing through its centre?
Explanation: For a thin uniform ring, every mass element is at a constant perpendicular distance $R$ from the central transverse axis, so $I = \int dm \, R^2 = MR^2$.
2. What is the moment of inertia of a thin ring of mass $M$ and radius $R$ about its diameter?
Explanation: By symmetry, two mutually perpendicular diameters in the plane have equal moments of inertia ($I_x = I_y$). By the perpendicular axes theorem, $I_z = I_x + I_y \implies MR^2 = 2I_{diameter}$, giving $I_{diameter} = \frac{1}{2}MR^2$.
3. What is the SI unit of moment of inertia?
Explanation: Moment of inertia is defined as $I = \sum m_i r_i^2$, which has units of mass ($\text{kg}$) multiplied by distance squared ($\text{m}^2$). Thus, the SI unit is $\text{kg}\cdot\text{m}^2$.
4. What is the dimensional formula for moment of inertia?
Explanation: The formula $I = \sum m_i r_i^2$ involves mass ($[M]$) and distance squared ($[L^2]$), with time not involved ($[T^0]$). Hence, the dimensional formula is $[M L^2 T^0]$.
5. What is the radius of gyration of a thin ring of radius $R$ about its central transverse axis?
Explanation: Radius of gyration $k$ is given by $I = Mk^2$. Since $I = MR^2$ for the central axis, $Mk^2 = MR^2 \implies k = R$.
6. What is the radius of gyration of a thin ring of radius $R$ about a diametrical axis?
Explanation: For a diametrical axis, $I = \frac{1}{2}MR^2$. Equating this to $Mk^2$ gives $Mk^2 = \frac{1}{2}MR^2 \implies k = \frac{R}{\sqrt{2}}$.
7. What is the moment of inertia of a ring about a tangent lying in the plane of the ring?
Explanation: Using the parallel axes theorem with the diametrical axis, $I = I_{cm} + Md^2 = \frac{1}{2}MR^2 + M(R)^2 = \frac{3}{2}MR^2$.
8. What is the moment of inertia of a ring about a tangent perpendicular to the plane of the ring and passing through its edge?
Explanation: Using the parallel axes theorem with the central transverse axis ($I_{cm} = MR^2$) and separation $d = R$, we get $I = MR^2 + M(R)^2 = 2MR^2$.
9. Which of the following theorems is applicable only to planar (two-dimensional) bodies?
Explanation: The perpendicular axes theorem relates the moment of inertia about three mutually orthogonal axes where two lie in the plane of the body and one is perpendicular to it. Thus, it is strictly valid only for 2D planar bodies.
10. Does the moment of inertia of a thin ring depend on its angular velocity?
Explanation: Moment of inertia is purely a geometric and mass distribution property of a rigid body for a chosen axis of rotation, and does not depend on kinematic variables like angular velocity or angular acceleration.
11. If the mass of a ring is doubled while keeping its radius constant, what happens to its moment of inertia about the central transverse axis?
Explanation: Since $I = MR^2$, moment of inertia is directly proportional to mass $M$. Doubling the mass doubles the moment of inertia.
12. If the radius of a ring is doubled while keeping its mass constant, what happens to its moment of inertia about a diametrical axis?
Explanation: For a diametrical axis, $I = \frac{1}{2}MR^2$. Doubling the radius $R$ increases $R^2$ by a factor of 4, hence the moment of inertia is quadrupled.
13. Using integration, consider a differential element of mass $dm$ on a ring of radius $R$. The integral $\int dm$ evaluated over the entire ring equals:
Explanation: The sum of all differential mass elements $dm$ over the entire ring equals the total mass $M$ of the ring.
14. A ring of mass $M$ and radius $R$ rolls without slipping on a horizontal surface. What is its total kinetic energy?
Explanation: Total kinetic energy is $K = K_{trans} + K_{rot} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$. For a ring, $I = MR^2$ and for rolling without slipping $v = R\omega$, so $K = \frac{1}{2}Mv^2 + \frac{1}{2}(MR^2)(\frac{v}{R})^2 = \frac{1}{2}Mv^2 + \frac{1}{2}Mv^2 = Mv^2$.
15. What is the moment of inertia of a thin circular strip of mass $m$ and radius $r$ about its central transverse axis?
Explanation: A thin circular strip behaves just like a thin ring of mass $m$ and radius $r$, so its moment of inertia about the central transverse axis is $mr^2$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. The moment of inertia of a thin ring of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is:
Explanation: By definition and direct derivation, every mass element of a thin ring is at distance $R$ from the central transverse axis, giving $I = \int dm \, R^2 = MR^2$.
2. If $I_1$ is the moment of inertia of a ring about its diameter and $I_2$ is its moment of inertia about a tangent in its plane, the ratio $I_1 : I_2$ is:
Explanation: Moment of inertia about a diameter is $I_1 = \frac{1}{2}MR^2$. Moment of inertia about a tangent in the plane is $I_2 = \frac{3}{2}MR^2$. Therefore, $I_1 : I_2 = \frac{1}{2} : \frac{3}{2} = 1 : 3$.
3. The radius of gyration of a circular ring of radius $R$ about a transverse axis passing through its centre is:
Explanation: For the central transverse axis, $I = MR^2$. Setting $I = Mk^2$ gives $Mk^2 = MR^2$, which yields $k = R$.
4. What is the moment of inertia of a thin uniform ring about a tangent perpendicular to the plane of the ring?
Explanation: Using the parallel axes theorem, $I = I_{cm} + Md^2$, where $I_{cm} = MR^2$ and $d = R$, giving $I = MR^2 + MR^2 = 2MR^2$.
5. Two rings of the same mass and radius are placed such that their centres coincide and their planes are mutually perpendicular. What is the moment of inertia of the system about an axis passing through their common centre and lying in the plane of one of the rings?
Explanation: Let Ring 1 lie in the xy-plane and Ring 2 in the yz-plane. The chosen axis is along the z-axis (say). For Ring 1 (in xy-plane), the z-axis is its central transverse axis, so $I_1 = MR^2$. For Ring 2 (in yz-plane), the z-axis lies along its diameter, so $I_2 = \frac{1}{2}MR^2$. Total moment of inertia $I = I_1 + I_2 = MR^2 + \frac{1}{2}MR^2 = \frac{3}{2}MR^2$.
6. A ring of mass $M$ and radius $R$ has its moment of inertia about an axis passing through its edge and perpendicular to its plane equal to $I$. What is $I$?
Explanation: An axis passing through the edge and perpendicular to the plane of the ring is a tangent perpendicular to the plane. By parallel axes theorem, $I = MR^2 + MR^2 = 2MR^2$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. If the radius of gyration of a ring about a certain axis is equal to its radius $R$, the axis must be:
Explanation: Since $k = R$, $Mk^2 = MR^2$, which corresponds to the central transverse axis where $I = MR^2$.
2. What is the ratio of the radius of gyration of a ring about its diametrical axis to that about its central transverse axis?
Explanation: Radius of gyration about a diameter is $k_1 = R/\sqrt{2}$. Radius of gyration about the central axis is $k_2 = R$. Thus, $k_1 / k_2 = 1/\sqrt{2}$.
3. A thin ring of mass $2\text{ kg}$ and radius $0.5\text{ m}$ rotates about its central transverse axis. Its moment of inertia is:
5. What is the work done in increasing the angular velocity of a ring of mass $M$ and radius $R$ from zero to $\omega$?
Explanation: Work done equals the rotational kinetic energy gained: $W = \Delta K = \frac{1}{2}I\omega^2$. Since $I = MR^2$, $W = \frac{1}{2}MR^2\omega^2$.
6. According to the perpendicular axes theorem $I_z = I_x + I_y$, if $I_z$ is the moment of inertia about the central transverse axis of a ring, what are $I_x$ and $I_y$?
Explanation: For a ring lying in the xy-plane, $I_z = MR^2$, and $I_x$ and $I_y$ are moments of inertia about any two mutually perpendicular diameters lying in the plane.
7. A ring and a disc have the same mass and radius. The ratio of their moments of inertia about their respective central transverse axes is:
Explanation: Moment of inertia of a ring is $I_{ring} = MR^2$, and of a disc is $I_{disc} = \frac{1}{2}MR^2$. The ratio $I_{ring} : I_{disc} = MR^2 : \frac{1}{2}MR^2 = 2 : 1$.
8. If a uniform ring of mass $M$ and radius $R$ is reshaped into another ring of radius $2R$ keeping mass constant, the moment of inertia about the central transverse axis becomes:
Explanation: $I = MR^2$. When radius becomes $2R$, $I' = M(2R)^2 = 4MR^2$, which is four times the initial value.
9. Which axis has the maximum moment of inertia for a thin circular ring among the standard choices?
Explanation: Comparing the values: diameter is $\frac{1}{2}MR^2$, central transverse is $MR^2$, tangent in plane is $\frac{3}{2}MR^2$, and tangent perpendicular to plane is $2MR^2$. The maximum is $2MR^2$.
10. What is the radius of gyration of a ring about a tangent perpendicular to its plane?
Explanation: For a tangent perpendicular to the plane, $I = 2MR^2$. Equating to $Mk^2$ gives $Mk^2 = 2MR^2 \implies k = R\sqrt{2}$.