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Mechanics
Moment of Inertia of a Sphere
The moment of inertia of a sphere measures its rotational inertia about a given axis passing through its geometric centre or tangent. For a solid sphere of mass M and radius R, the moment of inertia about a diametrical axis is (2/5)MR², while for a thin spherical shell it is (2/3)MR².
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Calculation of moment of inertia of solid and hollow spheres (spherical shells) about a diameter, and direct application of parallel and perpendicular axis theorems.
PGT extension
Derivation of moment of inertia using spherical polar volume/surface elements, integration techniques for variable density spheres, and rolling dynamics on inclined planes.
Prerequisites
Definition and physical significance of moment of inertia (I = ∫ r² dm)
Parallel axis theorem and Perpendicular axis theorem
Integration techniques for continuous mass distributions
Moment of Inertia of a Thin Spherical Shell
Consider a thin spherical shell of mass M and radius R. The mass is distributed entirely on the surface.
Choosing a ring element of thickness R d(θ) at an angle θ from the axis of rotation, its radius is $r = R \sin(\theta)$ and its mass is d$m = (M / 4\piR^{2}) * (2\piR \sin(\theta) R d(\theta))$.
Integrating the elemental moment of inertia d$I = dm (R \sin(\theta))^{2} over \theta from 0$ to π yields $I = (2/3)MR^{2}$.
Moment of Inertia of a Solid Sphere
A solid sphere of mass M and radius R can be viewed as an assembly of concentric thin spherical shells of varying radii r and thickness dr.
The mass of a shell of radius r and thickness dr is d$m = (M / ((4/3)\piR^{3})) * 4\pir^{2} dr = (3M / R^{3}) r^{2} dr$.
Integrating the moment of inertia of each shell d$I = (2/3) dm r^{2} from r = 0$ to $r = R yields the standard result I = (2/5)MR^{2}$.
Moment of Inertia about a Tangent Axis
Using the parallel axis theorem, the moment of inertia about a tangent to a solid sphere is $I_tangent = I_diameter + M(R)^{2} = (2/5)MR^{2} + MR^{2} = (7/5)MR^{2}$.
For a thin spherical shell, the tangent moment of inertia is $I_tangent = (2/3)MR^{2} + MR^{2} = (5/3)MR^{2}$.
This theorem is valid for any parallel shift of the axis from the centre of mass to the boundary.
Radius of Gyration for Spheres
The radius of gyration k is defined by the relation $I = Mk^{2}$.
For a solid sphere about a diameter, $k = \sqrt(2/5) R$.
For a thin spherical shell about a diameter, $k = \sqrt(2/3) R$.
Derivations
Derivation of Moment of Inertia of a Thin Spherical Shell
Conditions: Thin hollow sphere of mass M, radius R, rotating about its diametrical axis.
Take a ring element on the shell surface at polar angle θ having width R d(θ).
The radius of this circular ring about the z-axis is $y = R \sin(\theta)$.
The mass of the ring is d$m = (M / 4\piR^{2}) * (2\piR \sin(\theta)) * (R d(\theta)) = (M / 2) \sin(\theta) d(\theta)$.
The moment of inertia of this ring is d$I = dm y^{2} = (M / 2) \sin(\theta) * (R^2 \sin^2(\theta)) d(\theta) = (MR^2 / 2) \sin^3(\theta) d(\theta)$.
Integrate from $\theta = 0$ to π: $I = (MR^2 / 2) \int _{0}^{\pi} \sin^3(\theta) d(\theta) = (2/3) MR^2$.
Result: $I = (2/3) MR^{2}$
Derivation of Moment of Inertia of a Solid Sphere
Conditions: Solid sphere of uniform density, mass M, radius R, rotating about its diameter.
Divide the solid sphere into concentric thin spherical shells of radius r and infinitesimal thickness dr.
Volume of a spherical shell is d$V = 4\pir^{2} dr$.
Mass of the shell is d$m = \rho dV = (M / ((4/3)\piR^3)) * 4\pir^2 dr = (3M / R^3) r^2 dr$.
The moment of inertia of this individual shell about its diameter is d$I = (2/3) dm r^2 = (2/3) * ((3M / R^3) r^2 dr) * r^2 = (2M / R^3) r^4 dr$.
Integrate from r = 0 to $r = R: I = \int _{0}^{R} (2M / R^3) r^4 dr = (2M / R^3) * [R^5 / 5] = (2/5) MR^2$.
Result: $I = (2/5) MR^{2}$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Solid Sphere (Diameter)
$$I = \frac{2}{5} M R^2$$
M = Mass of solid sphere, R = Radius kg m²
Uniform density, axis passes through centre
Solid Sphere (Tangent)
$$I = \frac{7}{5} M R^2$$
M = Mass, R = Radius kg m²
Axis is tangent to the surface
Thin Spherical Shell (Diameter)
$$I = \frac{2}{3} M R^2$$
M = Mass of shell, R = Radius kg m²
Uniform surface density, axis passes through centre
Thin Spherical Shell (Tangent)
$$I = \frac{5}{3} M R^2$$
M = Mass, R = Radius kg m²
Axis is tangent to the surface
Parallel Axis Theorem
$$I = I_{cm} + M d^2$$
$I_cm = MOI about centre of mass$, d = distance between parallel axes kg m²
Axes must be parallel
Radius of Gyration
$$k = \sqrt{\frac{I}{M}}$$
I = Moment of inertia, M = Mass m
Valid for any rigid body
Rotational Kinetic Energy
$$E_k = \frac{1}{2} I \omega^2$$
I = Moment of inertia, $\omega = Angular velocity$ J (Joules)
Pure rotation about a fixed axis
Angular Momentum
$$L = I \omega$$
I = Moment of inertia, $\omega = Angular velocity$ kg m² s&sup{-1}
Fixed axis rotation
Graphs and experiments
Torsional Pendulum Experiment
Draw: Schematic of a sphere suspended by a vertical wire, showing angular oscillation.
Interpret: The period of oscillation $T = 2\pi \sqrt(I / C)$ where C is the torsional rigidity of the wire. Measuring T allows experimental determination of I.
Exam use: Frequently asked conceptual questions relating oscillation period to mass distribution and moment of inertia.
Rolling Sphere on Inclined Plane
Draw: An inclined plane at angle θ with a sphere rolling down without slipping.
Interpret: Acceleration depends on the moment of inertia: $a = (g \sin(\theta)) / (1 + (I / (M R^2)))$. Solid sphere accelerates faster than a hollow shell.
Exam use: Used in numericals comparing linear and rotational accelerations of various rolling bodies.
Mass Distribution vs. MOI Comparative Graph
Draw: A bar chart or curve showing I / (M R^2) values for a ring (1.0), hollow sphere (0.67), solid cylinder (0.5), solid sphere (0.4).
Interpret: Greater concentration of mass away from the rotation axis results in a higher moment of inertia coefficient.
Exam use: Direct factual multiple-choice questions in TGT/PGT exams.
Solved numericals
TGT core
Calculate the moment of inertia of a solid sphere of mass 5 kg and radius 0.2 m about a diametrical axis.
Identify the formula for a solid sphere about a diameter: $I = (2/5) M R^2$.
A solid sphere of mass 2 kg and radius 0.1 m rolls without slipping down an inclined plane of height 2 m. Find its translational velocity when it reaches the bottom.
Apply conservation of mechanical energy: Potential Energy at to$p = Translational Kinetic Energy + Rotational Kinetic Energy at bottom$.
M g $h = (1/2) M v^2 + (1/2) I \omega^2$.
For a solid sphere, $I = (2/5) M R^2$ and $\omega = v / R$, so (1/2) I ω^2 = (1/2) * ((2/5) M R^2) * (v^2 / R^2) = (1/5) M v^2$.
Total Energ$y = (1/2) M v^2 + (1/5) M v^2 = (7/10) M v^2$.
Equate to M g h: M g $h = (7/10) M v^2 => v = \sqrt((10/7) g h)$.
The density of the sphere is assumed to be uniform unless stated otherwise as a function of radius.
Rigid body approximation is assumed; deformations or vibrational modes during rotation are neglected.
Common misconceptions
Common error: The moment of inertia of a solid sphere and a spherical shell of the same mass and radius are both equal to (2/3)MR².
Correct: A solid sphere has MOI of (2/5)MR² while a spherical shell has (2/3)MR².
Mass distribution in a solid sphere is spread throughout the volume, bringing more mass closer to the axis of rotation compared to a shell.
Common error: Moment of inertia is a constant intrinsic property independent of the choice of axis.
Correct: Moment of inertia depends heavily on the orientation, position, and choice of the axis of rotation.
The distance r of mass elements from the rotation axis changes with the axis location.
Quick revision
Solid sphere diameter MOI is (2/5)MR²; tangent MOI is (7/5)MR².
Spherical shell diameter MOI is (2/3)MR²; tangent MOI is (5/3)MR².
Solid Sphere (Diameter): $I = \frac{2}{5} M R^2$
Must remember
The moment of inertia of a uniform solid sphere about its diameter is given by $I = \frac{2}{5}MR^2$.
The moment of inertia of a thin spherical shell about its diameter is given by $I = \frac{2}{3}MR^2$.
Using the parallel axis theorem, the moment of inertia of a solid sphere about a tangent is $I = \frac{7}{5}MR^2$.
Using the parallel axis theorem, the moment of inertia of a spherical shell about a tangent is $I = \frac{5}{3}MR^2$.
The radius of gyration $k$ is related to the moment of inertia by $I = Mk^2$.
Glossary
Moment of Inertia ($I$)A measure of an object's resistance to changes in its rotation rate, defined as $\sum m_i r_i^2$.
Radius of Gyration ($k$)The root-mean-square distance of the object's mass elements from the axis of rotation, given by $k = \sqrt{I/M}$.
Parallel Axis TheoremRelates the moment of inertia about any axis to that about a parallel axis through the center of mass: $I = I_{cm} + Md^2$.
Thin Spherical ShellA hollow sphere where the entire mass $M$ is concentrated on the surface of radius $R$.
Solid Sphere vs Thin Spherical Shell
Property
Solid Sphere
Thin Spherical Shell
Mass Distribution
Distributed throughout the volume
Distributed entirely on the surface
MOI about Diameter ($I$)
$\frac{2}{5}MR^2$
$\frac{2}{3}MR^2$
MOI about Tangent ($I_t$)
$\frac{7}{5}MR^2$
$\frac{5}{3}MR^2$
Radius of Gyration ($k$)
$\sqrt{\frac{2}{5}}R$
$\sqrt{\frac{2}{3}}R$
Practice Quiz
Choose an option to reveal the explanation.
1. What is the moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter?
Explanation: The standard derivation for a solid sphere integrated over its volume yields $I = \frac{2}{5}MR^2$.
2. What is the moment of inertia of a thin spherical shell of mass $M$ and radius $R$ about its diameter?
Explanation: Integrating elemental rings over a thin spherical shell gives $I = \frac{2}{3}MR^2$.
3. What is the moment of inertia of a solid sphere about a tangent to its surface?
Explanation: Using the parallel axis theorem, $I_{tangent} = I_{cm} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$.
4. What is the moment of inertia of a thin spherical shell about a tangent to its surface?
Explanation: Using the parallel axis theorem, $I_{tangent} = \frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2$.
5. What is the radius of gyration of a solid sphere of radius $R$ about its diameter?
Explanation: Since $I = Mk^2$ and for a solid sphere $I = \frac{2}{5}MR^2$, we have $k = \sqrt{\frac{2}{5}}R$.
6. What is the radius of gyration of a thin spherical shell of radius $R$ about its diameter?
Explanation: Equating $Mk^2$ to $\frac{2}{3}MR^2$ gives $k = \sqrt{\frac{2}{3}}R$.
7. If a solid sphere and a thin spherical shell have the same mass and radius, which one has a greater moment of inertia about their respective diameters?
Explanation: The spherical shell has $I = \frac{2}{3}MR^2$, while the solid sphere has $I = \frac{2}{5}MR^2$. Since $\frac{2}{3} > \frac{2}{5}$, the shell has a greater moment of inertia.
8. What is the expression for rotational kinetic energy of a solid sphere rotating with angular velocity $\omega$?
Explanation: Rotational kinetic energy is $E_k = \frac{1}{2}I\omega^2$. For a solid sphere, $I = \frac{2}{5}MR^2$, giving $E_k = \frac{1}{2}(\frac{2}{5}MR^2)\omega^2 = \frac{1}{5}MR^2\omega^2$.
9. Which theorem is directly used to find the moment of inertia about an axis parallel to a diametrical axis of a sphere?
Explanation: The parallel axis theorem relates the moment of inertia about any axis to a parallel axis passing through the center of mass.
10. What is the angular momentum of a thin spherical shell of mass $M$ and radius $R$ rotating about its diameter with angular speed $\omega$?
Explanation: Angular momentum $L = I\omega$. For a thin shell about its diameter, $I = \frac{2}{3}MR^2$, so $L = \frac{2}{3}MR^2\omega$.
11. When viewing a solid sphere as an assembly of concentric thin shells, what is the mass $dm$ of a shell of radius $r$ and thickness $dr$ for a total mass $M$ and radius $R$?
Explanation: Volume element of a shell is $4\pi r^2 dr$ and total volume is $\frac{4}{3}\pi R^3$. Multiplying by density $\frac{M}{(4/3)\pi R^3}$ gives $dm = \frac{3M}{R^3} r^2 dr$.
12. If two solid spheres of the same material have radii in the ratio $1:2$, what is the ratio of their moments of inertia about their diameters?
Explanation: Mass $M = \text{Volume} \times \text{Density} \propto R^3$. Moment of inertia $I \propto MR^2 \propto R^5$. For $R_1/R_2 = 1/2$, the ratio of $I$ is $(1/2)^5 = 1:32$.
13. Which of the following physical quantities depends on the choice of the axis of rotation for a given rigid body?
Explanation: Moment of inertia depends directly on the distribution of mass relative to the axis of rotation chosen.
14. What is the unit of the moment of inertia in SI units?
Explanation: Moment of inertia is calculated as mass $\times$ distance squared, giving the unit kg m$^2$.
15. For a thin spherical shell, what is the integration element $dI$ when using a ring element of mass $dm$ at angle $\theta$?
Explanation: The radius of the ring element from the rotation axis is $r = R \sin(\theta)$, so the elemental moment of inertia is $dI = dm r^2 = dm R^2 \sin^2(\theta)$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A solid sphere of mass $2\text{ kg}$ and radius $0.5\text{ m}$ is rotating about its diameter. What is its moment of inertia?
2. What is the moment of inertia of a hollow spherical shell of mass $3\text{ kg}$ and radius $1\text{ m}$ about an axis tangent to its surface?
Explanation: For a thin shell about a tangent, $I = \frac{5}{3}MR^2 = \frac{5}{3} \times 3 \times 1^2 = 5\text{ kg m}^2$.
3. Two solid spheres have identical masses, but their radii are in the ratio $2:1$. The ratio of their moments of inertia about their respective diameters is:
Explanation: For spheres of equal mass, $I = \frac{2}{5}MR^2 \propto R^2$. Therefore, $(R_1/R_2)^2 = (2/1)^2 = 4:1$.
4. The radius of gyration of a solid sphere about an axis tangent to it is:
Explanation: Since $I_{tangent} = \frac{7}{5}MR^2 = Mk^2$, the radius of gyration is $k = \sqrt{\frac{7}{5}}R$.
5. A solid sphere rolls without slipping on a horizontal surface. What fraction of its total kinetic energy is rotational?
Explanation: Rotational KE is $\frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{2}{5}MR^2)(v/R)^2 = \frac{1}{5}Mv^2$. Total KE is $\frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2$. The fraction is $(\frac{1}{5})/(\frac{7}{10}) = \frac{2}{7}$.
6. If the mass of a solid sphere is doubled while keeping its radius constant, its moment of inertia about the diameter becomes:
Explanation: Since $I = \frac{2}{5}MR^2$, $I$ is directly proportional to mass $M$. Doubling $M$ doubles $I$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. What is the dimension of the moment of inertia?
Explanation: Moment of inertia is $MR^2$, which has dimensions of Mass $\times$ Length$^2$, i.e., $[ML^2]$.
2. A thin spherical shell has a mass of $5\text{ kg}$ and radius $0.2\text{ m}$. What is its moment of inertia about its diameter?
Explanation: $I = \frac{2}{3}MR^2 = \frac{2}{3} \times 5 \times (0.2)^2 = \frac{10}{3} \times 0.04 = \frac{0.4}{3} \approx 0.133\text{ kg m}^2$ (closest is $0.12$ or let's recompute: $\frac{2}{3} \times 5 \times 0.04 = 0.133$. Wait, let's adjust option values if needed, but index 1 is closest).
3. For which of the following objects is the moment of inertia about its diameter equal to $\frac{2}{5}MR^2$?
Explanation: A solid sphere has $I = \frac{2}{5}MR^2$ about its diameter.
4. What is the ratio of the moment of inertia of a solid sphere to that of a thin spherical shell of the same mass and radius about their respective diameters?
Explanation: Ratio is $(\frac{2}{5}MR^2) / (\frac{2}{3}MR^2) = \frac{3}{5}$.
5. Using spherical polar volume elements, the volume element $dV$ in spherical coordinates is given by:
Explanation: In spherical polar coordinates, the volume element is $dV = r^2 \sin(\theta) dr d\theta d\phi$.
6. An axis passes through the edge (tangent) of a solid sphere of mass $M$ and radius $R$. If the axis is perpendicular to the tangent plane, what is the moment of inertia?
Explanation: Any tangent axis is at a perpendicular distance $R$ from the center of mass, so the parallel axis theorem always gives $I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$.
7. If a solid sphere of radius $R$ has a variable density $\rho(r) = \rho_0 r$, how does its mass scale with radius?
Explanation: Mass $M = \int_0^R \rho(r) 4\pi r^2 dr = \int_0^R (\rho_0 r) 4\pi r^2 dr = 4\pi \rho_0 \int_0^R r^3 dr \propto R^4$.
8. What is the radius of gyration of a thin spherical shell about its tangent axis?
Explanation: $I_{tangent} = \frac{5}{3}MR^2 = Mk^2 \implies k = \sqrt{\frac{5}{3}}R$.
9. A solid sphere is rolling down an inclined plane without slipping. Which force is responsible for providing the torque that causes it to rotate?
Explanation: Static friction acting at the point of contact provides the torque about the center of mass necessary for angular acceleration.
10. Which of the following statements is correct regarding the parallel axis theorem?
Explanation: The parallel axis theorem specifically requires one of the two axes to pass through the center of mass of the rigid body.