The motion of a rigid body is characterized by fixed distances between any two constituent particles under applied forces and torques. It is analyzed by decomposing the motion into translational motion of the centre of mass and rotational motion about the centre of mass.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Kinematics and dynamics of rotation about a fixed axis, moment of inertia, parallel and perpendicular axes theorems, torque, and conservation of angular momentum.
PGT extension
General general three-dimensional rigid body motion, Euler's equations of motion, principal moments of inertia, and gyroscope precession.
Prerequisites
Newton's laws of motion
Centre of mass and mass distribution
Vector calculus and cross products
Kinematics of Rigid Body Motion
A rigid body is defined as a system of particles in which the separation distance between every pair of particles remains constant over time.
General motion of a rigid body can be expressed as a superposition of translation of its centre of mass and rotation about an axis passing through the centre of mass (Chasles' theorem).
The velocity of any arbitrary point on the rigid body is given by $v = v_{cm} + \omega \times r$, where ω is the angular velocity vector.
Moment of Inertia and Radius of Gyration
Moment of inertia I measures the rotational inertia of a rigid body, defined as $I = \sum m_{i} r_{i}^{2}$ for discrete systems and $I = \int r^{2} dm$ for continuous bodies.
The SI unit of moment of inertia is kg m2, and its dimensional formula is [M L2].
Radius of gyration k is the root-mean-square distance of the mass particles from the axis of rotation, related by $I = Mk^{2}$.
Theorems of Moment of Inertia
Perpendicular Axes Theorem: Iz = Ix + Iy, valid strictly for plane laminar (two-dimensional) bodies where axes x, y, and z are mutually perpendicular and intersect at a common point.
Parallel Axes Theorem: $I = I_{cm} + Md^{2}$, where Icm is the moment of inertia about a parallel axis passing through the centre of mass, and d is the perpendicular separation between the axes.
These theorems allow the computation of moment of inertia for complex geometries without evaluating volume integrals directly.
Rotational Dynamics and Angular Momentum
The rotational analogue of Newton's second law is $\tau = I\alpha$, where τ is the net external torque and α is the angular acceleration.
Angular momentum L of a rigid body rotating about a fixed axis is given by $L = I\omega$.
In the absence of external torques (τext = 0), the total angular momentum of the rigid body is conserved (L = constant).
Derivations
Perpendicular Axes Theorem Derivation
Conditions: Planar lamina lying in the xy-plane
Consider a lamina in the xy-plane. Let a mass element dm be located at coordinates (x, y) from the origin.
The moment of inertia about the z-axis (perpendicular to the plane) is $I_z = \int (x^2 + y^2) dm$.
Recognize that int y^2 dm is the moment of inertia about the x-axis (I_x) and int x^2 dm is the moment of inertia about the y-axis (I_y).
Result: $I_z = I_x + I_y$
Parallel Axes Theorem Derivation
Conditions: Any rigid body, relating axis through centre of mass to a parallel axis at distance d
Let I_cm be the moment of inertia about an axis passing through the centre of mass, and I be the moment of inertia about a parallel axis at distance d.
The position of mass element dm relative to the centre of mass axis is r_i, and relative to the new axis is (r_i + d).
Write the moment of inertia: $I = \int (r_i + d)^2 dm = \int r_i^2 dm + d^2 \int dm + 2d \int r_i dm$.
Since the axis passes through the centre of mass, the first moment of mass int r_i dm vanishes by definition. Also int r_i^2 d$m = I_cm$ and int dm = M.
Result: $I = I_cm + Md^2$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Moment of Inertia (General)
$$I = \sum m_i r_i^2$$
$m_i = mass of particle$, $r_i = perpendicular distance from axis$ kg m^2
Discrete mass distribution
Radius of Gyration
$$k = \sqrt{\frac{I}{M}}$$
I = moment of inertia, M = total mass m
Rigid body of mass M
Parallel Axes Theorem
$$I = I_{cm} + Md^2$$
$I_{cm} = MOI about COM axis$, d = distance between parallel axes kg m^2
Any rigid body
Perpendicular Axes Theorem
$$I_z = I_x + I_y$$
I_x, I_y, $I_z = MOI about x$, y, and z axes kg m^2
Two-dimensional planar lamina
Rotational Kinetic Energy
$$K = \frac{1}{2} I \omega^2$$
I = moment of inertia, omega = angular velocity J
Fixed axis rotation
Torque and Angular Acceleration
$$\tau = I \alpha$$
\tau = external torque, I = moment of inertia, alpha = angular acceleration N m
Constant moment of inertia about fixed axis
Angular Momentum
$$L = I \omega$$
I = moment of inertia, omega = angular velocity kg m^2 s^-1
Use the parallel axes theorem to shift the axis from the diameter to the tangent in the plane: $I = I_d + M R^2$.
Calculate total MOI: $I = 0.125 + (2)(0.5)^2 = 0.125 + 0.5 = 0.625 kg m^2$.
Answer: 0.625 kg m^2
TGT core
A solid sphere of mass 1 kg and radius 0.1 m rolls without slipping on a horizontal surface with a linear velocity of 2 m/s. Calculate its total kinetic energy.
Total kinetic energy of a rolling body is the sum of translational and rotational kinetic energies: $K = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2$.
For a solid sphere, the moment of inertia about its diameter is $I = \frac{2}{5} M R^2$.
Substitute omeg$a = v / R into the kinetic energy equation: K = \frac{1}{2} M v^2 + \frac{1}{2} (\frac{2}{5} M R^2) (\frac{v}{R})^2 = \frac{1}{2} M v^2 + \frac{1}{5} M v^2 = \frac{7}{10} M v^2$.
A uniform rod of length L and mass M is free to rotate in a vertical plane about a smooth horizontal axis passing through one of its ends. If it is released from rest from the horizontal position, find its angular velocity when it reaches the vertical position.
The moment of inertia of a uniform rod about an axis through its end is $I = \frac{1}{3} M L^2$.
Loss in potential energy as the centre of mass falls by a vertical distance $h = L / 2$ is Delta U = M g (\frac{L}{2}).
By conservation of mechanical energy, loss in potential energy equals gain in rotational kinetic energy: M g \frac{L}{2} = \frac{1}{2} I omega^2.
Substitute $I = \frac{1}{3} M L^2: M g \frac{L}{2} = \frac{1}{2} (\frac{1}{3} M L^2) \omega^2$.
Simplify and solve for omega: omeg$a^2 = \frac{3g}{L}$, giving omega = sqrt{\frac{3g}{L}}.
Answer: sqrt{\frac{3g}{L}}
Assumptions and limitations
The body is assumed perfectly rigid, meaning internal deformations under applied forces are negligible.
The parallel and perpendicular axes theorems apply strictly to rigid bodies with time-invariant mass distribution.
Frictional forces during pure rolling are assumed non-dissipative in terms of mechanical energy transfer (no micro-sliding).
Common misconceptions
Common error: The moment of inertia of a body is a universal constant independent of the choice of axis.
Correct: Moment of inertia depends heavily on the orientation and location of the chosen axis of rotation.
MOI sums the mass-distance squared product (mr^2); changing the axis alters every particle's distance r.
Common error: The perpendicular axes theorem can be applied to any three-dimensional object like a sphere or a cube.
Correct: The perpendicular axes theorem is valid exclusively for two-dimensional planar laminar bodies.
The derivation relies on z-coordinates being identically zero across the entire mass distribution (sum z_i^2 dm = 0).
Common error: Friction always opposes motion and performs negative work on a rolling body.
Correct: Static friction acting during pure rolling does zero work because the point of contact is instantaneously at rest.
Work done is force multiplied by displacement of the point of application, which is zero for rolling without slipping.
Quick revision
Chasles' theorem states that general rigid body motion is a combination of translation of COM and rotation about COM.
Moment of Inertia (General): $I = \sum m_i r_i^2$
Radius of Gyration: k = sqrt{\frac{I}{M}}
Must remember
The general motion of a rigid body is a superposition of translation of its centre of mass and rotation about an axis passing through the centre of mass (Chasles' theorem).
The velocity of any arbitrary point on a rigid body is given by $v = v_{cm} + \omega \times r$.
Moment of inertia $I = \sum m_i r_i^2$ for discrete systems and $I = \int r^2 dm$ for continuous bodies, with SI unit $\text{kg m}^2$ and dimensional formula $[M L^2]$.
Radius of gyration $k$ relates moment of inertia and mass via $I = Mk^2$.
Perpendicular Axes Theorem states $I_z = I_x + I_y$, valid strictly for two-dimensional planar laminar bodies.
Parallel Axes Theorem states $I = I_{cm} + Md^2$, valid for any rigid body.
Rotational analogues of Newton's laws include $\tau = I\alpha$, angular momentum $L = I\omega$, and rotational kinetic energy $K = \frac{1}{2}I\omega^2$.
In the absence of external torques ($\tau_{ext} = 0$), total angular momentum is conserved.
Static friction during pure rolling does zero net work because the instantaneous point of contact is at rest.
Acceleration of an object undergoing pure rolling down an inclined plane is given by $a_{cm} = \frac{g \sin\theta}{1 + \frac{k^2}{R^2}}$.
Glossary
Rigid BodyA system of particles in which the separation distance between every pair of particles remains constant over time.
Moment of Inertia ($I$)A measure of the rotational inertia of a body, representing its resistance to angular acceleration.
Radius of Gyration ($k$)The root-mean-square distance of the mass particles of a body from the axis of rotation.
Perpendicular Axes TheoremRelates the moment of inertia of a planar lamina about three mutually perpendicular axes: $I_z = I_x + I_y$.
Parallel Axes TheoremRelates the moment of inertia about any axis to that about a parallel axis through the centre of mass: $I = I_{cm} + Md^2$.
Torque ($\tau$)The rotational analogue of force, defined as the cross product of position vector and force.
Angular Momentum ($L$)The rotational analogue of linear momentum, given by $L = I\omega$ for fixed-axis rotation.
Chasles' TheoremStates that any general displacement of a rigid body is equivalent to a translation of a point (like the centre of mass) followed by a rotation.
Translational vs Rotational Motion
Parameter
Translational Motion
Rotational Motion
Displacement
Linear displacement $x$
Angular displacement $\theta$
Velocity
Linear velocity $v = \frac{dx}{dt}$
Angular velocity $\omega = \frac{d\theta}{dt}$
Inertia
Mass $M$
Moment of Inertia $I = \sum m_i r_i^2$
Force / Torque
Force $F$
Torque $\tau$
Newton's Second Law
$F = Ma$
$\tau = I\alpha$
Momentum
Linear momentum $p = Mv$
Angular momentum $L = I\omega$
Kinetic Energy
$K = \frac{1}{2} M v^2$
$K = \frac{1}{2} I \omega^2$
Perpendicular vs Parallel Axes Theorems
Feature
Perpendicular Axes Theorem
Parallel Axes Theorem
Applicability
Strictly 2D planar laminar bodies
Any rigid body (1D, 2D, or 3D)
Mathematical Relation
$I_z = I_x + I_y$
$I = I_{cm} + Md^2$
Axis Configuration
Three mutually intersecting axes in the same plane
Two parallel axes (one passing through COM)
Dimensional Constraint
Requires $z_i = 0$ for all mass elements
Requires only spatial separation $d$ between axes
Practice Quiz
Choose an option to reveal the explanation.
1. What is the dimensional formula of the moment of inertia $I$?
Explanation: Moment of inertia is defined as $I = \sum m_i r_i^2$, which has dimensions of mass $\times$ distance squared, i.e., $[M][L]^2 = [M L^2]$. It has no time dependence in its base dimensions.
2. Which of the following theorems is applicable strictly to two-dimensional planar laminar bodies?
Explanation: The Perpendicular Axes Theorem ($I_z = I_x + I_y$) is valid strictly for plane laminar bodies because its mathematical derivation assumes the z-coordinate of every particle is zero.
3. If the radius of gyration of a body of mass $M$ is $k$, its moment of inertia about the axis is given by:
Explanation: By definition, the radius of gyration $k$ satisfies the relation $I = Mk^2$, where $k$ represents the root-mean-square distance of the mass distribution from the rotation axis.
4. The rotational analogue of Newton's second law for a rigid body rotating about a fixed axis is:
Explanation: Just as force $F = ma$ governs translational acceleration, net external torque $\tau = I\alpha$ governs angular acceleration $\alpha$ for a constant moment of inertia.
5. What is the work done by static friction during pure rolling of a rigid body on a rough surface?
Explanation: During pure rolling, the point of contact with the surface is instantaneously at rest. Since displacement of the point of application of friction is zero, the work done by static friction is zero.
6. According to Chasles' theorem, the general motion of a rigid body can be expressed as:
Explanation: Chasles' theorem states that any arbitrary displacement of a rigid body is equivalent to a translation of its centre of mass plus a rotation about an axis passing through the centre of mass.
7. What is the SI unit of angular momentum?
Explanation: Angular momentum is given by $L = I\omega$. The unit of $I$ is $\text{kg m}^2$ and $\omega$ is $\text{rad s}^{-1}$, yielding $\text{kg m}^2 \text{ s}^{-1}$ (or $\text{J s}$).
8. If no external torque acts on a rotating rigid body ($\tau_{ext} = 0$), which of the following quantities remains conserved?
Explanation: By the principle of conservation of angular momentum, since $\tau_{ext} = \frac{dL}{dt}$, if $\tau_{ext} = 0$, then $L = \text{constant}$.
9. The acceleration $a_{cm}$ of a rigid body rolling down an inclined plane of inclination $\theta$ is given by:
Explanation: Using energy conservation or Newton's laws for rolling motion, the linear acceleration down an incline is $a_{cm} = \frac{g \sin\theta}{1 + \frac{k^2}{R^2}}$.
10. The velocity of any arbitrary point on a rigid body is expressed as:
Explanation: Kinematics of a rigid body gives the velocity of any point as $v = v_{cm} + \omega \times r$, combining translational velocity of the centre of mass and rotational velocity.
11. Which of the following statements about the moment of inertia is correct?
Explanation: Moment of inertia depends heavily on the mass, its distribution relative to the axis, and the orientation and location of the chosen axis of rotation.
12. What is the expression for rotational kinetic energy of a rigid body rotating with angular velocity $\omega$?
Explanation: Rotational kinetic energy is the rotational counterpart to translational kinetic energy, expressed as $K = \frac{1}{2} I \omega^2$.
13. If the parallel axes theorem is stated as $I = I_{cm} + Md^2$, what does '$d$' represent?
Explanation: In the parallel axes theorem $I = I_{cm} + Md^2$, $d$ represents the perpendicular distance between the parallel axis of rotation and the parallel axis passing through the centre of mass.
14. In three-dimensional rigid body dynamics (PGT extension), motion is often analyzed using principal moments of inertia and:
Explanation: Euler's equations of motion describe the rotation of a rigid body in a rotating reference frame fixed to the body's principal axes of inertia.
15. Why does a gyroscope exhibit precession when subjected to a torque perpendicular to its spin axis?
Explanation: Precession occurs because the applied external torque equals the rate of change of angular momentum ($\tau = \frac{dL}{dt}$), causing the angular momentum vector to rotate without changing its magnitude.
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 circular ring of mass $M$ and radius $R$ about a tangent in its own plane is:
Explanation: The MOI of a ring about its diameter is $I_{diam} = \frac{1}{2}MR^2$. Using the parallel axes theorem with distance $d = R$ to a tangent in its plane: $I = I_{cm} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
2. A solid sphere and a hollow sphere of the same mass and radius roll down an inclined plane from rest without slipping. Which one reaches the bottom first?
Explanation: Acceleration is $a = \frac{g \sin\theta}{1 + k^2/R^2}$. For a solid sphere, $k^2/R^2 = 2/5 = 0.4$. For a hollow sphere, $k^2/R^2 = 2/3 \approx 0.67$. Since the solid sphere has a smaller $k^2/R^2$, its acceleration is larger, so it reaches the bottom first.
3. If a person sitting on a rotating stool with their arms stretched suddenly folds their arms inward, what happens to their angular velocity?
Explanation: By conservation of angular momentum ($L = I\omega = \text{constant}$), folding arms decreases the moment of inertia $I$. To keep $L$ constant, the angular velocity $\omega$ must increase.
4. The radius of gyration of a uniform disc of radius $R$ about an axis perpendicular to the disc and passing through its centre is:
Explanation: The moment of inertia of a disc about its central perpendicular axis is $I = \frac{1}{2}MR^2$. Setting this equal to $Mk^2$ gives $Mk^2 = \frac{1}{2}MR^2$, which yields $k = \frac{R}{\sqrt{2}}$.
5. What is the ratio of rotational kinetic energy to total kinetic energy for a rolling solid cylinder?
Explanation: For a solid cylinder (or disc), $I = \frac{1}{2}MR^2$. Rotational KE is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{4}Mv^2$. Translational KE is $K_{trans} = \frac{1}{2}Mv^2$. Total KE is $K_{total} = \frac{3}{4}Mv^2$. The ratio is $\frac{\frac{1}{4}}{\frac{3}{4}} = 1:3$.
6. Euler's equations for a torque-free rigid body rotation about principal axes with principal moments $I_1, I_2, I_3$ and angular velocities $\omega_1, \omega_2, \omega_3$ include the term:
Explanation: Euler's dynamical equations for rigid body rotation in a principal body-fixed frame are given by $I_1 \dot{\omega}_1 - (I_2 - I_3)\omega_2 \omega_3 = 0$, leading to $I_1 \frac{d\omega_1}{dt} = (I_2 - I_3)\omega_2 \omega_3$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. A uniform rod of mass $M$ and length $L$ is rotated about an axis passing through its centre and perpendicular to its length. Its moment of inertia is:
Explanation: Integrating $I = \int x^2 dm$ from $-L/2$ to $L/2$ for a rod of linear mass density $M/L$ gives $I = \frac{1}{12}ML^2$.
2. Using the perpendicular axes theorem, if the moment of inertia of a square lamina about axes $x$ and $y$ in its plane are $I_x$ and $I_y$, its moment of inertia about the perpendicular axis $z$ is:
Explanation: The Perpendicular Axes Theorem states that $I_z = I_x + I_y$ for any plane laminar body where $x$ and $y$ are orthogonal axes in the plane of the lamina.
3. If a constant torque of $50 \text{ N m}$ is applied to a wheel with moment of inertia $10 \text{ kg m}^2$, what is its angular acceleration?
Explanation: Using $\tau = I\alpha$, we have $\alpha = \frac{\tau}{I} = \frac{50 \text{ N m}}{10 \text{ kg m}^2} = 5 \text{ rad s}^{-2}$.
4. Which of the following bodies has the maximum moment of inertia about an axis passing through its centre and perpendicular to its plane, given equal mass $M$ and radius $R$?
Explanation: For a ring, $I = MR^2$. For a disc, $I = \frac{1}{2}MR^2$. For a solid sphere (about diameter), $I = \frac{2}{5}MR^2$. For a spherical shell, $I = \frac{2}{3}MR^2$. The circular ring has the largest numerical coefficient ($1$).ဲ့
5. A particle moves with velocity $v = v_{cm} + \omega \times r$. If the rotation is pure rotation about a fixed axis passing through the origin ($v_{cm} = 0$), the velocity of a particle at position $r$ is:
Explanation: When $v_{cm} = 0$, the velocity reduces strictly to $v = \omega \times r$, which represents the linear velocity of a point at position vector $r$ in pure rotation.
6. The kinetic energy of a rigid body executing both translation and rotation is given by:
Explanation: Total kinetic energy of a rigid body in general motion is the sum of translational kinetic energy of its centre of mass and rotational kinetic energy about the centre of mass: $K = \frac{1}{2} M v_{cm}^2 + \frac{1}{2} I_{cm} \omega^2$.
7. What is the radius of gyration of a uniform circular ring of radius $R$ about an axis passing through its centre and perpendicular to its plane?
Explanation: For a ring, $I = MR^2$. Equating this to $Mk^2$ gives $Mk^2 = MR^2$, which implies $k = R$.
8. If the angular momentum of a rotating body is doubled while its moment of inertia remains constant, its rotational kinetic energy becomes:
Explanation: Rotational kinetic energy can be expressed in terms of angular momentum as $K = \frac{L^2}{2I}$. If $L$ is doubled ($2L$), $K$ becomes $\frac{(2L)^2}{2I} = 4 \left(\frac{L^2}{2I}\right)$, meaning it is quadrupled.
9. In advanced rigid body dynamics (PGT extension), principal moments of inertia $I_1, I_2, I_3$ are the eigenvalues of which mathematical object?
Explanation: The inertia tensor is a symmetric $3 \times 3$ matrix whose eigenvalues represent the principal moments of inertia of the rigid body.
10. A sphere rolls without slipping on a horizontal surface. What fraction of its total kinetic energy is translational?
Explanation: For a solid sphere, $I = \frac{2}{5}MR^2$. $K_{trans} = \frac{1}{2}Mv^2$ and $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{5}Mv^2$. Total KE is $\frac{7}{10}Mv^2$. The fraction that is translational is $\frac{\frac{1}{2}}{\frac{7}{10}} = \frac{5}{7}$.