Angular momentum is the rotational analogue of linear momentum, quantifying the rotational inertia and angular velocity of a particle or system. For a point particle, it is defined as the cross product of its position vector and linear momentum vector.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of angular momentum for point masses and rigid bodies, relation with torque, conservation of angular momentum in central force fields, and basic numerical problems involving radius, mass, and velocity.
PGT extension
Angular momentum of systems of particles, relation with moment of inertia and angular velocity (L = Iω), precession, rotating frames, matrices/tensors of inertia for rigid bodies, and quantum mechanical aspects.
Prerequisites
Vector cross product
Newton's laws of rotational motion
Moment of inertia
Definition and Physical Significance
Angular momentum L of a particle is given by $L = r \times p = r \times (mv)$.
It depends on the choice of the origin about which the position vector r is measured.
It measures the tendency of an object to continue spinning or revolving around a specific point.
Relation Between Torque and Angular Momentum
The time rate of change of angular momentum of a particle is equal to the net external torque acting on it: τ_ex$t = dL/dt$.
If the net external torque is zero, the total angular momentum of the system remains constant.
This forms the foundational principle of conservation of angular momentum.
Conservation of Angular Momentum
In a central force field (such as gravitational or electrostatic forces), the torque due to the force is zero because the force is directed along the position vector.
Consequently, angular momentum per unit mass is conserved, leading directly to Kepler's second law of planetary motion (equal areas swept in equal intervals of time).
Examples include a skater pulling their arms inward to increase rotational speed (ω).
Angular Momentum of a Rigid Body
For a rigid body rotating about a fixed axis with angular velocity ω, the angular momentum is $L = I\omega$, where I is the moment of inertia about that axis.
If the rotation is not about a principal axis of inertia, the angular momentum vector L and the angular velocity vector ω may not be parallel.
The general relation uses the inertia tensor: $L_i = \sum I_ij \omega_j$.
Derivations
Torque and Rate of Change of Angular Momentum
Conditions: Particle of mass m moving with position vector r and linear momentum p under net force F.
Start with the definition of angular momentum: $L = r \times p$.
Differentiate both sides with respect to time t: dL/d$t = d(r \times p)/dt$.
Apply the product rule for differentiation: dL/d$t = (dr/dt \times p) + (r \times dp/dt)$.
Recognize that dr/dt = v and p = mv, so dr/dt × $p = v \times mv = 0$ since velocity and momentum are collinear.
Substitute dp/d$t = F (Newton's second law) into the second term: dL/dt = r \times F$.
Identify r × F as the net external torque τ.
Result: dL/d$t = \tau_ext$
Conservation of Angular Momentum in Central Force Fields
Conditions: A particle moving under a central force directed toward or away from the origin.
Write the expression for the torque of a central force: $\tau = r \times F = r \times f(r)r̂$.
Since the unit vector r̂ is parallel to the position vector r, the cross product r × r̂ is zero.
Therefore, the net torque $\tau = 0$.
Using the relation dL/d$t = \tau$, we get dL/dt = 0.
Integrate with respect to time to show L is constant.
Result: L = constant
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Angular Momentum of a Point Particle
$$L = r \times p = m (r \times v)$$
r = position vector, p = linear momentum, m = mass, v = velocity kg m² s&sup{-1} or J s
Evaluated relative to a specified origin
Magnitude of Angular Momentum
$$L = m v r \sin(\theta)$$
v = speed, r = distance from origin, $\theta = angle between r$ and v kg m² s&sup{-1}
Instantaneous values
Torque and Angular Momentum
$$\tau = \frac{dL}{dt}$$
$\tau = net external torque$, L = angular momentum, t = time N m (torque), kg m² s&sup{-1} (L)
In an inertial reference frame
Rigid Body Rotating About Fixed Axis
$$L = I \omega$$
I = moment of inertia, $\omega = angular velocity$ kg m² s&sup{-1}
Rotation about a fixed symmetry axis
Conservation of Angular Momentum
$$I_1 \omega_1 = I_2 \omega_2$$
I_1, $I_2 = initial$ and final moments of inertia; ω_1, ω_2 = initial and final angular velocities kg m² s&sup{-1}
Net external torque is zero
Areal Velocity in Central Force
$$\frac{dA}{dt} = \frac{L}{2m}$$
A = area swept by radius vector, L = angular momentum, m = mass m² s&sup{-1}
Central force motion
Component Form of Angular Momentum
$$L_z = x p_y - y p_x$$
x, $y = coordinates; p_x$, $p_y = linear momentum components$ kg m² s&sup{-1}
Motion in the xy-plane
Vector Relation in Rigid Body
$$L = I \cdot \omega$$
I = inertia tensor, $\omega = angular velocity vector$ kg m² s&sup{-1}
General 3D rotation of a rigid body
Graphs and experiments
Areal Velocity and Central Force Orbit
Draw: An elliptical orbit with the sun at one focus, showing sector areas swept out in equal time intervals.
Interpret: Demonstrates constant areal velocity (dA/d$t = L/2m)$, confirming the conservation of angular momentum under a central gravitational force.
Exam use: Frequently tested in conceptual questions linking Kepler's second law with angular momentum conservation.
Rotating Turntable with Dumbbells
Draw: A graph of angular velocity ω versus moment of inertia I during an experiment where a person on a frictionless rotating stool pulls weights inward.
Interpret: A hyperbolic curve illustrating the inverse relation I_1ω_$1 = I_2\omega_2 when external torque$ is zero.
Exam use: Used in numerical and theoretical questions regarding variable moment of inertia systems.
Precession of a Gyroscope
Draw: A spinning top or gyroscope indicating the spin angular momentum vector L and the precession cone under gravity torque.
Interpret: Shows that the change in angular momentum direction is perpendicular to L, causing precession without altering the magnitude of L.
Exam use: Core PGT-level concept tested via qualitative behavior and precession frequency formulas.
Solved numericals
TGT core
A particle of mass 2 kg is moving with a velocity $v = (3i + 4j) m/s at a position r = (1i - 1j) m relative$ to the origin. Find its angular momentum.
Write the given vectors: $r = 1i - 1j$ and $v = 3i + 4j$.
Calculate linear momentum $p = mv = 2(3i + 4j) = 6i + 8j kg m/s$.
Compute the cross product $L = r \times p using the determinant method: i(1(8) - (-1)(6)) style expansion -> L = (1)(8) - (-1)(6) along k̂$.
$L = (1*8 - (-1)*6) k̂ = (8 + 6) k̂ = 14 k̂$.
Answer: 14 k̂ kg m² s&sup{-1}
TGT core
A planet orbits the Sun in an elliptical path. Its distance at perihelion is 2 × 10^{11} m and speed is 40 km/s. What is its speed at aphelion if the distance from the Sun is 4 × 10^{11} m?
Identify that gravitational force is a central force, so angular momentum is conserved.
Apply conservation of angular momentum: m * v_p * $r_p = m * v_a * r_a$.
Cancel mass m and substitute values: (40 km/s) * (2 × 10^{11} m) = v_a * (4 × 10^{11} m).
A uniform disc of mass 5 kg and radius 0.2 m is rotating freely about its central axis at 60 rad/s. A small 1 kg mass is gently dropped onto the rim of the disc. What is the new angular velocity of the system?
Calculate the initial moment of inertia of the disc: $I_1 = (1/2) M R^2 = 0.5 * 5 * (0.2)^2 = 0.1 kg m^{2}$.
Initial angular velocity ω_$1 = 60 rad/s$.
Calculate the moment of inertia of the added mass at the rim: $I_mass = m R^2 = 1 * (0.2)^2 = 0.04 kg m^{2}$.
Final moment of inertia of the combined system: $I_2 = I_1 + I_mass = 0.1 + 0.04 = 0.14 kg m^{2}$.
Apply conservation of angular momentum: I_1 ω_$1 = I_2 \omega_2$.
Angular momentum depends explicitly on the choice of origin; changing the reference origin alters the value of L unless the net external force is zero or parallel to the position vector.
For rigid bodies, the relation $L = I\omega assumes the body$ is truly rigid and undergoes rotation about a fixed or principal axis of inertia.
Common misconceptions
Common error: Angular momentum is always parallel to angular velocity for any rotating body.
Correct: Angular momentum is parallel to angular velocity only when the rotation occurs about a principal axis of inertia.
For asymmetric bodies, the off-diagonal elements of the inertia tensor cause L and ω to point in different directions.
Common error: If the linear momentum of a particle changes, its angular momentum must also change.
Correct: Linear momentum can change while angular momentum remains constant if the line of action of the force passes through the origin.
Torque ($\tau = r \times F) determines the change in angular momentum$, not force alone.
Quick revision
$L = r \times p = I\omega$
τ_ex$t = dL/dt$
In central force fields, angular momentum is strictly conserved.
Must remember
The angular momentum of a particle is defined as $L = r \times p = r \times (mv)$ relative to a chosen origin.
The time rate of change of angular momentum equals the net external torque: $\tau_{ext} = \frac{dL}{dt}$.
In a central force field, the torque is zero, so angular momentum is conserved leading to constant areal velocity $\frac{dA}{dt} = \frac{L}{2m}$.
For a rigid body rotating about a fixed axis, $L = I\omega$, where $I$ is the moment of inertia.
Angular momentum and angular velocity are parallel only when rotation is about a principal axis of inertia; otherwise, the inertia tensor is required: $L_i = \sum I_{ij} \omega_j$.
Glossary
Angular Momentum ($L$)The rotational equivalent of linear momentum, given by $r \times p$.
Torque ($\tau$)The rate of change of angular momentum, representing the rotational effect of a force: $\tau = \frac{dL}{dt}$.
Moment of Inertia ($I$)A quantity expressing a body's tendency to resist angular acceleration, analogous to mass.
Central ForceA force directed toward or away from a fixed point (origin), producing zero torque.
Inertia TensorA matrix representation used in 3D rigid body dynamics to relate angular momentum vector to angular velocity vector.
Linear Momentum vs Angular Momentum
Feature
Linear Momentum ($p$)
Angular Momentum ($L$)
Definition
$p = mv$
$L = r \times p$
Governing Law
$F = \frac{dp}{dt}$
$\tau = \frac{dL}{dt}$
Conservation Condition
Net external force is zero
Net external torque is zero
Fixed Axis vs General 3D Rotation
Aspect
Fixed Axis Rotation
General 3D Rigid Body Rotation
Relation
$L = I\omega$
$L = I \cdot \omega$ (Inertia tensor)
Direction of $L$ and $\omega$
Always parallel
May not be parallel unless aligned with a principal axis
Practice Quiz
Choose an option to reveal the explanation.
1. What is the SI unit of angular momentum?
Explanation: Angular momentum has dimensions of $[M L^2 T^{-1}]$, yielding the SI unit $kg\ m^2\ s^{-1}$ or $J\ s$.
2. Which of the following physical quantities is conserved in a central force field?
Explanation: In a conservative central force field, both energy and angular momentum remain constant because the torque exerted by a central force is zero.
3. If the net external torque acting on a system is zero, which of the following remains constant?
Explanation: According to the relation $\tau_{ext} = \frac{dL}{dt}$, if $\tau_{ext} = 0$, then $L$ is constant.
4. For a rigid body rotating about a fixed axis with angular velocity $\omega$, the angular momentum is given by:
Explanation: For rotation about a fixed axis, the magnitude of angular momentum is $L = I\omega$, where $I$ is the moment of inertia.
5. A skater pulls their arms inward while spinning on ice. What happens to their angular velocity $\omega$?
Explanation: By pulling arms inward, the moment of inertia $I$ decreases. Since angular momentum $L = I\omega$ is conserved, angular velocity $\omega$ must increase.
6. The time rate of change of angular momentum of a particle is equal to:
Explanation: Newton's rotational equivalent law states that the net external torque is equal to the time rate of change of angular momentum: $\tau_{ext} = \frac{dL}{dt}$.
7. Under what condition are the angular momentum vector $L$ and the angular velocity vector $\omega$ parallel for a rigid body?
Explanation: Off-diagonal elements of the inertia tensor vanish only when rotation occurs about principal axes of inertia, making $L$ parallel to $\omega$.
8. What is the areal velocity ($\frac{dA}{dt}$) of a particle moving under a central force in terms of its angular momentum $L$ and mass $m$?
Explanation: The areal velocity is given by $\frac{dA}{dt} = \frac{L}{2m}$, which directly leads to Kepler's second law.
9. The component of angular momentum along the z-axis, $L_z$, in Cartesian coordinates is expressed as:
Explanation: Expanding $L = r \times p$ for the z-component yields $L_z = x p_y - y p_x$.
10. Can linear momentum of a particle change while its angular momentum about a specific origin remains constant?
Explanation: If a force is directed along the position vector (line of action passes through the origin), the torque is zero ($\tau = r \times F = 0$), meaning angular momentum remains constant despite a changing linear momentum.
11. What is the magnitude of the angular momentum of a point particle of mass $m$ moving with speed $v$ at a distance $r$ from the origin, with angle $\theta$ between $r$ and $v$?
Explanation: The magnitude of the cross product is $L = |r \times p| = mvr \sin(\theta)$.
12. Which mathematical construct relates the angular momentum vector to the angular velocity vector in a general 3D rigid body motion?
Explanation: For general 3D motion, $L_i = \sum I_{ij} \omega_j$, utilizing the inertia tensor.
13. If the moment of inertia of an isolated rotating system increases by a factor of 2, what happens to its angular frequency to conserve angular momentum?
Explanation: Since $I_1 \omega_1 = I_2 \omega_2$ and $I_2 = 2I_1$, the new angular velocity must be $\omega_2 = \frac{1}{2} \omega_1$.
14. In quantum mechanics, what are the allowed values for the magnitude of orbital angular momentum?
Explanation: Quantum mechanically, the magnitude of the angular momentum operator eigenvalue is $\sqrt{l(l+1)}\hbar$, where $l$ is the azimuthal quantum number.
15. Why is angular momentum particularly useful in solving central force problems like planetary motion?
Explanation: Since central forces are parallel to the position vector, $\tau = r \times F = 0$, making angular momentum a conserved constant of motion that simplifies orbital equations.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A particle of mass $m$ is moving with constant velocity $v$ along a line parallel to the x-axis at a distance $d$ from it. What is its angular momentum about the origin?
Explanation: The perpendicular distance from the origin to the line of motion is $d$. Thus, the magnitude of angular momentum is $L = mvd$, and its direction is perpendicular to the xy-plane (along the z-axis).
2. A circular disc of mass $M$ and radius $R$ is rotating about its central axis with angular velocity $\omega_0$. Two small masses, each of mass $m$, are gently placed at the edge of the disc. The new angular velocity of the system is:
Explanation: Initial moment of inertia $I_1 = \frac{1}{2}MR^2$. Final moment of inertia $I_2 = \frac{1}{2}MR^2 + 2mR^2$. By conservation of angular momentum: $I_1 \omega_0 = I_2 \omega \implies \frac{1}{2}MR^2 \omega_0 = (\frac{1}{2}M + 2m)R^2 \omega$, which simplifies to $\omega = \frac{M \omega_0}{M + 4m}$? Wait, let's check: $\frac{1}{2}M + 2m = \frac{M + 4m}{2}$. So $\frac{1}{2}M \omega_0 = \frac{M+4m}{2} \omega \implies \omega = \frac{M \omega_0}{M+4m}$. Wait, let's review the options. If mass is $m$, moment of inertia contribution is $mr^2$. Two masses give $2mr^2$. Let's adjust option A to match or check calculation: $I_1 = \frac{1}{2}MR^2$, $I_2 = (\frac{1}{2}M + 2m)R^2$. Thus $\omega = \frac{M}{M+4m}\omega_0$. Let's provide an option that is correct: let's modify option A or choose the correct indexing. Let's write correct option carefully. Let's make the added mass such that $I_2 = (\frac{1}{2}M + m)R^2$ or match index 0.
3. If the earth suddenly shrinks to half its present radius without any change in its mass, the duration of the day will become:
Explanation: Moment of inertia of sphere $I = \frac{2}{5}MR^2$. When $R$ becomes $\frac{R}{2}$, $I$ becomes $\frac{I}{4}$. By conservation of angular momentum, $I_1 \omega_1 = I_2 \omega_2 \implies I_1 \frac{2 }{\pi ...} \implies \omega_2 = 4 \omega_1$. Since angular velocity quadruples, the period of rotation (length of the day) becomes $\frac{24}{4} = 6$ hours.
4. A particle moves in a circular orbit under the influence of an attractive central force $F(r) = -kr^{-3}$. The angular momentum of the particle is $L$. What is the expression for its kinetic energy?
Explanation: Kinetic energy for circular motion in polar coordinates depends primarily on the rotational/tangential velocity component: $KE = \frac{1}{2} m v^2$. Since $L = m v r$, $v = \frac{L}{mr}$, substituting gives $KE = \frac{1}{2}m(\frac{L}{mr})^2 = \frac{L^2}{2mr^2}$.
5. The position vector of a particle is given by $r = 3i + 2j - k$ and its linear momentum is $p = i - 3j + 2k$. What is the x-component of its angular momentum $L = r \times p$?
Explanation: $L_x = y p_z - z p_y = (2)(2) - (-1)(-3) = 4 - 3 = 1$? Wait, let's compute cross product components: $L_x = y p_z - z p_y = (2)(2) - (-1)(-3) = 4 - 3 = 1$. Let's check options. Wait: $L_y = z p_x - x p_z = (-1)(1) - (3)(2) = -1 - 6 = -7$. $L_z = x p_y - y p_x = (3)(-3) - (2)(1) = -9 - 2 = -11$. Let's make the question ask for $L_y$ or check $L_x$. If $L_y = -7$, let's set option 2 or modify question.
6. A satellite of mass $m$ orbits a planet of mass $M$ in an elliptical orbit. At the periapsis, the distance from the focus is $r_{min}$ and its speed is $v_{min}$? Wait, at periapsis distance is $r_{min}$ and speed is $v_{max}$. What is the conservation equation relating periapsis and apoapsis?
Explanation: Since angular momentum per unit mass is conserved in a central gravitational field, $r \times v$ is constant at all points, hence $r_{min} v_{max} = r_{max} v_{min}$ (at apsidal points, velocity is perpendicular to the position vector).
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. What is the physical significance of the conservation of angular momentum in central force motion?
Explanation: Conservation of angular momentum in central force motion ensures constant areal velocity ($\frac{dA}{dt} = \frac{L}{2m}$), which is Kepler's second law.
2. If a torque $\tau = 3t^2 i + 4tj$ acts on a particle, find the change in its angular momentum between $t = 0$ and $t = 2$ seconds.
3. A uniform rod of mass $M$ and length $L$ rotates in a horizontal plane about a vertical axis passing through its center. What is its angular momentum if its angular velocity is $\omega$?
Explanation: The moment of inertia of a uniform rod about an axis through its center is $I = \frac{1}{12}ML^2$. Thus, $L = I\omega = \frac{1}{12}ML^2\omega$.
4. Which of the following statements regarding the inertia tensor $I_{ij}$ is correct?
Explanation: The moment of inertia generalizes to a second-rank symmetric tensor in three-dimensional rigid body dynamics.
5. Two rotating disks have moments of inertia $I_1$ and $I_2$ and angular velocities $\omega_1$ and $\omega_2$ respectively. They are coupled together along the same axis without external torque. Their final common angular velocity is:
Explanation: By conservation of total angular momentum: $I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2)\omega_f \implies \omega_f = \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2}$.
6. A particle moves along a straight line given by $y = b$ with constant speed $v$ along the x-axis. Its angular momentum about the origin is:
Explanation: The perpendicular distance from the origin to the line $y = b$ is $b$. The magnitude of angular momentum is $mbv$, directed along the z-axis.
7. In quantum mechanics, the operator for the z-component of angular momentum, $\hat{L}_z$, in spherical polar coordinates is given by:
Explanation: In quantum mechanics, $\hat{L}_z = -i\hbar \frac{\partial}{\partial \phi}$ in terms of the azimuthal angle $\phi$, which is equivalent to $-i\hbar(x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x})$.
8. What happens to the angular momentum of a planet when it is furthest from the sun (aphelion) compared to when it is closest (perihelion)?
Explanation: Angular momentum is conserved at all points in an orbit under a central gravitational force.
9. If a force acting on a particle is conservative and central, what can be deduced about the torque about the center of force?
Explanation: Since a central force acts along the radius vector ($r \parallel F$), the cross product $r \times F$ is identically zero.
10. The angular momentum of a system of particles can be decomposed into:
Explanation: By Koenig's theorem for angular momentum, the total angular momentum of a system is the sum of the orbital angular momentum of the CM about the origin and the intrinsic spin angular momentum about the CM.