Angular Momentum — Revision Notes

⚡ 30-Second Revision

Point Particle: — $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$
Magnitude (Point Particle): — $L = rp\sin\theta = rmv\sin\theta$
Rigid Body (Fixed Axis): — $L = I\omega$
Relation to Torque: — $\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}$
Conservation of Angular Momentum: — If $\vec{\tau}_{ext} = 0$, then $\vec{L}_{total} = \text{constant}$ (i.e., $I_1\omega_1 = I_2\omega_2$)
Units: — kg\cdot m^2/s or J\cdot s
Direction: — Right-hand rule for $\vec{r} \times \vec{p}$

2-Minute Revision

Angular momentum is the rotational analogue of linear momentum. For a point particle, it's defined as $\vec{L} = \vec{r} \times \vec{p}$, where $\vec{r}$ is the position vector from the origin and $\vec{p}$ is the linear momentum.

Its magnitude is $L = rmv\sin\theta$, and its direction is given by the right-hand rule. For a rigid body rotating about a fixed axis, angular momentum is simpler: $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular velocity.

The most crucial concept is the conservation of angular momentum: if the net external torque ($\vec{\tau}_{ext}$) acting on a system is zero, its total angular momentum ($\vec{L}_{total}$) remains constant.

This means if a system's moment of inertia changes, its angular velocity must adjust to keep $I\omega$ constant (e.g., a figure skater). The rate of change of angular momentum is equal to the net external torque: $\vec{\tau}_{ext} = d\vec{L}/dt$.

Remember that angular momentum is origin-dependent.

5-Minute Revision

Angular momentum, $\vec{L}$, is a fundamental vector quantity describing rotational motion. For a point particle, it's defined as $\vec{L} = \vec{r} \times \vec{p}$, where $\vec{r}$ is the position vector from a chosen origin and $\vec{p} = m\vec{v}$ is its linear momentum.

The magnitude is $L = rmv\sin\theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{v}$. The direction is found using the right-hand rule. For a rigid body rotating about a fixed axis, the angular momentum simplifies to $L = I\omega$, where $I$ is the moment of inertia about that axis and $\omega$ is the angular velocity.

The SI unit for angular momentum is kg\cdot m^2/s or J\cdot s.\n\nThe most important principle is the conservation of angular momentum. If the net external torque ($\vec{\tau}_{ext}$) acting on a system is zero, its total angular momentum remains constant.

This means $I_1\omega_1 = I_2\omega_2$ if the moment of inertia changes. For example, when a figure skater pulls her arms in, her moment of inertia ($I$) decreases, so her angular velocity ($\omega$) increases to conserve $L$.

Another key relation is that the net external torque is equal to the rate of change of angular momentum: $\vec{\tau}_{ext} = d\vec{L}/dt$. If torque is constant, $\Delta L = \tau \Delta t$. Always remember angular momentum is defined with respect to an origin, and this origin must be consistent throughout a problem.

\n\nWorked Example: A uniform rod of mass $M$ and length $L$ is rotating about its center with angular speed $\omega$. Two point masses, each of mass $m$, are attached to its ends. What is the new angular speed if the masses are moved to $L/4$ from the center?

\nSolution: \n1. **Initial Moment of Inertia ($I_1$):** Rod about center: $I_{rod} = \frac{1}{12}ML^2$. Two masses at ends ($L/2$ from center): $I_{masses} = m(L/2)^2 + m(L/2)^2 = 2m(L^2/4) = \frac{1}{2}mL^2$.

\n$I_1 = \frac{1}{12}ML^2 + \frac{1}{2}mL^2 = (\frac{M}{12} + \frac{m}{2})L^2$. \n2. **Final Moment of Inertia ($I_2$):** Rod remains same: $I_{rod} = \frac{1}{12}ML^2$. Two masses at $L/4$ from center: $I'_{masses} = m(L/4)^2 + m(L/4)^2 = 2m(L^2/16) = \frac{1}{8}mL^2$.

\n$I_2 = \frac{1}{12}ML^2 + \frac{1}{8}mL^2 = (\frac{M}{12} + \frac{m}{8})L^2$. \n3. Conservation of Angular Momentum: $I_1\omega_1 = I_2\omega_2$. \n$(\frac{M}{12} + \frac{m}{2})L^2 \omega = (\frac{M}{12} + \frac{m}{8})L^2 \omega_2$.

\n$\omega_2 = \frac{(\frac{M}{12} + \frac{m}{2})}{(\frac{M}{12} + \frac{m}{8})}\omega = \frac{(M + 6m)}{(M + 3m/2)}\omega = \frac{2(M + 6m)}{(2M + 3m)}\omega$.

Prelims Revision Notes

Angular Momentum (L)

Definition for Point Particle: — $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$

* Magnitude: $L = rp\sin\theta = rmv\sin\theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{p}$. * Direction: Perpendicular to the plane of $\vec{r}$ and $\vec{p}$, given by the right-hand rule. * Units: kg\cdot m^2/s or J\cdot s. Dimensions: $[ML^2T^{-1}]$.

Definition for Rigid Body (Fixed Axis): — $L = I\omega$

* $I$: Moment of inertia (rotational inertia, depends on mass distribution and axis). * $\omega$: Angular velocity.

Relation to Torque: — $\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}$

* Net external torque causes a change in angular momentum. * If $\vec{\tau}_{ext}$ is constant, $\Delta \vec{L} = \vec{\tau}_{ext} \Delta t$ (Angular Impulse-Momentum Theorem).

Conservation of Angular Momentum

Principle: — If the net external torque acting on a system is zero ($\vec{\tau}_{ext} = 0$), then the total angular momentum of the system remains constant (conserved).

* Mathematically: $\vec{L}_{total} = \text{constant}$. * For a rigid body or system with changing moment of inertia: $I_1\omega_1 = I_2\omega_2$.

Key Applications/Examples:

* Figure Skater: Pulling arms in decreases $I$, increases $\omega$. * Diver: Tucking body decreases $I$, increases $\omega$ for more somersaults. * Planetary Motion: Angular momentum of a planet about the Sun is conserved (Kepler's 2nd Law). * Rotating Platforms: Person moving towards/away from center changes $I$, thus $\omega$.

Conditions: — Only external torques matter. Internal torques do not change the total angular momentum of the system.

Important Points for NEET

Origin Dependence: — Angular momentum is always defined with respect to a specific origin. Be consistent.
Vector Nature: — Remember $\vec{L}$ is a vector. Its direction is crucial.
Moment of Inertia: — Be familiar with $I$ formulas for common shapes (disc, ring, rod, sphere) and parallel/perpendicular axis theorems.
Distinguish: — Don't confuse angular momentum with linear momentum or rotational kinetic energy ($KE_{rot} = \frac{1}{2}I\omega^2 = \frac{L^2}{2I}$).

Vyyuha Quick Recall

To remember the conservation of angular momentum: 'I Will Always Conserve'

I — Moment of Inertia
W — Angular Welocity ($\omega$)
A — Always
C — Conserve

This reminds you that $I\omega$ is conserved when external torque is zero. It's a simple way to recall the core principle $I_1\omega_1 = I_2\omega_2$.