Angular Momentum and General Rotation Study Notes

Chapter 11: Angular Momentum and General Rotation

11-1 Angular Momentum—Objects Rotating About a Fixed Axis

• Conceptual Overview: In Chapter 10, rotational motion was analyzed for objects with a fixed axis of rotation in an inertial reference frame. Chapter 11 extends these concepts to general rotational motion, where the axis might not be fixed.
• Definition of Angular Momentum ($L$): For an object with a moment of inertia $I$ rotating about a fixed axis through the center of mass (CM), the angular momentum is the rotational analog of linear momentum ($p = mv$).
  • Equation: $L = I\omega$
  • where $\omega$ is the angular velocity.
  • SI Units: The unit for $L$ is $\text{kg} \cdot \text{m}^2/\text{s}$. There is no special name for this unit.
• Rotational Equivalent of Newton's Second Law: Newton's second law for linear motion ($\sum F = \frac{dp}{dt}$) has a rotational counterpart:
  • Equation: $\sum \tau = \frac{dL}{dt}$
  • This equation is valid even if the moment of inertia ($I$) changes over time. It is also valid for a system of objects rotating about a fixed axis where $\sum \tau$ is the net external torque. It applies to moving objects if the rotation is about an axis passing through the center of mass.
• Law of Conservation of Angular Momentum:
  • Condition: The total angular momentum of a rotating object remains constant if the net external torque ($\sum \tau$) acting on it is zero.
  • Equation: If $\sum \tau = 0$, then $\frac{dL}{dt} = 0$ and $L = I\omega = \text{constant}$.
  • Application: $I_0 \omega_0 = I\omega$. If a part of the object changes position (changing $I$), $\omega$ must change to keep the product constant.

Examples of Angular Momentum Conservation

• Figure Skater: A skater spinning on ice illustrates this law.
  • When arms are outstretched, the radius ($R$) of the mass distribution is large, making $I = \sum mR^2$ large and $\omega$ small.
  • When the skater pulls her arms in, $R$ decreases, lowering $I$. Consequently, $\omega$ increase significantly. A reduction of $I$ by a factor of $2$ results in a doubling of $\omega$.
• The Diver: A diver leaving the board has initial angular momentum about her CM.
  • By curling into a "tuck" position, she reduces her moment of inertia (sometimes by a factor of $3$), causing her to rotate quickly.
  • Stretching out before entering the water increases $I$, reducing $\omega$ to a small value.
  • Gravity and Force: While gravity acts on the diver (net force is not zero), the net torque about the CM is zero because gravity is considered to act at the center of mass.
• Rotating Platform (Frictionless): A person standing on a circular platform initially at rest walks counterclockwise along the edge.
  • The person's angular momentum ($L = I\omega = (mR^2)(v/R)$) points upward.
  • To conserve the initial angular momentum of zero, the platform must rotate in the opposite direction (downward momentum), such that $\mathbf{L}_{\text{platform}} = -\mathbf{L}_{\text{person}}$.

11-2 Vector Cross Product; Torque as a Vector

• Vector Cross Product Definition: The cross product of two vectors $\mathbf{A}$ and $\mathbf{B}$ is a third vector $\mathbf{C} = \mathbf{A} \times \mathbf{B}$.
  • Magnitude: $C = |\mathbf{A} \times \mathbf{B}| = AB \sin(\theta)$, where $\theta$ is the angle ($< 180^{\circ}$) between the tails of the vectors.
  • Direction: Determined by the Right-Hand Rule (RHR). Point fingers along $\mathbf{A}$, curl them toward $\mathbf{B}$; the thumb points in the direction of $\mathbf{C}$. $\mathbf{C}$ is perpendicular to the plane of $\mathbf{A}$ and $\mathbf{B}$.
• Component Form: Using the determinant memory aid:
  • $\mathbf{A} \times \mathbf{B} = \begin{pmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{pmatrix} = (A_y B_z - A_z B_y) \mathbf{\hat{i}} + (A_z B_x - A_x B_z) \mathbf{\hat{j}} + (A_x B_y - A_y B_x) \mathbf{\hat{k}}$
• Properties of the Cross Product:
  • $\mathbf{A} \times \mathbf{A} = 0$
  • $\mathbf{A} \times \mathbf{B} = -(\mathbf{B} \times \mathbf{A})$ (Anti-commutative)
  • $\mathbf{A} \times (\mathbf{B} + \mathbf{C}) = (\mathbf{A} \times \mathbf{B}) + (\mathbf{A} \times \mathbf{C})$ (Distributive)
  • $\frac{d}{dt} (\mathbf{A} \times \mathbf{B}) = \frac{d\mathbf{A}}{dt} \times \mathbf{B} + \mathbf{A} \times \frac{d\mathbf{B}}{dt}$
• The Torque Vector: Defined as the cross product of the position vector $\mathbf{r}$ and the force vector $\mathbf{F}$.
  • Equation: $\mathbf{\tau} = \mathbf{r} \times \mathbf{F}$
  • The torque is calculated relative to a specific point (the origin $O$).
  • For a system of particles: $\mathbf{\tau}_{\text{net}} = \sum (\mathbf{r}_i \times \mathbf{F}_i)$.

11-3 Angular Momentum of a Particle

• Definition: The angular momentum $\mathbf{L}$ of a particle with mass $m$ and momentum $\mathbf{p}$ relative to origin $O$ is:
  • Equation: $\mathbf{L} = \mathbf{r} \times \mathbf{p} = \mathbf{r} \times m\mathbf{v}$
  • Magnitude: $L = rp \sin(\theta) = rp_{\perp} = r_{\perp} p$.
• Newton's Second Law for a Particle:
  • $\frac{d\mathbf{L}}{dt} = \frac{d}{dt} (\mathbf{r} \times \mathbf{p}) = (\mathbf{v} \times \mathbf{p}) + (\mathbf{r} \times \frac{d\mathbf{p}}{dt})$
  • Since $\mathbf{v}$ is parallel to $\mathbf{p}$, $\mathbf{v} \times \mathbf{p} = 0$.
  • Thus, $\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \sum \mathbf{F} = \sum \mathbf{\tau}$.
  • This is valid only in an inertial reference frame.

11-4 Angular Momentum and Torque for a System of Particles

• Total Angular Momentum: $\mathbf{L} = \sum \mathbf{L}_i$.
• Internal vs. External Torques: By Newton's third law, internal torques between particles in a system sum to zero. Therefore, the rate of change of total angular momentum depends only on external torques:
  • Equation: $\sum \mathbf{\tau}_{\text{ext}} = \frac{d\mathbf{L}}{dt}$
• Validity and Reference Points: The relationship $\sum \mathbf{\tau} = \frac{d\mathbf{L}}{dt}$ holds if torques and angular momentum are calculated about:
  1. A point fixed in an inertial reference frame.
  2. A point moving uniformly in an inertial frame.
  3. The Center of Mass (CM) of the system, even if the CM is accelerating.
• CM Derivation Requirement: The derivation proves $\frac{d\mathbf{L}_{\text{CM}}}{dt} = \sum \mathbf{\tau}_{\text{CM}}$ by defining the position of a particle relative to the CM as $\mathbf{r}_i = \mathbf{r}_{\text{CM}} + \mathbf{r}_i^*$.

11-5 Angular Momentum and Torque for a Rigid Object

• Component along Axis of Rotation ($z$): For any rigid object rotating about a fixed direction axis, the $\mathbf{z}$-component of angular momentum is:
  • Equation: $L_z = I\omega$
• Symmetry Axis: If the rotation axis is a symmetry axis passing through the CM, then $\mathbf{L}$ is parallel to $\mathbf{\omega}$, and the vector relation $\mathbf{L} = I\mathbf{\omega}$ holds.
• Non-Symmetry Axis Exceptions: If the axis is not a symmetry axis, $\mathbf{L}$ and $\mathbf{\omega}$ are not necessarily parallel.
  • Example: Two masses on a tilted rod. As the rod rotates, the direction of $\mathbf{L}$ changes even if $\omega$ is constant.
• Rotational Imbalance: When the direction of $\mathbf{L}$ changes ($\frac{d\mathbf{L}}{dt} \neq 0$), an external torque must be present. In a car, an unbalanced wheel causes $\mathbf{L}$ to wobble, requiring the bearings to exert forces that create vibrations and wear. "Dynamic balancing" adds weights to align $\mathbf{L}$ with $\mathbf{\omega}$, making $\sum \tau_{\text{ext}} = 0$.

11-7 The Spinning Top and Gyroscope

• Precession: The motion of a spinning top where the rotation axis sweeps out a cone around the vertical ($z$) axis. This occurs because gravity exerts a torque perpendicular to the angular momentum vector.
• Torque on a Top: $\mathbf{\tau} = \mathbf{r} \times M\mathbf{g}$.
  • Magnitude: $\tau = rMg \sin(\phi)$.
  • The change in angular momentum is $d\mathbf{L} = \mathbf{\tau} dt$. Because $\mathbf{\tau}$ is horizontal and perpendicular to $\mathbf{L}$, the magnitude of $\mathbf{L}$ stays constant, but its direction rotates.
• Precession Angular Velocity ($\Omega$):
  • Equation: $\Omega = \frac{d\theta}{dt} = \frac{\tau}{L \sin(\phi)}$
  • Substituting $\tau = rMg \sin(\phi)$: $\Omega = \frac{Mgr}{L}$
  • Using $L = I\omega$: $\Omega = \frac{Mgr}{I\omega}$
  • Observation: Faster spin (higher $\omega$) leads to slower precession (lower $\Omega$).

11-8 & 11-9 Rotating Frames of Reference and the Coriolis Effect

• Noninertial Reference Frames: A rotating platform is a noninertial frame where Newton's laws do not hold directly. Observers inside see objects accelerate without an applied force.
• Pseudoforces (Fictitious Forces): To use $\sum F = ma$ in a noninertial frame, we add "fictitious" forces:
  • Centrifugal Force: Acts radially outward with magnitude $m\omega^2 r$.
  • Coriolis Force: Acts sideways on an object moving relative to the rotating frame.
• Coriolis Acceleration ($a_{\text{cor}}$):
  • Equation: $a_{\text{cor}} = 2\omega v$
  • Calculated by the displacement $s = \omega v t^2$ and comparing it to the kinematic formula $s = \frac{1}{2}at^2$.
• Terrestrial Manifestations:
  • Weather Patterns: In the Northern Hemisphere, moving air is deflected to the right. This creates counterclockwise winds around low-pressure areas (cyclones) and clockwise winds around high-pressure areas (anticyclones). In the Southern Hemisphere, the directions are reversed.
  • Falling Objects: An object dropped from a high tower is deflected slightly to the east because the top of the tower has a higher tangential speed than the base.

Questions & Discussion

• Exercise A: When a figure skater pulls in her arms, her moment of inertia decreases and angular velocity increases to conserve angular momentum. Does her rotational kinetic energy increase? If so, where does the energy come from?
  • Answer Logic: Yes, $K_{\text{rot}} = \frac{1}{2}I\omega^2$. Since $L = I\omega$ is constant, we can write $K_{\text{rot}} = \frac{L^2}{2I}$. As $I$ decreases, kinetic energy increases. This energy comes from the chemical energy in the skater's muscles as she performs work to pull her arms inward against the "centrifugal force."
• Diver and External Forces: Why is angular momentum conserved for a diver even though gravity is acting?
  • Answer Logic: While the net force is gravity ($Mg$), if we calculate torque about the diver's center of mass, the lever arm for gravity is zero. Thus, $\sum \tau_{\text{CM}} = 0$, ensuring $L_{\text{CM}}$ is conserved.