#AP Physics 1: Angular Momentum - The Night Before 🚀

Hey! Let's get you prepped for the exam with a super focused review of angular momentum. We'll break it down, make it stick, and get you feeling confident. Let's do this!

#5.E: Conservation of Angular Momentum

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• What it is: Angular momentum is like the rotational version of linear momentum. It's all about how much 'rotational motion' something has.
• The Rule: If there's no outside twisting force (torque) acting on a system, its total angular momentum stays the same. Think of it like a spinning figure skater – when they pull their arms in, they spin faster, but the total angular momentum doesn't change.

• Formula: $L = I\omega$ where:
  • $L$ is angular momentum (kg⋅m²/s)
  • $I$ is the moment of inertia (how hard it is to rotate something)
  • $\omega$ is angular velocity (how fast something is spinning)

#Skater Example

• Initial State (Arms Out): High moment of inertia ($I$) because mass is far from the axis of rotation, lower angular velocity ($\omega$).
• Final State (Arms In): Lower moment of inertia ($I$) because mass is closer to the axis of rotation, higher angular velocity ($\omega$).
• Key Point: Angular momentum ($L$) stays constant!

#Planetary Motion

• Closed Orbits: Planets in closed orbits have constant angular momentum.
• Kepler's 2nd Law: Planets sweep out equal areas in equal times. This means their angular velocity is constant, even as their linear velocity changes.
• Why? Gravitational force is a central force, so there's no external torque on the planet. Angular momentum is conserved!

#Example Problem: Disk and Rod Collision

Let's tackle a classic problem that combines these ideas:

#Part A: Before the Collision

• Disk's Rotational Inertia: $I_{disk} = m_{disk}x^2$
• Disk's Angular Momentum: $L_{disk} = m_{disk}v_0x$

#Part B: After the Collision

• Conservation of Angular Momentum: $L_{initial} = L_{final}$ $m_{disk}v_0x = (I_{rod} + m_{disk}x^2)\omega$
• Solving for Final Angular Speed: $\omega = \frac{m_{disk}v_0x}{I_{rod} + m_{disk}x^2}$

#Part C: Bouncing vs. Sticking

• Key Idea: If the disk bounces back, it transfers more angular momentum to the rod.
• Why? The change in angular momentum of the disk is greater when it reverses direction than when it just stops. So, the rod must gain more angular momentum to conserve the total angular momentum of the system.
• Result: The rod's angular speed will be greater when the disk bounces off.

#Final Exam Focus 🎯

#High-Priority Topics:

• Conservation of Angular Momentum: Understand when and how angular momentum is conserved.
• Relating Linear and Angular Quantities: Know how linear velocity and angular velocity relate, especially in circular motion.
• Moment of Inertia: Be able to calculate and use moment of inertia for different shapes.
• Problem-Solving: Practice setting up and solving conservation of angular momentum problems.

#Common Question Types:

• Multiple Choice: Conceptual questions about conservation, changes in angular velocity, and moment of inertia.
• Free Response: Problems involving collisions, rotating objects, and planetary motion. Expect to set up conservation equations and solve for unknowns.

#Last-Minute Tips:

• Time Management: Don't spend too long on any one question. If you're stuck, move on and come back later.
• Units: Always include units in your calculations and answers.
• Check your Work: Make sure your answers make sense physically. If you get a huge or tiny number, double-check your calculations.
• Stay Calm: You've got this! Take a deep breath and trust your preparation.

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Practice Question

Practice Questions

#Multiple Choice Questions

1. A spinning skater pulls their arms inward. Which of the following statements is true? (A) The skater's angular velocity decreases. (B) The skater's moment of inertia increases. (C) The skater's angular momentum increases. (D) The skater's angular momentum remains constant.

2. A planet moves in an elliptical orbit around a star. At which point in the orbit is the planet's angular momentum the greatest? (A) At the point closest to the star. (B) At the point farthest from the star. (C) The angular momentum is the same at all points in the orbit. (D) The angular momentum is zero at all points in the orbit.

3. A uniform rod is rotating about one end on a frictionless horizontal surface. A small piece of clay is dropped vertically onto the rod and sticks to it. Which of the following is true about the system of the rod and clay? (A) Both the angular momentum and the kinetic energy of the system is conserved. (B) The angular momentum of the system is conserved but the kinetic energy is not conserved. (C) The kinetic energy of the system is conserved but the angular momentum is not conserved. (D) Neither the angular momentum nor the kinetic energy of the system is conserved.

#Free Response Question

A uniform disk of mass $M$ and radius $R$ is rotating with an initial angular velocity $\omega_0$ about a frictionless axle through its center. A small block of mass $m$ is dropped vertically onto the disk at a distance $R/2$ from the center and sticks to the disk.

(a) Derive an expression for the moment of inertia of the disk.

(b) Derive an expression for the angular velocity of the disk-block system after the collision.

(c) If the block of mass $m$ was dropped at the edge of the disk, would the final angular velocity be greater than, less than, or equal to the angular velocity calculated in part (b)? Explain your reasoning.

(d) Is the kinetic energy of the system conserved in the collision? Justify your answer.

#Scoring Guidelines

(a) Moment of Inertia of the Disk (2 points)

• 1 point: Correctly stating the moment of inertia of a disk about its center: $I_{disk} = \frac{1}{2}MR^2$
• 1 point: Correctly stating the moment of inertia of a point mass: $I_{block} = m(\frac{R}{2})^2 = \frac{1}{4}mR^2$

(b) Angular Velocity After Collision (4 points)

• 1 point: Recognizing that angular momentum is conserved: $L_{initial} = L_{final}$
• 1 point: Correctly expressing the initial angular momentum: $L_{initial} = I_{disk}\omega_0 = \frac{1}{2}MR^2\omega_0$
• 1 point: Correctly expressing the final angular momentum: $L_{final} = (I_{disk} + I_{block})\omega_f = (\frac{1}{2}MR^2 + \frac{1}{4}mR^2)\omega_f$
• 1 point: Correctly solving for the final angular velocity: $\omega_f = \frac{\frac{1}{2}MR^2\omega_0}{\frac{1}{2}MR^2 + \frac{1}{4}mR^2} = \frac{2M\omega_0}{2M+m}$

(c) Angular Velocity with Block at Edge (2 points)

• 1 point: Stating that the final angular velocity would be less than in part (b).
• 1 point: Providing a correct explanation that the moment of inertia of the block would be greater when dropped at the edge, resulting in a lower final angular velocity for the same initial angular momentum.

(d) Kinetic Energy Conservation (2 points)

• 1 point: Stating that kinetic energy is not conserved.
• 1 point: Providing a correct justification that the collision is inelastic, and some kinetic energy is converted to other forms of energy (e.g., heat).

Good luck, you've got this! 💪