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AP®PhysicsCollege BoardPhysics 1: Algebra-BasedStudy GuidesUnit 6: Energy & Momentum of Rotating SystemsAngular Momentum & Angular ImpulseAngular Momentum

Angular Momentum (College Board AP® Physics 1: Algebra-Based): Study Guide

Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

Updated on

Angular momentum

  • Angular momentum is the rotational equivalent of linear momentum

  • The magnitude of the angular momentum of a rigid system about a specific axis can be described by the equation:

L space equals space I omega

  • Where:

    • L = angular momentum, in kg times straight m squared divided by straight s

    • I = rotational inertia, in kg times straight m squared

    • omega = angular velocity, in rad divided by straight s

Angular Momentum of a Particle

  • Objects traveling in a straight line can possess both linear and angular momentum

  • The magnitude of this angular momentum depends on the particle's position relative to an axis of rotation

    • When a particle moves on a line relative to the axis, it will cause rotation and therefore has angular momentum

    • When a particle moves on a line that passes through the axis, it will not cause rotation and therefore has zero angular momentum

  • Therefore, an axis can be selected to change an object's angular momentum

The angular momentum of a particle depends on its position relative to an axis of rotation. Three scenarios are shown: 1. If there is no specified axis, we can't say if the particle has angular momentum 2. The particle causes the rod to rotate, so it must have angular momentum relative to the axis 3. If the particle hits the axis, it will not cause any rotation
The angular momentum of a particle depends on its position relative to an axis of rotation, as this determines whether or not it could cause a rotation

  • The magnitude of an object's angular momentum depends on:

    • the distance between a chosen reference point and the object

    • the mass of the object

    • the linear speed of the object

    • the angle between the radial distance and the velocity of the object

  • The rotational inertia of a particle at a distance r from an axis of rotation is equal to:

I space equals space m r squared

  • The angular velocity of the particle is given by:

omega space equals space v over r

  • Therefore, the angular momentum of a particle about a given reference point is:

L space equals space I omega space equals space open parentheses m r squared close parentheses open parentheses v over r close parentheses space equals space m v r

  • Where:

    • m = mass of the object, in kg

    • v = linear speed of the object, in straight m divided by straight s

    • r = distance between the reference point and the object, in straight m

  • This equation only applies when the axis is perpendicular to the plane of the object's motion

Angular momentum of a particle moving in a circular path

A particle of mass m moving with velocity v in a circular path. The distance between the particle and the rotational axis is r. The velocity and distance arrows are at right angles to each other.
The angular momentum of a particle moving in a circular path is given by the product of its mass, linear velocity and distance from the rotational axis

  • When the axis is not perpendicular, the magnitude of the angular momentum is:

L space equals space m v r space sin space theta

  • Where:

    • theta = the angle between the distance r and the object's velocity v, in rad

Angular momentum of a particle moving at an angle

A particle of mass m moving with velocity v in a straight line. The distance between the particle and the rotational axis is r. The velocity and distance arrows are at an angle of θ to each other.
The angular momentum of a particle moving in a straight line is given by the product of its mass, linear velocity, distance from the rotational axis and angle the angle between the radial distance and the velocity of the particle

Worked Example

A horizontal rigid bar is pivoted at its center so that it is free to rotate. A particle of mass 3 M is attached at one end of the bar and a container is attached at the other end, both are at a distance R from the central pivot. With respect to the pivot, the rigid bar has rotational inertia I space equals space 1 over 6 M R squared.

A particle of mass M moves with velocity v perpendicular to the rod as shown in the diagram.

Diagram showing a rotating rigid bar with a pivot at the center and masses 3M and M at opposite ends. The mass 3M is on the left, and a container with mass M is on the right, each a distance R from the pivot.

The particle collides with the container and stays within it as the system starts to rotate about the vertical axis with angular velocity omega.

Which of the following expressions correctly represents the angular momentum of the system about the vertical axis just after the collision?

A      4 M R squared omega

B      4 M R v

C      25 over 6 M R squared omega

D      25 over 6 M R v

The correct answer is C

Answer:

Step 1: Analyze the scenario

  • Before the collision:

    • the rigid bar, container, and particle of mass 3 M are all stationary

    • the particle of mass M moves with linear velocity v

Diagram of a rigid bar with the midpoint fixed on an axis and the other end marked with mass M. Distance from the axis to M is labeled R, and velocity v is indicated as perpendicular to R.

  • After the collision, the whole system rotates with an angular velocity of omega

Diagram showing a rigid bar with masses 3M and M at each end, rotating around an axis in the center. Distance between each mass and the axis labeled R. Both masses rotate clockwise with angular velocity omega.

Step 2: Eliminate any incorrect answers

  • The total angular momentum of the system is provided by the particle of mass M and is equal to M R v

  • The question asks about the angular momentum after the collision, which is also equal to M R v (due to the conservation of angular momentum)

  • This isn't one of the options, so, the final answer must be in terms of omega

    • This eliminates options B and D

Step 3: Determine the total rotational inertia of the system

  • The rotational inertia of

    • the rigid bar is I space equals space 1 over 6 M R squared

    • the particle of mass M is I space equals space M R squared

    • the particle of mass 3 M is I space equals space 3 M R squared

  • The total rotational inertia of the system is the sum of all the rotational inertia components

I space equals space sum m r squared space equals space 1 over 6 M R squared space plus thin space M R squared space plus thin space 3 M R squared

I space equals space 25 over 6 M R squared

Step 4: Determine the angular momentum of the system

  • The angular momentum of the system is equal to:

L space equals space I omega

  • The angular momentum of the system after the collision is:

L space equals space 25 over 6 M R squared omega

  • Therefore, option C is correct

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    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

    Caroline Carroll

    Author: Caroline Carroll

    Expertise: Physics Subject Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.