The rotational equivalent of linear momentum. Angular momentum is often approximately conserved in collisions, and is usually conserved when external [torques] sum to zero. The action of purely internal forces can sometimes create a change in the system moment of inertia while conserving the angular momentum, which leads to interesting effects.
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Motivation for Concept
Imagine a satellite spinning in space. In the absence of external influences, the satellite will continue to spin. To stop the spinning would require the application of force. The faster the satellite is spinning, the more work the force will have to perform to halt the satellite. These concepts are completely analogous to the linear ideas of force and work. As such, there must be some quantity analogous to the linear momentum which is conserved in the absence of external influences. Angular momentum is that quantity.
Conservation of angular momentum is a more complicated topic than conservation of linear momentum, however. It is very rare for an everyday system to experience a significant change in mass. Thus, in physics problems involving conservation of linear momentum, the center of mass velocity of the system is usually constant. It is not difficult, however, for a system to alter its moment of inertia. In a case where angular momentum is conserved, such an alteration must produce an inverse alteration in the rotational speed. This is easily seen in sporting events like figure skating and diving. When an athlete is rotating essentially free of external forces (such as a figure skater in a spin on slippery ice or a diver rotating in mid-air) they can still affect a dramatic change in their rotational rate simply by tucking in their arms and legs or, conversely, by extending their arms and legs.
Conditions for One-Dimensional Angular Momentum
With the exception of certain specific systems like the gyroscope, angular momentum problems in introductory physics will be confined to one dimension. To ensure that a problem can be treated using 1-D angular momentum, the following conditions must hold.
Movement in a Plane
Rotational motion in introductory physics will be concerned only with systems where the center of mass of each constituent is confined to move in a plane (we will refer to this as the xy plane).
Rotation Perpendicular to the Plane
Each rigid body in the system that is rotating must only rotate such that their angular velocity and angular acceleration are directed perpendicular to the xy plane containing the centers of mass. (They must point in the +z or -z direction.)
Rotational Symmetry
Finally, each rigid body in the system must obey one of the following rotational symmetry constraints:
- The object has a rotational symmetry about the axis of rotation.
- The object has a reflection symmetry about the xy plane.
Each class of objects in the table of common moments of inertia and their parallel axis theorem tranformations possess the second symmetry (some also possess the first).
Valid Systems
Systems that can be treated as having one-dimensional angular momentum include:
- Ball rolling from rest down an inclined plane.
- A yo-yo.
- Non-spinning baseball moving horizontally strikes bat pivoted horizontally at the handle with ball center of mass at the same height as bat center of mass.
Invalid Systems
Systems that require more than 1-D angular momentum:
- Styrofoam cup (wider on the top than the bottom) rolling down an inclined plane (the cup will begin to turn as it rolls).
- Baseball moving horizontally with spin along a horizontal axis strikes bat pivoted horizontally at the handle with the ball center of mass at same height as bat center of mass (ball and bat angular momenta perpendicular).
- Baseball moving horizontally with no spin strikes bat pivoted horizontally at the handle with ball center of mass above or below the bat center of mass (ball center of mass and bat center of mass move in different planes).
Definition of Angular Momentum
Angular Momentum of a Point Particle
Beginning with Newton's 2nd Law for a point particle:
\begin
[ \sum \vec
= \frac{d(m\vec
)}
] \end
we take the cross product of the particle's position measured in the xy plane from the point of intersection with the chosen axis of rotation:
\begin
[ \sum \vec
\times\vec
= \vec
\times \frac{d(m\vec
)}
]\end
Assuming m is constant, we can pass it out of the derivative. We cannot assume that it is reasonable to pass r into the derivative, however, since it certainly has a time dependence (unless the velocity is zero). We can, however, prove that it is reasonable to pass r into the derivative. To see this, we evaluate:
\begin
[ \frac{d(\vec
\times\vec
)}
= \frac{d\vec
}
\times \vec
+ \vec
\times\frac{d\vec{v}}
= \vec
\times\vec
+ \vec
\times\frac{d\vec
}
]\end
Noting that the cross product of any vector with itself is zero, we then see that if the particle's mass is constant:
\begin
[ \sum \vec
\times\vec
= \frac{d(m\vec
\times\vec
)}
] \end
The left hand side is simply the net [torque] on the particle about the chosen axis. Thus, in the absence of net torque, the quantity:
\begin
[ m\vec
\times\vec
]\end
is constant in time. We therefore choose to define this quantity as the angular momentum of the point particle about the chosen axis.