The Angular Impulse-Angular Momentum Relation
We've already used the impulse-momentum relation to analyze situations involving translations through three-dimensional space. The relation is typically applied in its component form:
Arguing by analogy, if the change in momentum in the x-, y-, and z-directions is equal to the impulse applied to the object in each direction doesn't it seem plausible that a similar relation would hold in the " -direction"? If so, this relation should be constructed between the torques acting on an object, the time interval over which torques act, and the change in angular velocity of an object.1 The relation should be:
The product of rotational inertia and angular velocity is termed the angular momentum of the object, typically denoted L, and the product of torque and the time interval over which it acts is termed the angular impulse applied to the object.
Thus, if no angular impulse is applied to an object, its angular momentum will remain constant. This special case is referred to as angular momentum conservation. However, if an angular impulse is applied to the object, the angular momentum will change by an amount exactly equal to the angular impulse applied. Angular momentum is changed through angular impulse.
Incorporating Rotation in the Work-Energy Relation
Our previous encounter with the work-energy relation resulted in:
Recall from our previous discussion of work-energy that this is not a vector equation, meaning it is not applied independently in each of the coordinate directions. Generalizing this equation to include rotation will involving adding terms to this equation, not creating a seperate "angular" energy equation.
Recall that we model the motion of an arbitrary rigid body as a superposition of a pure translation of the CM and a pure rotation about the CM. Let's investigate the effect of this model on our calculation of the kinetic energy of the object.
The translaton portion of the motion is easy. We envision the object as a point particle, localized at the CM of the real object, traveling with the velocity of the CM of the object. Thus, this portion of the motion contributes a kinetic energy,
What about the kinetic energy due to the pure rotation of the object about the axis through the CM?
Every small chuck of the object, dm, moves in a circle around the CM. Each of these pieces of mass has a velocity magnitude given by
Thus, the kinetic energy of each piece is
Therefore, the total kinetic energy due to rotation of all the little pieces is:
Combining the kinetic energy due to rotation with the kinetic energy due to translation leads to a total kinetic energy of:
and a work-energy relation of:
Unless a scenario involves springs or other elastic material, I'll typically write this relationship as:
1 Please remember that this model, and hence the relationships derived under it, are restricted to rigid bodies and motions in which the rotation axis is perpendicular to the plane in which the center-of-mass moves. Since objects must be rigid, their rotational inertia must remain constant. In addition, motions must either be about a stationary rotation axis or, if not, the rotation axis is taken to be through the CM.