Rotation of Point Masses and Rigid Bodies:
Suppose you whirl a ball about your head (assume that the ball is tied to an essentially weightless cord). If the ball is not very close to your head, we might be able to approximate it as a point mass. Once it's in motion, only the tension in the cord is necessary to keep it in circular motion (ignoring air friction).
Let's think about how you get it started. You can't pull directly inward on the cord as you can after it's in motion; you have to pull partially forward to give the ball an acceleration so it acquires a speed. This acceleration is the part I want to examine more closely.
The force you pull forward with is shown as F in the diagram; the corresponding acceleration that F produces is a tangential acceleration:
Using Newton's Second Law, and multiplying both sides by r, perhaps you can see that the force F has a torque about A:
I've reorganized the terms so that you can see that the sum of the forces version of Newton's Second Law implies a sum of the torques version, with the linear acceleration replaced by the angular acceleration and the mass replaced by an mr2 term.
You can perhaps get a better feeling by considering this:
The net torque produces a rotational acceleration; the mr2 product represents a "resistance" to this angular acceleration. It could have been called "angular inertia". Because torques are also called "moments", we call the term mr2 the "moment of inertia" I.
The sum of the torques is
For simplicity, I've taken all forces perpendicular to the radii. If these masses are not moving independently but are rigidly connected so that all must have the same angular acceleration, the moment of inertia is
This is the formula you'll use to find the moment of inertia of a system of point masses. This presentation, I hope, has given you an understanding of why people have defined moment of inertia as they have.
Comment: The angular acceleration and the torque are vectors that point in the same direction. If you curl the fingers of the right hand in the direction of the angular acceleration, the thumb provides the vector's direction. You might infer from this that the moment of inertia I is a scalar. Although this is well beyond the level of this course, I is a nine-component thingy called a tensor. Scalars are zero order tensors; vectors are first order tensors. The moment of inertia is a tensor of the second order. You don't need to know any of this!
Example 1: Moment of inertia of a system of point masses
Example 2: The acceleration of a rigidly connected system of point masses
A rigid body is an assembly of an infinite number of point masses (dm) - you guessed it, we're talking calculus!
Since r is different for each differential piece of mass dm in the body, in practice r and dm have to be related in order to do the integration. The moments of inertia given in the table on page 286 were obtained by integration.
Example 3: The acceleration of a rigid body
Example 4: A weird professor raises an elephant using a pulley
The Parallel Axis Theorem:
As we progressed from considering point masses to rigid bodies, we were talking about the rotation of the mass or system about a point - an axis of rotation. The examples of rigid body rotation you just studied utilized axes which passed through the centers of mass. But suppose the axis of rotation is about an axis which does not pass through the center of mass? For example, the rod in Table 10.2 has a moment of inertia about its center of mass of (1/12)ML2, but a value four times larger about its end! The integration formula above can be used to find the I about a new axis but this is a lot of work! Instead, we can use the center of mass I and the distance d between the center of mass axis and the new axis to find I about the new axis. Serway mentions the parallel axis theorem on page 290:
Example 5: Rod and Cylinder
Example 6: Using Experimental Moments of Inertia
Example 7: Moment of Inertia of a System of Rigid Bodies