Moments of Inertia

For linear motion, Newton's second law gives the acceleration of a particle of mass m when subjected to a force. In symbols, F = m*a

For angular (rotating) motion, the angular acceleration felt by a particle depends not only on the applied force, but also on the distance between the axis of rotation and the place where the force is applied. To see this, consider trying to open a heavy door (like the door to a bank vault). If you push the door hard, but apply your force close to the hinge, then the door will barely budge. It is much easier to move the door if you push far away from the hinge. That is why door knobs are placed far away from the door hinges.

Group Discussion

1. Think of three examples from your own experience in which it was easier to rotate an object when you applied force far from the axis of rotation.
2. Does the direction of the applied force matter? In what direction should you push on a heavy door in order to move it most efficiently?

Torque induces an angular acceleration. For linear acceleration, the mass of a particle gives the tendency of the particle to resist acceleration. We say that the mass gives the particle linear inertia. For rotational acceleration, however, the rotational inertia of a particle depends not only on the mass of the particle, but also on its distance from the axis of rotation. From Newton's second law of motion, you can show that the relationship between the torque, T, and the angular acceleration, A, of a particle is given by T = (m r^2) A . The quantity m r^2 is called the moment of inertia for the particle and is typically denoted by I.

For a rigid configuration of particles, the moment of inertia is found by adding up all of the individual moments. For a continuous distribution of mass, we must chop our distribution into tiny elements of mass, and, for each element, add up the moment of inertia due to that mass. In general, this will result in double or triple integreals, but for symmetric objects, we can express the moments of inertia as ordinary one variable integrals. The following activities show how integrals can be used to compute the total mass and the moments of inertia for symmetric objects.

Suppose you have a single mass attached to a pivot point by a thin rod of length r and you apply a force, F, perpendicular to the rod. Then we define the torque to be the product T = r F.

Note that the torque is proportional to the applied force, and also proportional to the distance from the applied froce to the axis of rotation, so this definition matches our physical intuition.

Group Discussion

Index

More: Radially Symmetric Objects

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