Two rods of the same shape are held at their centers and rotated back and forth. The red one is much easier to rotate than the blue one. What is a possible explanation for this?

The rods actually have the same mass, but the mass is distributed differently. The red one has its mass concentrated toward the center, while the mass of the blue rod is mostly at the ends.

Newton's first law: an object at rest tends to remain at rest, and an object that is spinning tends to spin with a constant angular velocity, unless it is acted on by a nonzero net torque or there is a change in the way the object's mass is distributed.

The net torque is the sum of all the torques acting on an object.

The tendency of an object to maintain its state of motion is known as inertia. For straight-line motion mass is a good measure of inertia, but mass by itself is not enough to define rotational inertia.

How easy or hard it is to get something to spin, or to change an object's rate of spin, depends on the mass, the shape of the object and how its mass is distributed, and on the position of the axis of rotation. Rotational inertia, known as moment of inertia, accounts for all these factors.

The moment of inertia, I, is the rotational equivalent of mass.

For a simple object like a ball on a string being whirled in a circle, where all the mass can be considered to be the same distance away from the axis of rotation, the moment of inertia is:

For a point mass: I = mr²

For something more complicated, where mass is distributed at different distances from the rotation axis, the moment of inertia is determined by integrating:

I = ∫ r² dm

Example - a uniform rod of length L rotating about one end

How do we evaluate the moment of inertia integral:

I = ∫ r² dm

for a uniform rod of length L rotating about an axis passing through one end of the rod, perpendicular to the rod?

Align the rod with the x axis so it extends from 0 to L. Split the rod into little pieces of size dx. The mass of each little piece is:

dm = λ dx, where λ is the mass per unit length of the rod.

λ = M/L, where M is the rod's total mass.

The integral becomes:

I = ∫ x² λ dx, integrated from 0 to L

I = [(1/3) λ x³] with upper limit L and lower limit 0

I = (1/3) λ L³

Substituting M = λL gives:

I = (1/3) ML²

for a uniform rod rotating about one end.

We could carry out such integrals for all sorts of different shapes, although many of them are inetgrals over areas or volumes instead of over lengths. It's easier to look up the result in the table on page 304 in the book.

For a given rotation axis direction, the moment of inertia will always be minimized when the axis of rotation passes through the object's center-of-mass. The moment of inertia increases as the rotation axis is moved further from the center-of-mass.

Parallel-axis Theorem

For an object of mass M, the parallel-axis theorem states:

I = I_com + Mh²

where h is the distance from the center-of-mass to the current axis of rotation, and I_com is the moment of inertia for the object rotating about the axis through the center of mass that is parallel to the current axis.

Example

The parallel-axis theorem is usually used to calculate the moment of inertia about a second axis when I_com is known. Let's use it to go the other way, using the moment of inertia we just calculated for a rod rotating about one end.

What is the moment of inertia I_com for a uniform rod of length L and mass M rotating about an axis through the center, perpendicular to the rod?

I = I_com + Mh²

I_com = I - Mh²

We found that the moment of inertia when the rod rotates about a parallel axis passing through the end of the rod is:

I = (1/3) ML²

The distance from the end of the rod to the center is h = L/2. Therefore:

I_com = (1/3) ML² - M(L/2)²

I_com = (1/3) ML² - (1/4) ML²

I_com = (1/12) ML²