display | more...

A measurement of how resistant to angular (rotational) acceleration an object is. The moment of intertia depends on the mass, position of axis of rotation, and shape of the object. In general, however, the moment of intertia is dependant on the concentration of mass a distance from the axis of rotation. For example, if you have two discs, equal radius, depth, and mass, but one is hollow (like a ring), the ring will have a greater moment of intertia. Like mass in trasnlational kinematics, objects with a greater moment of intertia require more torque to have the same angular acceleration. However, if two objects are rotating at the same angular velocity, then the object with a greater moment of intertia will have more kenetic energy.

The moment of inertia is to rotational motion what mass is to linear motion, but rotational motion adds a bit more complexity to the situation. The mass of an object provides it with inertia when it moves through space. You can express that either through saying that a more massive object has more momentum or that it takes more force to change the motion of a more massive object. In linear motion these ideas are expressed in the equations

p = m v and F = m a

Analogously, the moment of inertia relates the angular velocity of a rigid object to its angular momentum, and it tells you how much torque is required to change the rotational motion.

Let's start out by talking about an object that is rotationally symmetric about one axis1. Specifically let's imagine a ring of mass M with radius R rotating about its axis of symmetry. In that case the moment of inertia can be written down simply as a number

I = M R2

This value is said to be the moment of inertia "about the axis" of symmetry. We can then write down equations for the angular momentum, L, the angular acceleration, α, and the torque, τ along that axis.

L = I ω and τ = I α

These equations look precisely analogous to the equations I wrote above for linear motion. The second equation is analogous to Newton's second law, and it shows us that the larger the moment of inertia is the more difficult it is to speed up or slow down the rotation. Intuitively, you'd probably guess that the more massive an object is, the larger it's moment of inertia will be, and that's true. It's also true that the further a mass is from the axis of rotation, the harder it is to get it rotating. This is because something far from the axis of rotation must make a very large circle during each rotation. When the angular velocity of rotation increases a little bit, it means that that mass now must be going around a lot faster in order to make it around the circle in one period; thus, small angular acceleration translates into large tangential acceleration, which requires a lot of force. These dependences are reflected in the equation for the ring I = M R2.

Now suppose we have some other object that is constrained to rotate around a certain axis. Imagine, for example, that we took some object and stuck an axle through it, so that now the object must rotate only around the axle. If we imagine this object is made up of a set of point masses, then each point mass of mass mq located a distance r⊥q from the axis of rotation, as measured perpendicular to the axis, contributes a moment of inertia about that axis mq r⊥q2 (where q is just an integer numbering each mass). We can get the total moment of inertia with the formula

I = Σq mq r⊥q2


I = ∫ r2 dm = ∫ ρ(r) r2 dV

for a continuous system, where ρ(r) is the mass density and dV is an element of volume. Using these formulae, here are the moments of inertia for some uniform density objects of mass M with common shapes:

  • Ring of radius R rotating about its axis of symmetry: I = M R2
  • Cylinder (or disk) of radius R rotating about its axis of symmetry: I = 1/2 M R2
  • Ring of radius R rotating about an axis perpendicular to the axis of symmetry: I = 1/2 M R2
  • Thin rod of length L rotating about one end around an axis perpendicular to the rod: I = 1/3 M R2
  • Thin rod of length L rotating about its center around an axis perpendicular to the rod: I = 1/12 M R2
  • Sphere rotating about a diameter: I = 2/5 M R2

For objects that are not rotating about an axis of symmetry and are not constrained to rotate around a certain axis, the situation becomes a lot more complex. In general, the moment of inertia is not just a number, it's a symmetric rank 2 tensor, so you can write it down as a 3 by 3 matrix. This has many implications, including that the angular momentum vector is not always parallel to the angular velocity vector, and the relationship between angular acceleration and torque is no longer so simple. Some of everyday situations where these effects come into play are in the precession of a top and the wobble of a football2 thrown with a spiral pass. In the general case where angular momentum and angular velocity are not aligned, the moment of inertia changes with time (when measured in terms of the axes of a stationary observer, sometimes called the "space fixed" axes), and the relationship of torque and angular acceleration becomes

τ = dL/dt = I α + dI/dt ω

One way of dealing with this is to change to the "body axes", meaning to solve the problem as though you're sitting on the spinning body. In that case the moment of inertia doesn't change in time, but now you're working in a non-inertial reference frame where there are pseudoforces, so you still don't have the simple equation relating torque and angular acceleration. Still, this can be a very useful technique.

We can derive the full form for the moment of inertia tensor from the definition of angular momentum for a point particle.

Lq = rq × pq

Adding this up over all particles (all q values) gives

L = Σq rq × pq


pq = mq ω × rq


L = Σq mq rq × (ω × rq) = Σq mq (ω (rq)2 - rq (ωrq))

using the identity for the vector triple product. If we write this component by component then with a little work we can obtain

I jk = Σq mq(δjk rq2 - rqj rqk)

Remember that q numbers all the masses, while j and k are just numbers 1 through 3 for the 3 dimensions. δjk is just the Kronecker delta, which is 1 when j = k and 0 when j ≠ k.

Because I is symmetric, it can be diagonalized by an orthogonal transformation, meaning that if we just rotate our coordinate axes we can find ones where I is diagonal. The three axes of this special coordinate system are called the principle axes of the body. As long as the body only rotates around one of those axes, you can just use the moment of inertia value corresponding to that element on the diagonal, known as the principle moment for that axis. In cases where two or three of the eigenvalues are degenerate, there may be more than one set of axes in which I is diagonal, as is the case for a sphere or a disk.

Considering the tensor nature of the moment of inertia can lead to interesting new effects like precession and nutation. These are important in understanding the behavior of a top, a gyroscope, the rotation of the earth, and the orbits of satellites. This is related to the concept of a moment, which is also used in statics. An important result about the moment of inertia that you use a lot in practice is the parallel axis theorem, and Euler angles are also very useful.

  1. I'm assuming that all these objects have uniform mass density.
  2. I'm talking about an American football here, not a soccer ball.


Log in or register to write something here or to contact authors.