Because the centroid of a shape is the geometric center of an area or volume, the average distance to any one point in a body is at a minimum. We then measured all distances from that point or axis, where the distances were the moment arms in our moment integrals. When we calculated the area and mass moments of inertia via integration, one of the first things we had to do was to select a point or axis we were going to take the moment of inertia about. Once the moments of inertia are adjusted with the Parallel Axis Theorem, then we can add them together using the method of composite parts.
We will use something called the Parallel Axis Theorem to adjust the moments of inertia so that they are all taken about some standard axis or point. Because each part has its own individual centroid coordinate, we cannot simply add these numbers. The moments of inertia in the table are generally listed relative to that shape's centroid though. Moments of inertia for the parts of the body can only be added when they are taken about the same axis. As discussed on the previous pages, the area and mass moments of inertia are dependent upon the chosen axis of rotation. This method is known as the method of composite parts.Ī key part to this process that was not present in centroid calculations is the adjustment for position. In this method we will break down a complex shape into simple parts, look up the moments of inertia for these parts in a table, adjust the moments of inertia for position, and finally add the adjusted values together to find the overall moment of inertia. You can see again that this quantity is proportional to r 2-like terms.Method of Composite Parts for Moments of Inertia and the Parallel Axis TheoremĪs an alternative to integration, both area and mass moments of inertia can be calculated via the method of composite parts, similar to what we did with centroids. Where I is a 3x3 matrix defined by the integral When dealing with extended rigid bodies rather than particles, all you have to do is "add up" (integrate over - sorry, calculus!) all of its constituent particles. Since angular velocity and angular momentum are obviously salient quantities for rotational problems, it follows that moment of inertia must be too. This equation L = mr 2? is a sort of rotational analogue of p = m v, except velocity v is replaced by angular velocity ?, and mass m is replaced by moment of inertia I = mr 2. If you choose sensible axes, so that r and ? are perpendicular, then you have that L = mr 2?, so the moment of inertia takes the simple form I = mr 2. You can see the r 2-ness of this quantity already. The 3x3 matrix m(r 2? - rr T) is the moment of inertia, and usually called I. Where ? is the 3x3 identity matrix and r T is the matrix transpose of the vector r, so that rr T is a 3x3 matrix. whose motion is purely due to a rotation about some axis), v = ? × r, where ? is angular velocity, represented as a pseudo-vector. Where r is the particle's position and p is its momentum.
(It's salient for the same sort of reasons that momentum is salient: it's conserved for force-free systems, and it's temporal derivative yields force-like quantities - in the rotational case, torque.) For a single particle, this quantity is given by You can make similar arguments about the angular momentum L = r x p = I ωĪ salient physical quantity for rotating systems is the angular momentum. There are more advanced concepts like the moment of inertia tensor which gives you the moment of inertia along an arbitrary axis of rotation, but this is the basic concept. I = Σ mi ri² (in the discrete case where the body consists of a discrete number of masses mi at radii ri) or ∫ r² dm in the continuous case. Its rotational kinetic energy will then be 1/2 I ω² again. Now if you have an extended body rotating at angular velocity ω (and you don't care how fast the point masses it consists of are moving individually - that varies depending how far they are from the rotational axis), you can calculate its total moment of inertia I by summing all the contributions mr² of all points masses it consists of. Now you define the moment of inertia of the point particle to be I = mr², so T = 1/2 I ω². You can express the velocity as v = rω and the kinetic energy can then be written as T = 1/2 m r² ω². The angular velocity is ω = v/r where r is the radius of that circle. What's the kinetic energy of a point particle going around (uniformly) in a circle? T = 1/2 m v² where v is the tangential velocity.