The distance between the \(y\) and \(y'\) axis is available from the diagram. The centroidal moment of inertia is calculated similarly using (10.2.10). The area of the ring is found by subtracting the area of the inner circle from the area of the outer circle. The best way to learn how to do this is by example. What can I say about the perpendicular axis theorem other than it's interesting. What if an object isn't being rotated about the axis used to calculate the moment of inertia? Apply the parallel axis theorem. We can also use the moment of inertia for a hollow sphere ( 2 3 m a 2 ) to calcul ate the moment of inertia of a nonuniform solid sphere in which the density varies as ( r). Where α is a simple rational number like 1 for a hoop, ½ for a cylinder, or ⅖ for a sphere. When you are done with all of this, you oftentimes end up with a nice little formula that looks something like this… These methods can be used to find the moment of inertia of things like spheres, hollow spheres, thin spherical shells and other more exotic shapes like cones, buckets, and eggs - basically, anything that might roll and that has a fairly simple mathematical description. Or this for stacked disks and washers I = Something like for nested, cylindrical shells… I = In physics and applied mathematics, the mass moment of inertia, usually denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass.Mass moments of inertia have units of dimension ML 2 (mass × length 2). When shapes get more complicated, but are still somewhat simple geometrically, break them up into pieces that resemble shapes that have already been worked on and add up these known moments of inertia to get the total.įor slightly more complicated round shapes, you may have to revert to an integral that I'm not sure how to write. This method can be applied to disks, pipes, tubes, cylinders, pencils, paper rolls and maybe even tree branches, vases, and actual leeks (if they have a simple mathematical description). The volume of each infinitesimal layer is then…įor many cylindrical objects, you basically start with something like this… I = Imagine a leek.Įach layer of the leek has a circumference 2π r, thickness dr, and height h. The other easy volume element to work with is the infinitesimal tube. Note that although the strict mathematical description requires a triple integral, for many simple shapes the actual number of integrals worked out through brute force analysis may be less. This is the way to find the moment of inertia for cubes, boxes, plates, tiles, rods and other rectangular stuff. Even it is close to solution, what mass to put in. Homework Equations The Attempt at a Solution I guess that subtracting the moment of inertia of the inner cube from the moment of inertia of the outer cube is wrong. When an object is essentially rectangular, you get a set up something like this… I = Homework Statement How to calculate moment of inertia of hollow cube. The resulting infinite sum is called an integral. The volume of each infinitesimal piece is… The moment of inertia of a point mass is given by I mr 2, but the rod would have to be considered to be an infinite number of point masses, and each must be multiplied by the square of its distance from the axis. The pieces are dx wide, dy high, and dz deep. The infinitesimal box is probably the easiest conceptually. In practice, this may take one of two forms (but it is not limited to these two forms). The infinitesimal quantity dV is a teeny tiny piece of the whole body. The hollow spheres mass is M, and the radius. In practice, for objects with uniform density ( ρ = m/ V) you do something like this… I =įor objects with nonuniform density, replace density with a density function, ρ( r). A moment of inertia of a hollow sphere will be the same as any axis passing through its center. You add up (integrate) all the moments of inertia contributed by the teeny, tiny masses ( dm) located at whatever distance ( r) from the axis they happen to lie. It works like mass in this respect as long as you're adding moments that are measured about the same axis.įor an extended body, replace the summation with an integral and the mass with an infinitesimal mass. For mass M kg and radius R cm the moment of inertia of a. Say it, kilogram meter squared and don't say it some other way by accident.įor a collection of objects, just add the moments. The moments of inertia of a mass have units of dimension ML 2 ( mass × length 2 ). The moment of inertia of a sphere about its central axis and a thin spherical shell are shown. It's a scalar quantity (like its translational cousin, mass), but has unusual looking units. Logic behind the moment of inertia: Why do we need this?
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |