## Chapter 2 – Excerpt 5

A mathematical excerpt, using integration to calculate moments of inertia of inanimate shapes, such as a cat flap door.

Moments of Inertia of Extended Regular Objects Extended objects can be considered to be composed of infinitely many point masses. Perhaps more easily we can imagine a very large number of very small masses. Let’s call each tiny chunk of mass Δm. To calculate the total rotational inertia of an object, we need to add up the contributions from all the pieces of mass:
To be accurate, we actually need to take limits, letting the tiny pieces of mass become infinitesimal pieces approaching size zero; this requires the calculus technique of integration – I = ∫r For example, for a hoop of mass M with a thickness that is very small compared to its radius R (see Figure 2-14), we can approximate the distance of each piece of mass from a perpendicular axis through its center by R. Hence, each piece of mass Δm contributes an amount equal to ΔmR For an object of uniform density ρ (the Greek letter rho is typically used for density), we have ρ = M/V and Δm = ρ ΔV for each piece of volume ΔV. In two-dimensional cases, these would be ρ = M/A and Δm = ρ ΔA, and in 1-dimension they’re ρ = M/L and Δm = ρ ΔL. Using infinitesimals, these become dm = ρ dV, ρ dA or ρ dL, and we have I = ∫∫∫r As a more interesting example, which Note that rotational inertia is not an In both cases we will put the origin where the axis of rotation meets the flag’s lower edge, with the x-axis along the lower edge of the flag, and the y-axis through the axis of rotation. So r = |x|. Then we have
This result makes intuitive sense. We would expect the rotational inertia for the former to be four times as large as the latter because, on the average, the mass of the Wildcats flag is twice as far away from its axis of rotation as the mass of the Halloween flag is from its axis. For round objects, these calculations are easier to do with polar coordinates. As an example, let’s calculate the moment of inertia of a cylindrical can of catfood about an axis through the center of its circular top and bottom. See Figure 2-17. We’ll call the height of the can H, its radius R, and its mass M. Then with the origin at the center of the bottom of the can, we have Note that this result is independent of the height of the can. For a sphere, around an axis through its center, it turns out that the rotational inertia formula is I = 2/5 MR Imagine you are unwinding a spherical ball of yarn of radius R by pulling on it with a constant force F along the floor, as shown in Figure 2-18:. Its rotational speed will increase with a rotational acceleration of . As it gets unwound, R decreases, of course. But the torque τ decreases All of the above moment of inertia formulas, along with a few other common ones, are tabulated in Figure 2-19 for reference. Note that the more symmetric cases tend to have a lower rotational inertia.
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