Fizyx for Felines: A Physics Textbook for the Curious Cat

Chapter 2 – Excerpt 5

Posted in Uncategorized by skonabrittain on 16 April, 2010

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:

(5) I = Σ Δmr2

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 = ∫r2dm – because the distance r may not be constant for a whole piece Δm, no matter how small it is. However, for simple objects with simplifying approximations, we can perform the addition without using calculus.

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 ΔmR2 to the hoop’s rotational inertia. Adding up all the pieces of rotational inertia just involves adding up all the Δm’s, which have a sum equal to the total mass of the hoop, M (i.e. ΣΔm = M). This gives a total moment of inertia of MR2 for the hoop.


Figure 2-14. A thin hoop of radius R and mass M
has a moment of inertia about its axis of symmetry equal to MR2.


 

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 = ∫∫∫r2ρdV or ∫∫r2ρdA or ∫r2ρdL.

As a more interesting example, which does require calculus, let’s calculate the moment of inertia of a cat flap door. Say that the flap is constructed of a rectangular piece of material, such as rubber, of uniform density ρ, and has a total mass of M, a height of H, and a width of W. Then ρ = M/A = M/(WH). Putting the origin at the upper left corner of the flap (see Figure 2-15), the axis of rotation is the positive x-axis, so the distance of a piece of mass dm from the axis is (-y). So we have dm = ρ dA = (M/WH) dA = (M/WH) dxdy and hence


Figure 2-15. For a uniform rectangular sheet about its top, I = 1/3 MH2.


Note that rotational inertia is not an intrinsic property of an object. It is only defined with respect to an axis of rotation. For example, let’s compare the moment of inertia of a stiff rectangular flag around its flagpole (see Figure 2-16(a)) to that of one around an axis through its center in its own plane (see Fig 2-16(b)). To illustrate these different axes, we have arbitrarily chosen a team flag and a holiday flag, respectively.



Figure 2-16. (a) The flag of the Northwestern University athletic teams (all known as the Wildcats)
(b) A version of this Halloween flag with a more aesthetic aspect ratio is available from flags.com.


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

vs.

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.


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 MR2.
Q. What is the moment of inertia of the earth about its axis of rotation?
A. Since the quantity is being squared, we need to use a more precise value for the radius than we did in the previous chapter. By the above formula, we have I = 2/5 MR2≈ (2/5)(6.0 × 1024kg)(6.37×106m)2≈ 9.7 ×1037 kg m2.

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 linearly with R, whereas the moment of inertia I decreases as the square of R. So the rotational acceleration increases inversely proportional to R. (We must admit that this example is oversimplified: as the yarn unwinds, the point of application of your pulling force would vary wildly rather than remain at the bottom of the ball.)

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.


Figure 2-19. Some Common Moment of Inertia Formulas
Thin Hoop
around center
Thin Spherical Shell
around center
Cylinder
around center
Sphere
around center
Rectangle around side Rectangle
around center
           
I = MR2 I = (2/3)MR2 I = (1/2)MR2 I = (2/5)MR2 I = (1/3)MR2 I = (1/12)MR2


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