Find the moment of inertia of a disk about a diameter by using thin rods, moment of inertia integral
00:00 In this challenging physical integration problem, we find the moment of inertia of a disk about a diameter by using thin rods. When we say the rotation axis is a diameter of the disk, we mean the disk is flopping over rather than spinning on the axis perpendicular to the disk and passing through the center. There are many ways to approach this moment of inertia integral, but our choice in this problem is to break the disk into thin rods rotating about their centers, and we're going to use the previous result that the moment of inertia of a thin rod rotating about its center is given by 1/12ML^2.
00:31 Diagram a thin rod increment: we draw a thin rod contributing the infinitesimal mass dm to the disk and located at a vertical coordinate of y from the horizontal reference line cutting the disk in half. We use trigonometry to express the length of the thin rod in terms of y.
01:24 Reminder on area density: we get a quick reminder of the definition of area density as sigma=mass/area. We can turn this around to express any mass as sigma*area.
02:00 Express dm in terms of y: we express the mass of the thin rod dm in terms of y by writing dm=sigma*dA and expressing the infinitesimal area dA as the length of the rod multiplied by its thickness, dy.
02:40 Express the moment of inertia contribution of the thin rod entirely in terms of y: now we use the previous result for moment of inertia of a thin rod to write the moment of inertia increment as dI=1/12*dm*L^2. We sub in dm in terms of y and the length of the rod in terms of y to obtain an expression for dI entirely in terms of y.
04:07 Final setup for the moment of inertia integral: now that we have an expression for dI in terms of y, we use an integral to sum the moment of inertia contributions. We take care to include the limits of integration, as y goes from -R to +R (the bottom of the disk to the top). This results in a difficult integral 2/3*sigma integral_-R^R (R^2-y^2)^(3/2)dy.
04:56 Make a trig substitution to transform the integral to theta space: to solve the integral, we let y=Rsin(theta) and dy=Rcos(theta)d(theta). The limits of integration transform to -pi/2 to pi/2. We begin to simplify the integrand using the pythagorean identity 1-sin^2(theta)=cos^2(theta). The integrand simplifies to cos^4(theta), and this is an even function, allowing us to do only half the integral, then double the result.
07:30 Integrate cos^4(theta): we apply the power reducing identity cos^2(theta)=1/2(1+cos(2*theta)) two times to reduce powers in the integrand. When we expand the result, we still have one problematic term cos^2(2*theta), and we apply the identity one last time. Now all the pieces of the integrand have guessable antiderivatives by using the chain rule backwards.
08:49 Final expression for the moment of inertia: we guess antiderivatives and sub in the limits of integration to obtain a simple expression for the moment of inertia in terms of the mass of the disk, M, the radius of the disk, R, and the area density, sigma. We still need to eliminate sigma by expressing it as the mass of the disk divided by the area, M/pi*R^2. After making this substitution, we arrive at the final expression for the moment of inertia of the disk rotating about a diameter: 1/4*MR^2, and we're done!
