Derivation: moment of inertia of a thin rod rotating about its center, moment of inertia integral.
00:00 In this moment of inertia integral, we compute the moment of inertia of a thin rod rotating about its center. We are asked to break the thin rod into point mass contributions and apply the formula for the moment of inertia of a point mass in order to set up the moment of inertia integral.
00:30 Start by drawing the mass increment dm, located at a distance of x from the rotation axis, which is where we locate the origin. We note that the lowest value of x is -L/2 and the highest value of x is +L/2, and those will be the limits of integration once our moment of inertia integral is set up.
01:16 Reminder of linear density: it's handy to use linear density lambda=mass/length in order to express dm in terms of dx, the width of the infinitesimal increment, dx. So dm=lambda*dx.
02:25 Moment of inertia contribution of the infinitesimal slice: now we apply the moment of inertia formula for a point mass to write the contribution dI as dm*x^2 = lambda*x^2*dx.
03:03 Use integration to sum the moment of inertia contributions: now that dI is entirely written in terms of the variable x, we can sum the contributions to moment of inertia by using an integral. We write I=integral(dI) and replace dI in terms of x. We integrate from -L/2 to L/2. We take advantage of symmetry here: this is an integral of an even function on an interval symmetric about the origin, so we can do half the integral and double the result. We compute the integral using the power rule and we have an expression for the moment of inertia of the thin rod in terms of the linear density, lambda.
04:47 Substitute for linear density in terms of mass and length: we replace lambda with M/L and simplify the result, completing the derivation. Moment of inertia of a thin rod rotating about its center is 1/12ML^2!
