Moment of inertia using calculus

In physics books,the moment of inertia of a uniform rod with negligible thickness is determined using calculus.They take an element $$dx$$ $$x$$ distance from the center of the rod and calculuate moment of inertia as $$dI=\frac{M}{l}x^2dx$$,where $$M$$ is the mass of whole rod and $$l$$ is the total length.Then they integrate to find the total moment of inertia.My question is how do we express the integral as the limit of riemann sum or an approximation using $$\Delta x$$ instead of $$dx$$?

• Do you know the definition of a Riemann sum? Feb 14 '21 at 4:42

You know that the moment of inertia of a point mass $$m$$ at distance $$x$$ from the axis of rotation is $$mx^2$$.

The rod is rotated about the axis perpendicular to the rod passing through its center. So, we can consider only the momentum of inertia of half of the rod and take the double at the end.

This half rod can now be approximated by $$n$$ equally distributed point masses with mass $$\frac{M}{l}\Delta x^{(n)}$$ at distance $$x_i=i\cdot \Delta x^{(n)}$$ where $$\Delta x^{(n)} = \frac l2\cdot \frac 1n$$ and with a total momentum of inertia of

$$\frac Ml\sum_{i=1}^nx_i^2\Delta x^{(n)}\stackrel{n\to\infty}{\longrightarrow}\frac Ml\int_0^{\frac l2}x^2\,dx$$

So, for the whole rod you then get

$$I = 2\frac Ml\int_0^{\frac l2}x^2\,dx = \frac{Ml^2}{12}$$

• Please pardon me,but what do we mean by n equally distributed point masses?Also if the rod is treated as a $1d$ object,why do we take $dx$?A point is supposed to be of 0 length,right?Why do we then take $dx$ element instead of 0 though it is just a point?Please explain it to me. Feb 14 '21 at 7:49
• What about drawing a line segment and dividing it into $n$ pieces of equal length? Now, you consider each of these small pieces as a point mass concentrated at the end of each small piece as an approximation. Calculus tell you then if you send $n$ to infinity, the sum I wrote in my answer will converge to an integral. If you haven't heard about point masses, you should first study this and then go back to solids which are usually approximated by an increasing number of point masses which then lead to plenty of integrals in the mechanics of solids. Feb 14 '21 at 8:03
• Do you mean we take some finite line segments and when $n$ approaches infinity then the lines will look like points?If so what is the benefit of using lines instead of points?If i am wrong kindly explain and give me a resourcr for learning about point mass Feb 14 '21 at 9:16
• When I google "point mass" I get plenty of hits. So, you can do the same. Concerning your questions I will give you a question back: Each little part of the rod adds to the moment of intertia. So, how would you model "each little part"? Feb 14 '21 at 9:31
• That is exactly i don't understand,could you kindle shed some light on it? Feb 14 '21 at 9:44

Integrating your given expression, we get $$I=\int_a^b\frac{M}{l}x^2dx$$. We note that $$M$$ and $$l$$ are constants. Now, we examine the Riemann sum definition of the integral which states that $$\int_a^b f(x)dx=\lim \limits_{n \to \infty} \sum_{i=1}^n f(x_i)*\frac{b-a}{n}$$. We have that $$f(x) = \frac{M}{l}x^2$$, so our Reimann sum will be $$\lim \limits_{n \to \infty} \sum_{i=1}^n \frac{M}{l}{x_i}^2*\frac{b-a}{n}$$.