1
$\begingroup$

I am currently following Taylor's "Classical Mechanics" and I am trying to understand creating the correct integrals to solve some problems related to the inertia of various shapes. I am to find the moment of inertia of a uniform disc of mass M and radius R about its axis, replacing the equation:

$$ I = \sum m_{\alpha}\ r_{\alpha}^2$$

with the appropriate integral and doing the integral in polar coordinates.

My answer was as follows:

$$ I = \sum m_{\alpha}\ r_{\alpha}^2\\ \tag{1}\label{1} \Rightarrow \int_0^M \ dm \int^R_0 r dr $$ Uniform disk so:
$$ dm = \rho dV \\ $$ where $$ \int dV = \int \int\int dxdydz\\ = \int^R_0 r dr \int^{2\pi}_0 d\phi \int^D_0dz\\ \Rightarrow dm = \rho 2\pi D\frac{R^2}{2} $$ I then formulate the equation in terms of the integral of the volume elements instead of mass elements: $$ I = \int_0^M \ dm \int^R_0 r dr\\ = \rho \pi DR^2\int^R_0 r dr\\ = \rho \pi D\frac{R^4}{2} $$ Finally:
$$ \rho = \frac{M}{V_{disc}}\\ = \frac{M}{\pi R^2D}\\ \Rightarrow I = \frac{1}{2} MR^2 $$

I read the ideal answer for this and had some questions comparing my answer with it. In \eqref{1}, the ideal answer is: $$ I = \sum m_{\alpha}\ \rho_{\alpha}^2\\ \Rightarrow \int_0^M r^2 dm $$ Q.1) Why is it only the mass which is changed to a mass element and the distance variable kept intact in the ideal answer? I thought it would have to become an integral as I'm summing over it?
Q.2) In my answer, the r variable is evaluated in two separate integrals then multiplied together, whereas in the correct answer it would be one integral with $r^3$ as the variable, so does this give the same answer in general?
Q.3) Lastly, I understand going from $m_{\alpha}$ and $r_{\alpha}$ to dm and dr as it's like summing infinitely many infinitely small masses and radii but the intuition behind leaving an r in the integral in \eqref{1} kinda confused me, like why does only one $r_{\alpha}$ become dr?

I know these are very basic questions but I would really like to have a strong understanding of using integrals in problems like this as it's a weak point of mine, and the intuition will be useful later. Thanks for any help in advance!

$\endgroup$
0
2
$\begingroup$

The key when changing discrete sums to integrals is to ask oneself what things in the integral need to be small and which ones can be large. In the case of the moment of inertia around an axis, the formula

$$I=\sum_{i=1}^Nm_i r_i^2$$

is valid for the case of $N$ point particles, with $m_i$ the mass of the ith particle and $r_i$ its distance to the axis. If one wants now to consider a continuous body like a disk, the strategy is to think of the disk as being made of infinitely many small pieces, and integrating over the volume of the disk.

Now imagine one small piece of the disk. Since it's infinitesimally small it must have infinitesimally little mass, so the change $m_i\to dm$ is appropriate. But now, that small chunk of mass could be anywhere in the disk; it could be very close to the axis or as far as $r=R$ from the axis. And in fact, as we integrate over the whole volume of the disk, we will be considering different pieces in the body, every one of them with a small mass, but each one of them at different distances from the axis, so the change $r\to dr$ is not appropriate. Write then the integrand as $r^2dm$ and substitute $dm=\rho\ dV=\rho\ rdr\,d\varphi\, dz$. You'll end up with just one integral in cylindrical polar coordinates

$$I=\int \rho r^3 dr d\varphi dz.\tag{1}$$

As to why your method works for this particular case, it's just a coincidence. Note that if one were to write the moment of inertia as $$\int_0^M \ dm \int^R_0 r dr,\tag{*}$$ then it would always be equal to $MR^2/2$, regardless of the mass distribution $\rho(\mathbf r)$ in the disk. As an exercise, you can calculate the moment of inertia of a disk with a mass distribution $\rho(\mathbf r)=\rho_0 r$ that grows linearly with the distance to the axis, using $(1)$ and $(*)$, and check that they are different. Choose $\rho_0$ so that the total mass of the disk is $M$.

$\endgroup$
2
  • $\begingroup$ So, because the values of mass in each small section are always infinitesimally small, but the distance from the axis of these "masses" can be large, we don't take r→dr? Also, thank you very much for the explanation, it was well explained! $\endgroup$ – Andrew Jan 18 at 23:14
  • $\begingroup$ Yes, that's right. I'm glad it helped. $\endgroup$ – Urb Jan 18 at 23:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.