I have a cone (vertex at $(x,y,z)=0$) with height $H$ and radius $a$ (radius of the base). Here is a picture:

enter image description here

I want to find the moment of inertia around the $z$-axis using calculus. I know the integral I have to evaluate is :$$I_z={\int \int \int}_V \vec{r}^2 dm$$

I decided to switch to cylindrical coordinates and my integrals becomes:

$$I_z={\int_0^H \int_0^{2\pi} \int_0^{\color{red}{?}}} r^3 dr d \theta dz$$

Since my radius is changing with $z$ I need to parameterize $r$ in terms of $z$. By similar triangles we have $$\frac{a}{H}=\frac{r(z)}{z} \iff r(z)=\frac{a z}{H}$$

When I tried to do this exercise without any help I substituted $r^3$ with $r(z)^3=\left( \frac{az}{H}\right)^3$:

$$\implies I_z={\int_0^H \int_0^{2\pi} \int_0^{\color{red}{a}}} \left( \frac{az}{H}\right)^3 dr d \theta dz$$

However, this seems to be wrong and I know the correct way to write the integral is this:

$$I_z={\int_0^H \int_0^{2\pi} \int_0^{\color{red}{\frac{az}{H}}}} r^3 dr d \theta dz$$

I don't understand what's wrong with my method and I don't really understand why the upper bound of the integral w.r.t $r$ is $\frac{az}{H}$? Why is it not just the radius $a$ of the base? What am I getting wrong here?


For a differential mass $dm$, the moment of inertia is $r^2dm$ or $\rho r^2dV$. In cylindrical coordinates $dV=r \ dr\ d\varphi \ dz$. So:

$$dI=\rho r^3\ dr\ d\varphi \ dz$$

$$I=\rho\int_0^{2\pi} d\varphi\int_0^H dz\int_0^{az \over H} r^3\ dr$$

Why do you have to put $\frac{az}{H}$ as the upper limit? If you put just $a$ you would actually calculate the moment inertia of a cylinder of radius $a$! Look at the range of values that $r$ can take when you move $dm$ inside the cone. At level $z=0$ the maximum distance of $dm$ from the vertical axis is 0. At level $z=H$ the value of $r$ varies between 0 and $a$. But to calculate the integral properly, you need limits for $r$ in an arbitrary position defined by an arbitrary value of $z$ (next variable in the order of integration). For arbitrary height $z$ the distance of $dm$ from the vertical axis varies from 0 to $az/H$ and that is the upper limit of the third integral..

  • $\begingroup$ Ah of course! That makes perfect sense! Thank you very much for your answer. $\endgroup$ – Nullspace Dec 9 '18 at 18:37

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.