# Thread: Find z-coordinate of center of mass

1. ## Find z-coordinate of center of mass

Hi

I wasn´t quite sure where to post this, but it involves integrals so I chosed calculus.

I have attached a picture.

Problem:
Find the z-coordinate of center of mass of the object.

Not really sure how to start.

But wont a differential slice of the cone be something like $\pi \cdot r_{0}^{2} \cdot dz$ ?

But $r_{0}^{2}=x^{2}+y^{2}$ , for each different slice. But the radius $r_{0}$ is obviously changing with z.

Should I try to express x and y as functions of z ?

Thx

2. Originally Posted by Twig
Hi

I wasn´t quite sure where to post this, but it involves integrals so I chosed calculus.

I have attached a picture.

Problem:
Find the z-coordinate of center of mass of the object.

Not really sure how to start.

But wont a differential slice of the cone be something like $\pi \cdot r_{0}^{2} \cdot dz$ ?

But $r_{0}^{2}=x^{2}+y^{2}$ , for each different slice. But the radius $r_{0}$ is obviously changing with z.

Should I try to express x and y as functions of z ?

Thx
Let $R$ be the radius of circular base of the cone.

and $r$ be the radius of "slice" at distance $z$ from the vertex.

then, $\frac{R}{h}=\frac{r}{z}$

There is no need of x and y in the equation, only z is required(since by symmetry, centroid lies on z-axis)

3. Hi

Wow, nice thats what I did =)

So got $r=\frac{zR}{h}$

Thus..

$\int_{0}^{h} z\cdot \pi \cdot z^{2}\frac{R^{2}}{h^{2}} \; dz = \frac{\pi R^{2}h^{2}}{4}$

4. Originally Posted by Twig
Hi

Wow, nice thats what I did =)

So got $r=\frac{zR}{h}$

Thus..

$\int_{0}^{h} z\cdot \pi \cdot z^{2}\frac{R^{2}}{h^{2}} \; dz = \frac{\pi R^{2}h^{2}}{4}$
z co-ordinate:
$\frac{ \int_{0}^{h} z\cdot \pi \cdot z^{2}\frac{R^{2}}{h^{2}} \; dz}{\int_{0}^{h} \pi \cdot z^{2}\frac{R^{2}}{h^{2}} \; dz}$

5. Originally Posted by malaygoel
z co-ordinate:
$\frac{ \int_{0}^{h} z\cdot \pi \cdot z^{2}\frac{R^{2}}{h^{2}} \; dz}{\int_{0}^{h} \pi \cdot z^{2}\frac{R^{2}}{h^{2}} \; dz}$

Yes, alright thanks. Here you divide by the entire volume of the cone, right, since it evaluates to $\frac{\pi R^{2}h}{3}$ .

The 'formula' is $Vz_{T}=\int z_{c}\; dV$ ?

Where I let the subscripit T denote the center of mass and subscript C denote moment arm.

6. Originally Posted by Twig
Yes, alright thanks. Here you divide by the entire volume of the cone, right, since it evaluates to $\frac{\pi R^{2}h}{3}$ .

The 'formula' is $Vz_{T}=\int z_{c}\; dV$ ?

Where I let the subscripit T denote the center of mass and subscript C denote moment arm.

As in this question, use symmetry wherever possible

7. Hi!

Yes, I got $\frac{3h}{4}$

I think I understand decently.
Yes, the book also says USE SYMMETRY! =)
makes life easier

The hardest part is definetely setting up the integral, or the geometry of the figure.

thanks a lot.