Mass of a Cone with varying Density

Homework Statement

Let a cone with height $h$ and base area $A$ have the density $\rho (x) = \rho_{0} \frac{3x^{2} + 2xh}{h^{2}}, 0 \leq x \leq h$

the relation between cone radius $r$ and distance from cone apex $x$ is given by:

$r = (\frac{B}{\pi h^{2}})^{\frac{1}{2}}x$

Find the total mass $M$ of the cone.

The Attempt at a Solution

Ok, so i've seen problems where the density varies, but this is the first one i've looked at where both the eare and the density vary.

I take a small slice of the cone, $V \approx A \Delta x$ (this works for a rod, but hopefully this is fine when I take the limit of delta x)

Mass of the slice, $m \approx \rho (x) A \Delta x$

Total mass of the cone, is the sum of the slices $= \Sigma A \Delta x \rho (x)$

$lim \Delta x \rightarrow 0$

$M = \int^{h}_{0} A \rho (x) dx$

The area of a thin slice is given by $A = \frac{r^{2} \pi h^{2}}{x^{2}}$

So my integral becomes $M = \int^{h}_{0} \frac{r^{2} \pi h^{2}}{x^{2}} \rho (x) dx$

Subbing in the density funcion,

$M = \int^{h}_{0} \frac{r^{2} \pi h^{2}}{x^{2}} \rho_{0} \frac{3x^{2} + 2xh}{h^{2}} dx$

$M = \int^{h}_{0} \frac{r^{2}\pi h^{2} \rho_{0} (3x^{2} + 2xh)}{x^2 h^2} dx$

Firstly, is my method correct?

and secondly, how do I go about performing this integral?

thanks!

Mark44
Mentor

Homework Statement

Let a cone with height $h$ and base area $A$ have the density $\rho (x) = \rho_{0} \frac{3x^{2} + 2xh}{h^{2}}, 0 \leq x \leq h$

the relation between cone radius $r$ and distance from cone apex $x$ is given by:

$r = (\frac{B}{\pi h^{2}})^{\frac{1}{2}}x$

Find the total mass $M$ of the cone.

The Attempt at a Solution

Ok, so i've seen problems where the density varies, but this is the first one i've looked at where both the eare and the density vary.

I take a small slice of the cone, $V \approx A \Delta x$ (this works for a rod, but hopefully this is fine when I take the limit of delta x)

Mass of the slice, $m \approx \rho (x) A \Delta x$

Total mass of the cone, is the sum of the slices $= \Sigma A \Delta x \rho (x)$

$lim \Delta x \rightarrow 0$

$M = \int^{h}_{0} A \rho (x) dx$

The area of a thin slice is given by $A = \frac{r^{2} \pi h^{2}}{x^{2}}$

So my integral becomes $M = \int^{h}_{0} \frac{r^{2} \pi h^{2}}{x^{2}} \rho (x) dx$

Subbing in the density funcion,

$M = \int^{h}_{0} \frac{r^{2} \pi h^{2}}{x^{2}} \rho_{0} \frac{3x^{2} + 2xh}{h^{2}} dx$

$M = \int^{h}_{0} \frac{r^{2}\pi h^{2} \rho_{0} (3x^{2} + 2xh)}{x^2 h^2} dx$

Firstly, is my method correct?
I don't think so. The density varies with x, so horizontal slices won't work, as you can't just multiply the area of a slice by its density. I haven't worked the problem through, but what I would do is to use cylindrical shells that divide up the cone kind of like the layers in an onion. For each shell, the thickness is ##\Delta x##, so the density will be nearly constant throughout the shell.

The volume of one of the shells is ##2\pi * \text{radius} * \text{height} * \Delta x##.
BOAS said:
and secondly, how do I go about performing this integral?

thanks!

LCKurtz
Homework Helper
Gold Member

Homework Statement

Let a cone with height $h$ and base area $A$ have the density $\rho (x) = \rho_{0} \frac{3x^{2} + 2xh}{h^{2}}, 0 \leq x \leq h$

the relation between cone radius $r$ and distance from cone apex $x$ is given by:

$r = (\frac{B}{\pi h^{2}})^{\frac{1}{2}}x$

Find the total mass $M$ of the cone.

What is ##B##? I'm guessing from the given above that you mean ##A## in that formula.

The Attempt at a Solution

Ok, so i've seen problems where the density varies, but this is the first one i've looked at where both the eare and the density vary.

I take a small slice of the cone, $V \approx A \Delta x$ (this works for a rod, but hopefully this is fine when I take the limit of delta x)

Mass of the slice, $m \approx \rho (x) A \Delta x$

Total mass of the cone, is the sum of the slices $= \Sigma A \Delta x \rho (x)$

$lim \Delta x \rightarrow 0$

$M = \int^{h}_{0} A \rho (x) dx$

The area of a thin slice is given by $A = \frac{r^{2} \pi h^{2}}{x^{2}}$

This is very confusing. You are using ##A## for two different things.

So my integral becomes $M = \int^{h}_{0} \frac{r^{2} \pi h^{2}}{x^{2}} \rho (x) dx$

Subbing in the density funcion,

$M = \int^{h}_{0} \frac{r^{2} \pi h^{2}}{x^{2}} \rho_{0} \frac{3x^{2} + 2xh}{h^{2}} dx$

$M = \int^{h}_{0} \frac{r^{2}\pi h^{2} \rho_{0} (3x^{2} + 2xh)}{x^2 h^2} dx$

Firstly, is my method correct?

and secondly, how do I go about performing this integral?

thanks!

Once you get it straightened out, you need to express ##r## in terms of ##x## in your integral, because ##r## depends on ##x##.

haruspex
Homework Helper
Gold Member
The area of a thin slice is given by $A = \frac{r^{2} \pi h^{2}}{x^{2}}$
Why isn't it just $A = \pi r^{2}$?
The density varies with x, so horizontal slices won't work
x is height here, so I think horizontal slices are fine.
You are using A for two different things.
BOAS defines it as the area of a horizontal slice. I agree that the algebraic expression given for its value is wrong, but I can't relate that to a changed definition. Maybe you're seeing something I'm not.

Chestermiller
Mentor
Why isn't it just $A = \pi r^{2}$?

BOAS defines it as the area of a horizontal slice. I agree that the algebraic expression given for its value is wrong, but I can't relate that to a changed definition. Maybe you're seeing something I'm not.
The first sentence in the problem statement states that A is the area of the base. Later on, in the equation for r vs x, it seems that B is the area of the base. What the Starter is calling A (the local cross section) is really equal to Bx2/h2.

Chet

haruspex
Homework Helper
Gold Member
The first sentence in the problem statement states that A is the area of the base.
OK, missed that. I was looking at the general expression for the radius, which implies it is B. Having been put right on that, I now see where the equation ##A = \frac{r^{2} \pi h^{2}}{x^{2}}## comes from. BOAS is here taking r to be the radius at the base, not a variable function of x. The error, then, is that (with that usage) the equation should read ##A = \frac{r^{2} \pi x^{2}}{h^{2}}##.

Hello,

I really appreciate the time you've put in already. But I hadn't anticipated that I was this far off course, and that I need to dedicate a decent amount of time to it. I have a more pressing (graded) lab report that I must finish, so won't be able to give this the attention it deserves until Thursday.

I think it's best that I take your comments on board and re-work the question with them in mind, and if I still encounter problems; create a new OP. I hope that's ok.

Thanks,

Jacob.

Mark44
Mentor
The density varies with x, so horizontal slices won't work
haruspex said:
x is height here, so I think horizontal slices are fine.
Guess I didn't read the OP close enough. Thanks!

LCKurtz
Homework Helper
Gold Member
Hello,

I really appreciate the time you've put in already. But I hadn't anticipated that I was this far off course, and that I need to dedicate a decent amount of time to it. I have a more pressing (graded) lab report that I must finish, so won't be able to give this the attention it deserves until Thursday.

I think it's best that I take your comments on board and re-work the question with them in mind, and if I still encounter problems; create a new OP. I hope that's ok.

Thanks,

Jacob.

OK. When you get back to it I suggest you begin with calling the area of the base ##B## and the radius of the base ##R##. Then get ##R## in terms of ##B##. Then let ##A## be the area of your variable cross section and ##r## be its radius. You should be able to express everything in your integral in terms of the variable ##x## and the constants ##B##, ##h## and ##\rho_0##. The integral will be a simple polynomial in ##x## to integrate. Also, as it turns out, there will be no ##\pi## in the answer.

BOAS
I think LCKurtz is correct but just to be clear, I do believe there will be a $\pi$ in the answer. It will be hidden inside $B$

LCKurtz
I think LCKurtz is correct but just to be clear, I do believe there will be a $\pi$ in the answer. It will be hidden inside $B$