Mass of a Cone with varying Density

• BOAS
In summary: I don't want to give it away.The average density is going to be the integral of the density function over the variable x, ##\rho(x)##, divided by the height, ##h##:##\bar{\rho} = \frac{1}{h} \int_0^h \rho(x) \,dx##So, if you want the total mass of the cone, you need to multiply the average density by the volume of the cone, which is ##\frac{1}{3} \pi r^2 h##.The problem is that the radius ##r## varies with ##x##, so you need to express it in terms of ##x## before you can evaluate the integral

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!

BOAS said:

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!

BOAS said:

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##.

BOAS said:
The area of a thin slice is given by $A = \frac{r^{2} \pi h^{2}}{x^{2}}$
Why isn't it just $A = \pi r^{2}$?
Mark44 said:
The density varies with x, so horizontal slices won't work
x is height here, so I think horizontal slices are fine.
LCKurtz said:
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.

haruspex said:
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

Chestermiller said:
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 said:
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!

BOAS said:
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 $B$

jimbo007 said:
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 $B$
Yes. That's why the ##\pi##'s in the calculation appear to disappear.

1. What is the formula for calculating the mass of a cone with varying density?

The formula for calculating the mass of a cone with varying density is M = (1/3) x π x r^2 x h x ρ, where M is the mass, r is the radius of the base, h is the height, and ρ is the density.

2. How do you determine the density of a cone with varying density?

The density of a cone with varying density can be determined by dividing the mass of the cone by its volume. The volume can be calculated using the formula V = (1/3) x π x r^2 x h, where V is the volume, r is the radius of the base, and h is the height.

3. Can the mass of a cone with varying density be negative?

No, the mass of a cone with varying density cannot be negative. Mass is a measure of the amount of matter in an object, and it is always a positive value.

4. How does the varying density affect the mass of a cone?

The varying density of a cone affects its mass because the density is a factor in the formula for calculating mass. As the density changes, the mass of the cone will also change.

5. What units are used to measure the mass of a cone with varying density?

The mass of a cone with varying density is typically measured in grams (g) or kilograms (kg). However, other units such as pounds (lbs) or ounces (oz) can also be used depending on the specific application.