Mass of a Cone with varying Density

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Homework Statement



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

the relation between cone radius [itex]r[/itex] and distance from cone apex [itex]x[/itex] is given by:

[itex]r = (\frac{B}{\pi h^{2}})^{\frac{1}{2}}x[/itex]

Find the total mass [itex]M[/itex] of the cone.

Homework Equations




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, [itex]V \approx A \Delta x[/itex] (this works for a rod, but hopefully this is fine when I take the limit of delta x)

Mass of the slice, [itex]m \approx \rho (x) A \Delta x[/itex]

Total mass of the cone, is the sum of the slices [itex]= \Sigma A \Delta x \rho (x)[/itex]

[itex]lim \Delta x \rightarrow 0[/itex]

[itex]M = \int^{h}_{0} A \rho (x) dx[/itex]

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

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

Subbing in the density funcion,

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

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

Firstly, is my method correct?

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

thanks!
 

Answers and Replies

  • #2
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Homework Statement



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

the relation between cone radius [itex]r[/itex] and distance from cone apex [itex]x[/itex] is given by:

[itex]r = (\frac{B}{\pi h^{2}})^{\frac{1}{2}}x[/itex]

Find the total mass [itex]M[/itex] of the cone.

Homework Equations




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, [itex]V \approx A \Delta x[/itex] (this works for a rod, but hopefully this is fine when I take the limit of delta x)

Mass of the slice, [itex]m \approx \rho (x) A \Delta x[/itex]

Total mass of the cone, is the sum of the slices [itex]= \Sigma A \Delta x \rho (x)[/itex]

[itex]lim \Delta x \rightarrow 0[/itex]

[itex]M = \int^{h}_{0} A \rho (x) dx[/itex]

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

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

Subbing in the density funcion,

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

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

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!
 
  • #3
LCKurtz
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Homework Statement



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

the relation between cone radius [itex]r[/itex] and distance from cone apex [itex]x[/itex] is given by:

[itex]r = (\frac{B}{\pi h^{2}})^{\frac{1}{2}}x[/itex]

Find the total mass [itex]M[/itex] of the cone.

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

Homework Equations




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, [itex]V \approx A \Delta x[/itex] (this works for a rod, but hopefully this is fine when I take the limit of delta x)

Mass of the slice, [itex]m \approx \rho (x) A \Delta x[/itex]

Total mass of the cone, is the sum of the slices [itex]= \Sigma A \Delta x \rho (x)[/itex]

[itex]lim \Delta x \rightarrow 0[/itex]

[itex]M = \int^{h}_{0} A \rho (x) dx[/itex]

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

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

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

Subbing in the density funcion,

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

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

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##.
 
  • #4
haruspex
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The area of a thin slice is given by [itex]A = \frac{r^{2} \pi h^{2}}{x^{2}}[/itex]
Why isn't it just [itex]A = \pi r^{2} [/itex]?
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.
 
  • #5
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Why isn't it just [itex]A = \pi r^{2} [/itex]?

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
 
  • #6
haruspex
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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}}##.
 
  • #7
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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.
 
  • #8
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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!
 
  • #9
LCKurtz
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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.
 
  • #10
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I think LCKurtz is correct but just to be clear, I do believe there will be a [itex]\pi[/itex] in the answer. It will be hidden inside [itex]B[/itex]
 
  • #11
LCKurtz
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I think LCKurtz is correct but just to be clear, I do believe there will be a [itex]\pi[/itex] in the answer. It will be hidden inside [itex]B[/itex]
Yes. That's why the ##\pi##'s in the calculation appear to disappear.
 

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