Find mass and center of mass of ice cream cone

In summary: Oh shoot. I missed the radius of the base of the cone. It should be 3, not 2. So the sphere equation should be: x^2 + y^2 + (z - 8)^2 = 4^2, or x^2 + y^2 + z^2 - 16z + 60 = 0. I'm still not clear on what your equations are supposed to represent. If you're trying to find the mass and center of mass, you don't need to do integrals. You need to find the volume of the cone and the sphere, and multiply each by the appropriate density function, and then add the two masses to get the total mass. For the
  • #1
Michels10
18
0

Homework Statement


A toy manufacturer wants to create a toy ice cream cone by fitting a sphere of radius 4 cm inside a cone with a height of 8 cm and radius of the base of 3 cm. The base of the cone is concave, but the rest of the cone is solid plastic so that with the sphere attached there is no hollow space inside. The sphere and cone are constructed from special materials so that the density of the sphere is proportional to the distance from the tip of the cone, with constant of proportionality 1.4, and the density of the cone is proportional to the distance from its vertical axis, with constant of proportionality 1.8. Find the mass and center of mass of the ice cream cone. (Hint: decide on a position for the whole ice cream cone. If the center of the sphere is the origin, then the tip of the cone cannot also be at the origin. It is a good idea to have the cone oriented vertically the way you would normally hold one.) Also, provide a plot showing the entire ice cream cone.

Note from professor: Instead of dealing with these
integrals, I will allow you to change the density functions given so
that instead the densities for both the cone and sphere are
proportional to the square of the distance described. With these
simpler density functions (without the square roots), the integrals
should be much simpler to solve.


Homework Equations



Sphere: z = x^2 + y^2 + z^2
Cone: z = sqrt(x^2 + y^2)
Vsphere = 4/3*pi*r^3
Vcone = (pi*h*r^2)/3
density = mass/volume

center of mass:
x(bar) = Myz/m
y(bar) = Mxz/m
z(bar) = Mxy/m

where M is the moment about a coordinate plane.

The Attempt at a Solution



Not exactly sure where to start... If I have no mass how can I begin to solve for this? Should I start by solving for the volume that the sphere+cone take up?

Any help is greatly appreciated!
Thanks,
Michels10
 
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  • #2
Michels10 said:

Homework Statement


A toy manufacturer wants to create a toy ice cream cone by fitting a sphere of radius 4 cm inside a cone with a height of 8 cm and radius of the base of 3 cm. The base of the cone is concave, but the rest of the cone is solid plastic so that with the sphere attached there is no hollow space inside.
Here is the information you need to use to get the mass.
Michels10 said:
The sphere and cone are constructed from special materials so that the density of the sphere is proportional to the distance from the tip of the cone, with constant of proportionality 1.4, and the density of the cone is proportional to the distance from its vertical axis, with constant of proportionality 1.8.
The mass of the sphere = density * volume, with the density as specified above. The mass of the cone can be computed similarly. Also, keep in mind the hint given by your professor.

In both, the density is not constant (i.e., is a function).
Michels10 said:
Find the mass and center of mass of the ice cream cone. (Hint: decide on a position for the whole ice cream cone. If the center of the sphere is the origin, then the tip of the cone cannot also be at the origin. It is a good idea to have the cone oriented vertically the way you would normally hold one.) Also, provide a plot showing the entire ice cream cone.

Note from professor: Instead of dealing with these
integrals, I will allow you to change the density functions given so
that instead the densities for both the cone and sphere are
proportional to the square of the distance described. With these
simpler density functions (without the square roots), the integrals
should be much simpler to solve.


Homework Equations



Sphere: z = x^2 + y^2 + z^2
Your equation above is incorrect. You have z as a function of x, y, and z. What you should have is x^2 + y^2 + z^2 = r^2.
Michels10 said:
Cone: z = sqrt(x^2 + y^2)
Vsphere = 4/3*pi*r^3
Vcone = (pi*h*r^2)/3
density = mass/volume

center of mass:
x(bar) = Myz/m
y(bar) = Mxz/m
z(bar) = Mxy/m

where M is the moment about a coordinate plane.

The Attempt at a Solution



Not exactly sure where to start... If I have no mass how can I begin to solve for this? Should I start by solving for the volume that the sphere+cone take up?

Any help is greatly appreciated!
Thanks,
Michels10
 
  • #3
Thanks for your reply Mark44!

Dont I still have two unknowns? density and mass? I guess my problem now is finding the equation for density. Do I need to set the equations equal to and find out where they intersect?

I've solved for both volumes:

Vcone = 24*pi
Vsphere = (32/3)*pi

Thanks,
Michels10
 
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  • #4
Your density functions will be in units of mass/volume. Due to the way the density functions are defined, the center of mass will be along the central axis of the cone+sphere, which I'm assuming you will orient so that the z-axis passes vertically through the center of the cone and sphere.

It's not enough to get the volume of each of the two solids. The formulas I gave before should be interpreted as incremental mass, [itex]\Delta m[/itex], which you will need to use in each integral.

Of course you need the equations of the cone and sphere. Hopefully you have already graphed them.
 
  • #5
how do these equations look:

cone- z2 = (64/9)(x2+y2)

sphere- 4 = x2+y2+z2

Cone:
Int(Int(Int(1.4*(64/9)(r2)*r,dr),d(theta)),dz)
r=0 to 3
theta = 0 to 2pi
z = 0 to 8

Sphere:
Int(Int(Int(1.8*p2*sin(phi),d(p),d(theta)),d(phi))
p = 0..2
theta = 0..2pi
phi = 0..pi
 
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  • #6
i've attached a picture of the cone/sphere graphed.
 

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  • #7
The equations in post #5 don't go with the graphs in post #6. Your equation for the sphere has its center at the origin. The graph of the sphere has its center at somewhere around (0, 0, 11).

Also, your equation for the sphere has a radius of 2. In the problem description, the radius is supposed to be 4.
 
  • #8
I updated one of my last posts with some integrals... But back to the graph..

The cone has a height of 8. The total height from the bottom of the cone to the center of the sphere is 8+sqrt(7). And the radius is still 4 for the sphere, and still 3 for the cone
 
  • #9
Read what I said in post #7.

As for your integrals, I think you are jumping the gun. It doesn't look to me like you have the density functions figured out yet.
 
  • #10
Michels10 said:
how do these equations look:

cone- z2 = (64/9)(x2+y2)

sphere- 4 = x2+y2+z2

Okay, then how about I change them...

sphere- 16 = x2+y2+(z - (8+sqrt(7)))2

cone- x2 + (y-3)2 = (z-8)2
 
  • #11
Where did the 8 + sqrt(7) come from?

Why is the y - 3 term in the equation for the cone?
 
  • #12
I think I've about given up on this one. I clearly have no idea what I'm doing. I was trying to rearrange the equations to fit my graph. I saw my professor, he gave me an equation of

z^2 = 64/9 * x^2 + y^2 for the cone

and said that the height from the center of the sphere to the bottom of the cone was 8+ sqrt(7). the radius of the cone is 3, the radius of the ball is 4... To find the distance from the center of the sphere to the cone-- that leaves us with a triangle that has 4^2 = x^2 + 3^2.

x= +- sqrt(7). Add this to the height of the cone and we get 8+sqrt(7)

I've got an example in my book that calculates the density+mass+center of mass, so I tried adapting that to this problem and that is why I 'jumped the gun' with the integrals.

If the tripple integral of the constant of proportionality multiplied by the formula of the equation doesn't get me the mass, then I am not sure what does.

my book reads:

p(x,y,z) = K*(x^2+y^2+z^2) for density.

where K is a constant

and

int(int(int( p(x,y,z) dx )dy)dz)

the tripple integral of the density equation for the mass.
 
  • #13
Michels10 said:
I think I've about given up on this one. I clearly have no idea what I'm doing. I was trying to rearrange the equations to fit my graph. I saw my professor, he gave me an equation of

z^2 = 64/9 * x^2 + y^2 for the cone
This looks fine. The assumption here is that the tip of the cone is at the origin.
Michels10 said:
and said that the height from the center of the sphere to the bottom of the cone was 8+ sqrt(7). the radius of the cone is 3, the radius of the ball is 4... To find the distance from the center of the sphere to the cone-- that leaves us with a triangle that has 4^2 = x^2 + 3^2.

x= +- sqrt(7). Add this to the height of the cone and we get 8+sqrt(7)
That's fine. I just didn't see where you got it, and now I do. This makes the equation of the sphere x2 + y2 + (z - (8 + sqrt(7))2 = 16.
Michels10 said:
I've got an example in my book that calculates the density+mass+center of mass, so I tried adapting that to this problem and that is why I 'jumped the gun' with the integrals.

If the tripple integral of the constant of proportionality multiplied by the formula of the equation doesn't get me the mass, then I am not sure what does.

my book reads:

p(x,y,z) = K*(x^2+y^2+z^2) for density.

where K is a constant
But not for this problem. In this problem, the density varies in different ways for the cone and the sphere.

Before setting up the triple (one p in triple) integrals, see if you can find formulas for the density functions as described in your first post, and using your instructor's suggestion. You'll need one function for the cone density and another for the sphere density.
Michels10 said:
and

int(int(int( p(x,y,z) dx )dy)dz)

the tripple integral of the density equation for the mass.
 
  • #14
Okay, I'm assuming it has something to do with the slope of the cone, and the increasing radius/size as you move up from the bottom of the cone... Since the radius is 3 and the height is 8, this gives us a slope of 8/3.

I'm assuming that I need to use that 8/3 in my equation, I'm not sure where though...

Cone:

[tex]
\int^{2pi}_{0}\int^{8}_{0}\int^{(8/3)x}_{0} (1.4*r^2) r dr dz dtheta
[/tex]

Not sure about the sphere, though...
 
  • #15
Double integrals should suffice for both the cone and sphere.

For the cone, the region over which integration is performed is {{r, theta) | 0 <= r <= 3, 0 <= theta <= 2pi}.

The density function is as you have it, 1.4*r^2 (using the note from your professor). The integral for the mass of the cone should look like this:
[tex]m = \int_{\theta = 0}^{2\pi} \int_{r = 0}^3 (z_{top} - z_{cone}) 1.4 r^2 r dr d\theta[/tex]

What I'm showing as ztop is the z value at the top of the cone, which is the z value on the lower half of the sphere. zcone is the z value at a point on the surface of the cone.
 
  • #16
Okay very cool. I suppose now I have to find out how deep the cone is in the center (along the z axis) due to concavity.

Thanks you for your help thus far. I am going to keep going at it! I'll post back soon with an update.
 
  • #17
Yes, that's basically it. The top surface of the cone is the bottom surface of the sphere. You don't have to find how deep the top surface of the cone is - you just need the z value, which will be the z value on the lower half of the sphere.

Still ahead, you need to find Mz, the moment about the xy plane, and the mass and moment for the sphere.
 
  • #18
The center of mass of the cone is somewhere along the Z axis. This yields a problem when dealing with double integrals doesn't it? I can't substitute a Z into the equation when dealing with polar coordinates.

I believe the equation is: 1/mass Int(Int((z*density)rdr)dtheta)

edit: I've been reading that the center of mass of a cone is simply h/4... But since my cone is concave, I'm not sure how to go about it.
 
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  • #19
Michels10 said:
The center of mass of the cone is somewhere along the Z axis. This yields a problem when dealing with double integrals doesn't it?

Yes, it does. You need a triple integral.

I can't substitute a Z into the equation when dealing with polar coordinates.

I believe the equation is: 1/mass Int(Int((z*density)rdr)dtheta)

edit: I've been reading that the center of mass of a cone is simply h/4... But since my cone is concave, I'm not sure how to go about it.

Just add the dz integral between the lower and upper limits (cone and lower sphere) as the inside integral (r dz dr dtheta).
 
  • #20
Okay, got that, now what should I do for the sphere? Finding the density of the sphere should be relatively easy... Its not just the triple integral of volume * density constant *radius?
 
  • #21
Michels10 said:
Okay, got that, now what should I do for the sphere? Finding the density of the sphere should be relatively easy... Its not just the triple integral of volume * density constant *radius?

No, it isn't. The density isn't constant. You have a constant of proportionality given. Before you worry about that, your first step should be to set up a triple integral in either cylindrical or appropriate spherical coordinates with an integrand of 1 to calculate just the volume of the sphere which should come out

[tex] V = \frac {4\pi r^3} 3 =\frac {256\pi} 3[/tex]

Once you have that working you can add density or moment arms etc.

And, out of curiosity, what did you get for the center of mass of the cone? Are you using Maple to do the heavy lifting in this problem?
 
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  • #22
I am indeed using maple. I've already found the volume of the sphere,using triple integrals-

int(int(int(vsp(rho, theta, Phi), rho = 0 .. 4), theta = 0 .. 2*Pi), Phi = 0 .. Pi) = 256Pi/3

for the mass of the cone and the center of mass i got:

f := ((4-sqrt(7))*1.4)*r^3
mass:=Int(Int(f, r = 0 .. 3), theta = 0 .. 2*Pi) = 241.2300214

cmass := 1.4*z*(4-sqrt(7))*r^3
centerOfMass:=(Int(Int(Int(cmass, z = 0 .. 8-sqrt(7)), r = 0 .. 3), theta = 0 .. 2*Pi))/(241.23) = 14.33399078
 
  • #23
Michels10 said:
I am indeed using maple. I've already found the volume of the sphere,using triple integrals-

int(int(int(vsp(rho, theta, Phi), rho = 0 .. 4), theta = 0 .. 2*Pi), Phi = 0 .. Pi) = 256Pi/3

That is the volume of a sphere and it is equal to the volume of your sphere because they are the same size. But you didn't calculate it using your sphere which is not at the origin. You are going to have to set it up using the sphere above the origin because where it is matters when you consider the density and hence the mass.

for the mass of the cone and the center of mass i got:

f := ((4-sqrt(7))*1.4)*r^3
mass:=Int(Int(f, r = 0 .. 3), theta = 0 .. 2*Pi) = 241.2300214

cmass := 1.4*z*(4-sqrt(7))*r^3
centerOfMass:=(Int(Int(Int(cmass, z = 0 .. 8-sqrt(7)), r = 0 .. 3), theta = 0 .. 2*Pi))/(241.23) = 14.33399078

Your cone extends from the origin up to z = 8. Does it bother you that your calculated center of mass is outside the cone?
 
  • #24
I have to leave for the rest of the afternoon now, so here are some additional comments about your cone.

Michels10 said:
for the mass of the cone and the center of mass i got:

f := ((4-sqrt(7))*1.4)*r^3
mass:=Int(Int(f, r = 0 .. 3), theta = 0 .. 2*Pi) = 241.2300214

What happened to the triple integral we were talking about earlier?

cmass := 1.4*z*(4-sqrt(7))*r^3
centerOfMass:=(Int(Int(Int(cmass, z = 0 .. 8-sqrt(7)), r = 0 .. 3), theta = 0 .. 2*Pi))/(241.23) = 14.33399078

I don't understand your density. Furthermore, the limits you have given in this last triple integral don't describe your cone. They would describe a cylinder. You haven't used either of the equations for the top and bottom surfaces of the cone. They matter, don't they? Back to the drawing board.
 
  • #25
LCKurtz said:
That is the volume of a sphere and it is equal to the volume of your sphere because they are the same size. But you didn't calculate it using your sphere which is not at the origin. You are going to have to set it up using the sphere above the origin because where it is matters when you consider the density and hence the mass.



Your cone extends from the origin up to z = 8. Does it bother you that your calculated center of mass is outside the cone?

Yes it bothers me. This assignment was due two days ago and clearly I do not understand it. My book examples don't help me at all and I've got nothing to go by. Intuition?

As for the "back to the drawing board" -- I don't thinks so. These repeated attempts post due-date are just brewing frustration. There is no way for me to understand how to do a problem like this because I've never seen one done before.

Thanks for your help,
Michels10
 
  • #26
Michels10 said:
Yes it bothers me. This assignment was due two days ago and clearly I do not understand it. My book examples don't help me at all and I've got nothing to go by. Intuition?

As for the "back to the drawing board" -- I don't thinks so. These repeated attempts post due-date are just brewing frustration. There is no way for me to understand how to do a problem like this because I've never seen one done before.

Thanks for your help,
Michels10

Is there no teacher involved in your course? Are you saying your teacher has never discussed similar problems? I am curious what sort of academic setting you are in.
 
  • #27
There is, but the semester is over. The problems we have discussed/were in the book did not involve varying densities(not to mention the concavity at the top of the cone[irregular shapes]). We have done simple double or triple integrals of the equation for density*the density proportionality constant. Honestly I don't understand why I am having such trouble with this problem. Just seems like whatever I throw out there is wrong. My professor answers with cryptic responses similar to those that I have received in this thread- "reread the problem" or "redo the problem with respect to the tip of the sphere". Things that clearly arent helping. Its probably just my lack of understanding though, being that the majority of you guys find this to be trivial.
 
  • #28
No, this is not a trivial problem, especially with the added complication of the density functions. To tackle a problem like this you need to lay the groundwork by getting a reasonable graph of the solid, equations for the two parts of the graph (taking into account where you are placing the origin), then getting expresssions for the typical mass element, determinining the region over which integration takes place, and then finally, setting up and evaluating the integrals involved.
 
  • #29
Michels10, at this point, if you would like, I can send you a pdf of a Maple worksheet for this problem for you to look at. Maybe it will help you. If you want me to do that, go to my PF profile and visit my home page. There is a link there to email me. If you do that I will reply with the pdf attached.
 

What is the mass of an ice cream cone?

The mass of an ice cream cone varies depending on the type and size of the cone, as well as the amount of ice cream it contains. It can range from a few grams to several hundred grams. To determine the exact mass, a scale can be used to weigh the cone and ice cream together.

How do you find the center of mass of an ice cream cone?

The center of mass of an ice cream cone can be found by balancing the cone on a pivot point, such as a pencil, and marking the point where it balances. This point is the center of mass. Alternatively, mathematical equations can be used to calculate the center of mass based on the shape and distribution of the cone and ice cream.

Does the shape of the ice cream cone affect its center of mass?

Yes, the shape of the ice cream cone can affect its center of mass. A taller and narrower cone will have a higher center of mass, while a shorter and wider cone will have a lower center of mass.

How does the amount of ice cream affect the center of mass of an ice cream cone?

The amount of ice cream in an ice cream cone can affect its center of mass. As more ice cream is added, the center of mass will shift towards the heavier side. This can be offset by balancing the cone and ice cream on a pivot point.

Why is it important to know the mass and center of mass of an ice cream cone?

Knowing the mass and center of mass of an ice cream cone is important for understanding its stability and how it will behave when held or placed on a surface. It can also be useful in ensuring that toppings and decorations are evenly distributed for a better eating experience.

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