Finding the Work to Move Chocolate

[DameLight]

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >[/color]

A tank in the shape of an inverted right circular cone has height 5 meters and radius 2 meters. It is filled with 4 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. The density of hot chocolate is δ = 1020 kg/m³. Your answer must include the correct units.

h = 5m
r = 2m
and I am assuming that up to 4m is filled with chocolate
δ = 1020 kg/m³

1. ΔW = F * D = F * y
F = mg
m = δ * V

2. ΔW = ((1020 * V)9.8) * y

Volume of a disk = πr²h = π((-5/2y + 2)1/2)²Δy
r = (my + b)1/2 = (-5/2y + 2)1/2
m = Δy/Δx = -5/2
b = 2

3. ΔW = ((1020 * π((-5/2y + 2)1/2)²Δy )9.8) * y

4. ∫₁⁵ = 1020*9.8*π*(-5/4y + 1)²*y*dy

All I want to know is if I am on the right track. I was having difficulties picking out the steps from my professor's examples. (am I approaching things in a logical order?) and obviously I want to know if I'm wrong.

Thank you for your time

 

[SteamKing]

A tank in the shape of an inverted right circular cone has height 5 meters and radius 2 meters. It is filled with 4 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. The density of hot chocolate is δ = 1020 kg/m³. Your answer must include the correct units.

h = 5m
r = 2m
and I am assuming that up to 4m is filled with chocolate
δ = 1020 kg/m³

1. ΔW = F * D = F * y
F = mg
m = δ * V

2. ΔW = ((1020 * V)9.8) * y

Volume of a disk = πr²h = π((-5/2y + 2)1/2)²Δy
r = (my + b)1/2 = (-5/2y + 2)1/2
m = Δy/Δx = -5/2
b = 2

3. ΔW = ((1020 * π((-5/2y + 2)1/2)²Δy )9.8) * y

4. ∫₁⁵ = 1020*9.8*π*(-5/4y + 1)²*y*dy

All I want to know is if I am on the right track. I was having difficulties picking out the steps from my professor's examples. (am I approaching things in a logical order?) and obviously I want to know if I'm wrong.

Thank you for your time

In problems such as this one, making a simple sketch can add much clarity to how you tackle the problem.

Also, it's not clear what the variable 'y' represents. You should always list any variables you introduce into your calculations, along with a short description of their meaning.

 

[DameLight]

I did draw a sketch and I put the object on the coordinate plane as this question calls for.

y represents all possible values between 0 and 5 along the y axis.

 

[SteamKing]

I did draw a sketch and I put the object on the coordinate plane as this question calls for.

This part of the question statement seems to have been omitted from the OP.

y represents all possible values between 0 and 5 along the y axis.

This is much too vague a description.

y can represent a general height coordinate, but you also seem to use it to represent the distance that a given amount of chocolate is raised to bring to to the top of the tank.

In mathematics, precision is key. You can't just make up a variable, cross your fingers, and hope things work out for the best.

For example, your function which describes the radius of the tank versus height:

r = (my + b)1/2 = (-5/2y + 2)1/2
m = Δy/Δx = -5/2
b = 2

Have you checked this function to see if it gives the correct radius for the tank as y changes? What does the factor "1/2" represent?

If y = 0, what is r?
If y = 2.5, what is r?
If y = 5, what is r?

 

[DameLight]

A tank in the shape of an inverted right circular cone has height 5 meters and radius 2 meters. It is filled with 4 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. The density of hot chocolate is δ = 1020 kg/m3. Your answer must include the correct units.

h = 5m
r = 2m
and I am assuming that up to 4m is filled with chocolate
δ = 1020 kg/m3
the y-axis is positive downwards from the top of the cylinder

These are the equations I used:

ΔW = F * D
F = mg
m = δ * V
V = πr²h for a typical slice of the cone parallel to the x axis

Now I start solving the equations from the bottom to the top:

1. V = πr²h

variable r is unknown
and the height is the small change in y

r must be the result of a linear relationship between the radius and the height so to solve:
at y = 0; r = 2
at y = 5; r = 0
Δr/Δy = -2/5
r = -2/5y + 2

plug this back into the equation:
1. V = π(-2/5y+2)²Δy

and then plug the equation into the next one:
2. m = δ * V
2. m = 1020 * π(-2/5y+2)²Δy

nothing to solve for so continuing to the next equation:
3. F = mg
3. F = 1020 * π(-2/5y+2)²Δy * 9.8

where g = 9.8 m/s²

Finally plug everything into the last equation:
4. ΔW = F * D
4. ΔW = 1020 * π(-2/5y+2)²Δy * 9.8 * y

where y = the distance from the top of the cone

Then solve for the integral from 1 to 5 since the cone is only 4/5 filled

∫ 1020 * π(-2/5y+2)²dy * 9.8 * y

rearrange:

∫ 1020 * 9.8 * π(-2/5y+2)²dy * y

use the integral property to pull out the constants:

(1020 * 9.8 * π) ∫ (-2/5y+2)²dy * y

now focusing on the integral:

∫ (-2/5y+2)²dy * y
∫ (4/25y² - 8/5y + 4)y * dy
∫ (4/25y³ - 8/5y² + 4y)dy

lastly find the antiderivative and solve

 

[DameLight]

It is my fault on all accounts I did not have a way to post a picture in the OP.

The coordinate system should have been elaborated further, but I think I described it better in my last post/edit.

I realized my mistake with the radius. I must have misread the question and lead myself to believe it represented the diameter hence the 1/2. I also went back and redid the formula for r.

 

[Nathanael]

Your post #5 looks good, nice job.

It should maybe be noted that you are assuming that the speed of the hot-chocolate as it goes over the edge is negligible (or else your "F" in eq.3 would be greater than mg). In other words, you are calculating the minimum work required.

Another way to approach the problem:
What if all of the hot chocolate were pumped onto a(n imaginary) flat platform at the height of the top of the cone (y=0). The work done would be the same as pumping it over the edge. So we can now just consider the initial and final states of the system and use the fact that the work done is equal to the change in energy. (Again, we would take the assumption that it makes it on to this imaginary platform with negligible velocity.)

The change in energy (= to the work done) would be (mgh) the weight of the entire mass of liquid multiplied by the change in height of the center of mass of the liquid. Since we know the center of mass ends up at a height of y=0 (because that's where all the mass ends up) the problem is then reduced to finding the initial height of the center of mass of the liquid.Maybe this approach seems more complicated but I thought it was worth mentioning.

 

[DameLight]

Thanks : )

I've seen that approach in my physics classes. Is that a physics approach or does that use calculus similar to what I used above?

 

[Nathanael]

It involves a touch of physics (the work-energy theorem) but yes the calculus involved (in finding the original center of mass of the chocolate) is pretty similar to what you used in your approach.

I like this method because somehow it seems more organized to me... I'm not sure why though, I just like breaking problems into pieces.