# Velocity of melting icecream

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A chocolate ice cream cone is filled with melted chocolate ice cream of
density 1.20g/cm^3 The cone is 10 cm long and has diameter of 6cm at the
larger top end No a small hole of diameter 1 mm is produced at the
narrow bottom part and the ice cream starts exiting out the bottom hole.
Ignoring the viscosity of the melted ice cream, find the amount of time
it takes the ice cream to run out through the bottom hole completely.
Assume the velocity of the chocolate ice cream to be zero at the larger
top end.

Please help i dont know where to begin some one told me its just sqrt(2gh) but i dont know where that formula comes a derivation would be nice to know my book dosnt seem to have this equation at all

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You can just use conservation of energy (kinetic + potential = constant)

(1/2)mv^2 + mgh = constant

dividing by volume this change the mass terms to density:

(1/2)pv^2 + pgh = constant

This should be constant at all places in the fluid so we consider the very top of the fluid as the first point. Here v = 0.

The second point we consider is the hole at the bottom of the cone where h = 0.

so we can now set the equation equal at the two points:

pgh = (1/2)pv^2

and just solve for v:

v^2 = 2gh

v = sqrt(2gh)

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If you google this it seems everyone tells you to use Bernoulli's equation, which looks very similar to what I had:

(1/2)pv^2 + pgh = constant

except with an additional P on the left side for the pressure at that point, and they say P is the same at both points so it cancels anyway....

But, honestly it seems wrong to me to use Bernoulli's equation for this. I don't see why P would be the same at both points, and also Bernoulli's equation is only supposed to be valid for two points along the same stream line, which doesn't seem to work in this case.

If anyone can explain to me why Bernoulli's equation would work here please do! Otherwise I'll just assume it's not appropriate.

Edited by Staysys
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Its been a long time I have to deal with stuff like this but i think the question is equivalent to calculate the time taken by a cylinder of 1 mm diameter and height such that it equals the volume of the cone. Having find the length of that "spaghetti", one can find the time as it was a falling object.

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Its been a long time I have to deal with stuff like this but i think the question is equivalent to calculate the time taken by a cylinder of 1 mm diameter and height such that it equals the volume of the cone. Having find the length of that "spaghetti", one can find the time as it was a falling object.

I don't think that works. If you can just treat it as a falling cylinder then after about 10 seconds the chocolate would be flowing out at a speed of near 100m/s. That clearly doesn't happen, because the chocolate slowly oozes it's way to the hole opening and then falls out at a low speed, it isn't being accelerated at freefall toward the hole.

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I don't think that works. If you can just treat it as a falling cylinder then after about 10 seconds the chocolate would be flowing out at a speed of near 100m/s. That clearly doesn't happen, because the chocolate slowly oozes it's way to the hole opening and then falls out at a low speed, it isn't being accelerated at freefall toward the hole.

Right. Then it must be the sum of falling spaghettis each one with length the height of the cone.
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Right. Then it must be the sum of falling spaghettis each one with length the height of the cone.

umm.. did you see anything wrong with the solution I posted in my first post?

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umm.. did you see anything wrong with the solution I posted in my first post?

You showed how to get v (seemed a bit complicated for me).

You still must find t.

I suggested that v would be that of a falling cylinder of 1 mm diam and height of the cone. If that is correct you need the formulas for

a falling body and for the volume of a cone & the volume of a cylinder.

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The other solution is to say that the melting chocolate will make a crust on the top and nothing will go down. ##### Share on other sites

You showed how to get v (seemed a bit complicated for me).

You still must find t.

I suggested that v would be that of a falling cylinder of 1 mm diam and height of the cone. If that is correct you need the formulas for

a falling body and for the volume of a cone & the volume of a cylinder.

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The other solution is to say that the melting chocolate will make a crust on the top and nothing will go down. oh right, I don't know how I missed that. I guess because I arrived at the same answer OP said it was supposed to be.

But, ya I think you are right about using a cylinder then.

I'm not sure if the OP is even still checking his thread though so I won't bother getting an actual number right now though.

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freeflight1 use what Staysys shows on the first post to find the expression of velocity as a function of height since ,the velocity is going to vary as time passed 'cause volume is diminishing so do height ,then set v = dh/dt then integrate to find the time

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