 Posers and Puzzles

1. 14 Mar '07 14:00
Last winter, I made two snowballs, one of which had twice as big a diameter as the other. Unfortunately, the weather didn't coöperate, and the snowballs soon started to melt. The melting only took place at the surface of the balls, so the speed with which the balls melted was proportional to the surface of the [remainder of the] balls.

How much was left of the small snowball, when half of the volume of the large snowball had melted?
2. 14 Mar '07 15:32
Originally posted by Mathurine
Last winter, I made two snowballs, one of which had twice as big a diameter as the other. Unfortunately, the weather didn't coöperate, and the snowballs soon started to melt. The melting only took place at the surface of the balls, so the speed with which the balls melted was proportional to the surface of the [remainder of the] balls.

[b]How much was left of the small snowball, when half of the volume of the large snowball had melted?
[/b]
From the problem statement we know that dV/dt = -k*(4*pi*r^2). Now:

dV/dt = d((4/3)*pi*r^3)/dt
= 4*pi*r^2 * (dr/dt)

Equating these two expressions, we get:

4*pi*r^2 * (dr/dt) = -k*(4*pi*r^2)

Therefore (dr/dt) = -k, which means that the radii decrease at a constant rate. Solving this differential equation for "r", we get:

r = r0 - kt

Now, the larger snowball decreases in volume by half. This means that:

V = (1/2)*V0

(4/3)*pi*r^3 = (1/2)*(4/3)*pi*r0^3

r^3 = (1/2)*r0^3

r = r0*(1/2)^(1/3)

r = r0*b (where b = (1/2)^(1/3), for brevity)

and

(r0-r) = r0 - r0*b = r0(1-b)

Since the radii decrease at a constant rate, this is the same change in the small snowball's radius.

rS = rS0 - r0(1-b)

rS = rS0 - 2rS0(1-b)

rS = rS0(1-2(1-b))

Solving for the new volume, we get:

VS = (4/3)*pi*(rS0*(1-2(1-b)))^3

VS = rS0^3 * (4/3)*pi*(1-2(1-b))^3

VS = rS0^3 * 0.849 (approx.)
3. 15 Mar '07 05:463 edits
Let the diameter of the smaller snowball be set to 1 unit in distance.

The larger snowball therefore has a diameter of 2 units, and will have half it's volume when the diameter is equal to 2^(2/3) units (cuberoot(4) units).

Now the smaller snowball will have a diameter of 1 less than the larger at any time, since the diameters shrink uniformly, which means it's diameter will be equal to cuberoot(4)-1 units, which is about 0.5874 units.

It's volume will have decreased from pi/6 (0.5263 units cubed) down to about 0.1061 units cubed, a mere 20.3% of its inital volume.

The exact volume is pi/2 * (cuberoot(4) - 2 * cuberoot(2) + 1) units cubed.
4. 15 Mar '07 09:421 edit
We seem to have a differance of opinion here...I put my money on PBE6🙂
5. 15 Mar '07 13:59
Originally posted by sonhouse
We seem to have a differance of opinion here...I put my money on PBE6🙂
Nah, I think I goofed. I thought 85% of original size was too big, but I couldn't find the mistake so I left it.
6. 15 Mar '07 21:58
My feeling is that the smaller will have 1/4 it's original volume, but I don't have any working to back it up. Just based on the power difference between Surface Area and Volume.
7. 15 Mar '07 22:15
Well, just for the heck of it i am going to say it will have 1/pi of its original volume
8. 17 Mar '07 11:563 edits
Well, PBE6 did provide the proof that the diameters shrink uniformly, whereas I merely accepted it as a given. (Proofs aren't my strong suit).

The percentage of shrinkage of the smaller isn't nice and neat, though. It does seem to be fairly close to 80% shrinkage though, which seems to be reasonable. (You'd expect that the smaller losing a greater percentage of it diameter, would also lose a greater percentage of it's overall volume.)

Hmm, let me see.. since the diameter of the small drops from 1 unit to cuberoot(4)-1 units, the ratio of the volume would be (cuberoot(4)-1)^3 or 3 - 6*cuberoot(2) + 3*cuberoot(4)
9. 22 Mar '07 17:277 edits
Originally posted by geepamoogle
Well, PBE6 did provide the proof that the diameters shrink uniformly, whereas I merely accepted it as a given. (Proofs aren't my strong suit).

The percentage of shrinkage of the smaller isn't nice and neat, though. It does seem to be fairly close to 80% shrinkage though, which seems to be reasonable. (You'd expect that the smaller losing a greater of the volume would be [b](cuberoot(4)-1)^3
or 3 - 6*cuberoot(2) + 3*cuberoot(4)[/b]
You guys make things way too complicated.

Snowballs melts at a uniform rate.

If it takes 1 hour for a 2 cubic centimeter snowball to melt down to a 1 cubic centimeter snowball, then a 1 cubic centimeter snowball will be completely melted after 1 hour.

In other words, the smaller snowball will lose just as much volume as the larger snowball.

So THEREFORE,
if X= original volume of bigger snowball, and
Y= original volume of smaller snowball,

then the answer is

leftover Volume of Y= (orignal volume of Y)-(X/2)

Subsititute any volume in there for the 2 snowballs and the answer will always be ZERO for this given question.
10. 23 Mar '07 10:18
Originally posted by uzless
Snowballs melts at a uniform rate.

If it takes 1 hour for a 2 cubic centimeter snowball to melt down to a 1 cubic centimeter snowball, then a 1 cubic centimeter snowball will be completely melted after 1 hour.
So the melting rate is just a constant then ?
Not "proportional to the surface of the [remainder of the] balls." as stated in the problem ?
11. 23 Mar '07 13:072 edits
Originally posted by geepamoogle
Well, PBE6 did provide the proof that the diameters shrink uniformly, whereas I merely accepted it as a given. (Proofs aren't my strong suit).

The percentage of shrinkage of the smaller isn't nice and neat, though. It does seem to be fairly close to 80% shrinkage though, which seems to be reasonable. (You'd expect that the smaller losing a greater of the volume would be [b](cuberoot(4)-1)^3
or 3 - 6*cuberoot(2) + 3*cuberoot(4)[/b]
The volume of the larger ball drops to R = R0 * (1/2) ^ (1/3) where R0 is the initial radius of the large ball. Following PBE8 p = (1/2) ^ (1/3). This means that the radius of the small ball is:

change in radius = (p - 1)R0 (note this is negative)

r = r0 + (p - 1)R0, R0 = 2*r0

r = r0 (2p - 1)

The volume of the small ball is therefore given by:

V = V0 (2p - 1)^3 = 0.2026 V0

The volume of the ball drops to about 20% of it's original volume. I agree with geepamoogle's calculation. PBE6 seems to have multiplied this by 4pi/3.
12. 23 Mar '07 14:533 edits
Originally posted by aging blitzer
So the melting rate is just a constant then ?
Not "proportional to the surface of the [remainder of the] balls." as stated in the problem ?
The melting rate depends on whether or not you have a constant supply of uniform heat.

For example, If you put the two snowballs in a box and did not allow new heat to be introduced into the box, then the snowballs would use up the heat in the box and melt slower over time.

If you put the snowballs outside, the melting rate for each snowball would be different again.
13. 23 Mar '07 15:27
Originally posted by DeepThought
The volume of the larger ball drops to R = R0 * (1/2) ^ (1/3) where R0 is the initial radius of the large ball. Following PBE8 p = (1/2) ^ (1/3). This means that the radius of the small ball is:

change in radius = (p - 1)R0 (note this is negative)

r = r0 + (p - 1)R0, R0 = 2*r0

r = r0 (2p - 1)

The volume of the small ball is therefore gi ...[text shortened]... olume. I agree with geepamoogle's calculation. PBE6 seems to have multiplied this by 4pi/3.
Yeah, just double checked on paper. I meant to divide Vsmall by Vsmall0, not just multiply the (4/3)*pi straight through:

Vsmall = (4/3)*pi*rS0^3 * (1-2(1-b))^3
Vsmall0 = (4/3)*pi*rS0^3

Vsmall/Vsmall0 = (1-2(1-b))^3 = 20.3% (approx)

as stated above.
14. 23 Mar '07 15:41
Originally posted by uzless
The melting rate depends on whether or not you have a constant supply of uniform heat.

For example, If you put the two snowballs in a box and did not allow new heat to be introduced into the box, then the snowballs would use up the heat in the box and melt slower over time.

If you put the snowballs outside, the melting rate for each snowball would be different again.
But would still depend on the surface area. The absolute rate of melting dosn't affect the calculation, the overall time it takes the larger ball to lose half it's volume might change, but the calculation given above by geepamoogle and PBE8 is unaffected by this.
15. 23 Mar '07 15:52
Originally posted by PBE6
Yeah, just double checked on paper. I meant to divide Vsmall by Vsmall0, not just multiply the (4/3)*pi straight through:

Vsmall = (4/3)*pi*rS0^3 * (1-2(1-b))^3
Vsmall0 = (4/3)*pi*rS0^3

Vsmall/Vsmall0 = (1-2(1-b))^3 = 20.3% (approx)

as stated above.
It's a ridiculously fiddly calculation - it took me three goes just to check what you and geepamoogle had done. It'd be nice if we could put equations in tags like you can in LaTeX, so that for example $\sum_0^\infinity x_i^2$ would come out as a nice pretty equation instead of the mess we end up with trying to express all this stuff with ascii.