# How to make a lead ball float

JuliusDarius

## Homework Statement

Today In class(1st day of physics) the teacher paired us up and told us to figure out how to make a lead ball float in water. The ball has a radius of 1 meter and is made of lead. To make it float we can hollow it out and fill the empty space with air. For this problem we need to figure out how thick the wall of the sphere needs to be so the lead sphere floats.

## Homework Equations

All I know is that density somehow plays into it.

None so far.

## Answers and Replies

Gold Member
2021 Award
Do you have any idea how to calculate the volume of a sphere?

JuliusDarius
Do you have any idea how to calculate the volume of a sphere?
Yes I do. I calculated the volume of a sphere with a radius of 1 to be about ~4.18879

Gold Member
2021 Award
Yes I do. I calculated the volume of a sphere with a radius of 1 to be about ~4.18879

Well, can you produce a generic equation for the volume of the shell, based on, say, the inner radius?

JuliusDarius
Well, can you produce a generic equation for the volume of the shell, based on, say, the inner radius?
That I'm not sure if I can do.

Antiphon
Volumes add and subtract like areas and weights.

If you hollow out a ball it's like subtracting a small ball from a bigger ball.

Now can you do it?

JuliusDarius
Volumes add and subtract like areas and weights.

If you hollow out a ball it's like subtracting a small ball from a bigger ball.

Now can you do it?
Would it just be V=4/3∏r^3-4/3∏r^3?

Antiphon
Would it just be V=4/3∏r^3-4/3∏r^3?

If the first r is the big ball radius and the second r is the little ball (or the hole) radius then yes.

You should write them r1 and r2 so that it's clearer that they are not the same r.

JuliusDarius
If the first r is the big ball radius and the second r is the little ball (or the hole) radius then yes.

You should write them r1 and r2 so that it's clearer that they are not the same r.
V=4/3∏r1^3 - 4/3∏r2^3. Looks much better. Now what should I do?

Antiphon
V=4/3∏r1^3 - 4/3∏r2^3. Looks much better. Now what should I do?

The basic principle of buoyancy is that the object will float if it weighs less than a blob of water of the same shape.

Q: how much does the lead ball weigh, how much does a water ball weigh, and how can you make them the same using your new formula?

JuliusDarius
The basic principle of buoyancy is that the object will float if it weighs less than a blob of water of the same shape.

Q: how much does the lead ball weigh, how much does a water ball weigh, and how can you make them the same using your new formula?
If mass=density*volume then water has a ball of water 2m in diameter is 4,188.79 kg/m^3? And then a ball of lead 2m in diameter would be 11340 kg/m^3? Now I'm not really sure how to use the equation though.

Samky
You're close to the answer. Antiphon's point was key, how can you make them the same?

Here's an in-between step. Can you write an equation for the mass of your lead ball using your new formula for volume?

Then, as he asked, how can you make sure the mass of your lead ball will not be greater than the mass of the water ball?

Antiphon
The one control you have is to hollow out the lead ball. You should hollow it out enough that the weight matches the water ball.

You now only need to write the formula for the weight of the lead ball minus the hollowed out part and set it equal to the weight of the water ball.

JuliusDarius
The one control you have is to hollow out the lead ball. You should hollow it out enough that the weight matches the water ball.

You now only need to write the formula for the weight of the lead ball minus the hollowed out part and set it equal to the weight of the water ball.
So,11340 kg/m^3-X=4,188.79 kg/m^3?

JuliusDarius
So if I remove 7152 kg/m^3 of lead the balls weight will equal the water's mass. But what about the air in the hole now? And how do I use that to determine how thick the wall needs to be?

azizlwl
It depends on what level of the ball you want it to float.
Apply Archimedes Principles- states that a body immersed in a fluid experiences a buoyant force equal to the weight of the fluid it displaces.

The volume is constant thus reducing the weight(removing inside mass) equal to weight displace is the only way.

Will you find the weight of final ball weight equal to weight of water it displaces?

A lead ball with a molecule thickness will surely floats.

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Samky
Yes, if you remove that much density then the ball would not sink.

As he said your only control is to hollow it out, so the question is how much would you have to hollow out to reduce the density that much?

It's quite likely that they will treat the mass of the air as negligible.

So to solve for your control variable (the smaller radius, r2) you need to write an equation that involves r2 and the desired density drop. I believe you already came up with a portion of it above when you found an expression for the volume.

JuliusDarius
Yes, if you remove that much density then the ball would not sink.

As he said your only control is to hollow it out, so the question is how much would you have to hollow out to reduce the density that much?

It's quite likely that they will treat the mass of the air as negligible.

So to solve for your control variable (the smaller radius, r2) you need to write an equation that involves r2 and the desired density drop. I believe you already came up with a portion of it above when you found an expression for the volume.
I understand what you mean, I'm just totally lost on how to make up the equation. Thanks for the help everyone so far by the way. Sorry for being so bad at math/physics!

Antiphon
Don't worry about the air. It weighs next to nothing but you are correct; the "true" answer requires pumping the air out of the hole or taking out a tiny bit more lead to offset the tiny weight of the air inside.

The equation is not hard if you think about what it's really doing. You want to say the water ball weighs the same as the hollowed-out ball.

(water weight) = (full ball weight) - (removed-ball weight)

JuliusDarius
Don't worry about the air. It weighs next to nothing but you are correct; the "true" answer requires pumping the air out of the hole or taking out a tiny bit more lead to offset the tiny weight of the air inside.

The equation is not hard if you think about what it's really doing. You want to say the water ball weighs the same as the hollowed-out ball.

(water weight) = (full ball weight) - (removed-ball weight)
So, (water weight, 4188)=(Full Pb ball,11340 kg/m^3)-(removed part, 7152 kg/m^3 I think?)

Samky
A quick tip that is a useful habit for nearly all problems, is to keep your work in variable form and only plug in the numbers at the very end when you're ready to solve.

Another great tip I like to use when I feel stuck is to see if I can write my unknown(s) in terms of knowns. Either you'll solve for it right away, or you'll get left with some other unknown you know how to find.

For example in Antiphon's equation you know the water weight. You know the full ball weight. But you don't know the removed ball weight (remember in terms of your variables, mass, volume, density, radius, that stuff).

So see if you can write your unknown: "removed ball weight" in terms of known variables. The answer is you already did. You said weight = density * volume. You known density, now volume is your unknown.

Does it help to write volume in terms of other variables? The only variable in the volume equation is r. But wait, that's your control, so you're all set.

That probably sounds pretty confusing after working on it so long, but it's like a puzzle. If a = b -c and c = de and e = f then you can just substitute and write

a = b - df What's the difference if all we did was substitute? Well if f is the thing your problem asks for, then you're set to solve! Recall it asks for the final thickness of the wall so that the ball will float.

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