# Sphere problem

1. Feb 4, 2008

### chocolatelover

[SOLVED] sphere problem

1. The problem statement, all variables and given/known data

One cubic meter of aluminum has a mass of 2.70 x 10^3 kg, and the same volume of iron has a mass of 7.86X 10^3 kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 1.80 cm on an equal-arm balance.

2. Relevant equations

volume of a sphere=4/3pir^3

3. The attempt at a solution

I would first need to convert everything into cm and then set 4/3pir^3=4/3pi1.80^3 and solve for r, right? I don't need to do anything with the mass, right?

2. Feb 5, 2008

### Kurdt

Staff Emeritus
You can work out the densities of each substance. Once you do this you can work out how much mass is in the iron sphere and the volume of an aluminium sphere that would be equal.

3. Feb 5, 2008

### chocolatelover

Thank you very much

I know that the formula for finding density is m/v

So, I got 2.70X10^3kg/m^3 and 7.86X10^3kg/m^3 Is that correct so far?

I don't understand the second part.

Woud I set 7.86X10^3= something

Don't I need the 1.80 cm for something?

Thank you

Last edited: Feb 5, 2008
4. Feb 5, 2008

### Kurdt

Staff Emeritus
Yes, you need to work out the volume of the iron sphere to work out what its mass is.

5. Feb 5, 2008

### chocolatelover

Well, I know that the formula for finding the volume of a sphere is 4/3pir^3 and the volume of both is 1.00 m^3 (as givin) so wouldn't I just solve for r?

4/3pir^3=4/3pi1.80
r=1.342

Thank you

6. Feb 5, 2008

### Kurdt

Staff Emeritus
The masses of those meter cubed volumes are only given so you can work out the density. A sphere of radius 1.8 cm doesn't have a volume of 1 meter cubed.

If you do it algebraically obviously you won't have to work out the volume and then the mass, you can just manipulate the variables and plug the numbers in at the end.

Last edited: Feb 5, 2008
7. Feb 5, 2008

### chocolatelover

Thank you very much

Could you please show me what you mean? How would you do it algebraically?

Thank you

8. Feb 5, 2008

### Kurdt

Staff Emeritus
You know the aluminium sphere must be the same mass as the iron sphere to balance. Therefore:

$$\rho_{al}\frac{4}{3} \pi r_1^3 = \rho_{fe}\frac{4}{3} \pi r_2^3$$

then one would solve for $r_1$.

9. Feb 5, 2008

### chocolatelover

Thank you very much

What is the pal and pfe? Don't I need to know what r2 is in order to solve for r1?

Thank you

10. Feb 5, 2008

### Kurdt

Staff Emeritus
$\rho_{al}$ and $\rho_{fe}$ are the densities of the aluminium and the iron respectively. r2 is just 1.8 cm for the iron sphere.

11. Feb 5, 2008

### chocolatelover

Do I need to do dimentional analysis or would 1.8 be the final answer?

Thank you

12. Feb 5, 2008

### Kurdt

Staff Emeritus
No. I think you're trying to make this a lot harder than it is. you have all the variables but the one you want to find. All you have to do is rearrange the equation and put the numbers in.

13. Feb 5, 2008

### chocolatelover

So, I just need to solve for r?

2.70X10^3kg(4/3pi)r^3=7.86(4/3)1.8^3

r=2.57016
r=2.57

14. Feb 5, 2008

### Kurdt

Staff Emeritus
Looks good to me. Its a lot easier if you rearrange the symbols then put the numbers in.

15. Feb 5, 2008

### chocolatelover

Thank you very much

Regards