Finding Radius of Aluminum Sphere

[chocolatelover]

[SOLVED] sphere problem

Homework Statement

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.

Homework Equations

volume of a sphere=4/3pir^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?

Thank you in advance

 

[Kurdt]

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.

 

[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

 

[Kurdt]

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

 

[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

 

[Kurdt]

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.

 

[chocolatelover]

Thank you very much

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.

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

Thank you

 

[Kurdt]

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 $r_1$.

 

[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

 

[Kurdt]

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

 

[chocolatelover]

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

Thank you

 

[Kurdt]

Do I need to do dimensional analysis?

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.

 

[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

 

[Kurdt]

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

 

[chocolatelover]

Thank you very much

Regards