- #1

- 1

- 0

## Main Question or Discussion Point

I need to find the mass of an aluminum rod without the use of a scale. I have all the dimensions of the rod but I just don't know the right equations etc. Thanks for the help!

- Thread starter SirSpanky0
- Start date

- #1

- 1

- 0

I need to find the mass of an aluminum rod without the use of a scale. I have all the dimensions of the rod but I just don't know the right equations etc. Thanks for the help!

- #2

- 9,705

- 939

Maybe you're stuck finding the volume of a "rod". A "rod" could have a lot of shapes, though the first one that springs to mind is that of a circular cylinder.

I think that if this is a homework problem you'd do better if you made some effort to solve it yourself, first (the "show your work" rule) - also, there is a special forum for homework problems.

- #3

- 2,946

- 0

You can accelerate it with a constant force and then divide the force exerted and the acceleration and that will give you the mass. Easy to say, hard to do for a lab experiment in school.SirSpanky0 said:I need to find the mass of an aluminum rod without the use of a scale. I have all the dimensions of the rod but I just don't know the right equations etc. Thanks for the help!

Pete

- #4

- 52

- 0

Yes, but how are you going to calculate the force with which you accelerate? you may still need a scale.pmb_phy said:You can accelerate it with a constant force and then divide the force exerted and the acceleration and that will give you the mass. Easy to say, hard to do for a lab experiment in school.

Pete

- #5

- 2,946

- 0

It depends on the particular way you choose to accelerate it. If you use two charges then you use Coulombs law. If you use a spring then you use the law for springs (the name evades me at the moment).amt said:Yes, but how are you going to calculate the force with which you accelerate? you may still need a scale.

Pete

- #6

- 782

- 1

Would this be Hooke's Law ([itex]F = kx[/itex])? Sorry if I'm wrong, I just like to think I know what's going on once in a while :tongue:pmb_phy said:[...]If you use a spring then you use the law for springs (the name evades me at the moment).

Last edited:

- Replies
- 2

- Views
- 608

- Last Post

- Replies
- 3

- Views
- 2K

- Replies
- 1

- Views
- 2K

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 21

- Views
- 5K

- Replies
- 7

- Views
- 4K

- Last Post

- Replies
- 80

- Views
- 91K

- Replies
- 5

- Views
- 3K

- Replies
- 19

- Views
- 39K

- Last Post

- Replies
- 15

- Views
- 2K