- #1
V0ODO0CH1LD
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I am actually going to post the problem that sparked the question I am about to ask, but I don't need help with the answer and that's why I didn't start this thread in the homework and coursework section. Actually, I just realized I read the problem statement in the wrong way, which makes my question pointless with regards to the problem. Anyway, I am still wondering about it independently of weather it helps with the problem or not. So here it it:
The problem stated that an acceleration has magnitude \mu(r+\frac{a^3}{r^2}), where a is the initial displacement and r is the distance from the origin. What the problem asked was to check what were the dimensions of \mu. When you read it right it's an easy problem, right?
But in the way I had originally read it, it stated: an acceleration has magnitude \mu(\frac{r + a^3}{r^2}). Which makes the thing inside the parenthesis look like \mu(\frac{1}{r}+\frac{a^3}{r^2}).
Well, \frac{a^3}{r^2} would just be something like \frac{(am)^3}{(rm)^2}, where m is some unit of distance and that would simplify to \frac{a^3}{r^2}m. Which is fine.
But what would \frac{1}{rm} mean? Like, a dimensionless something per unit of distance? How should I think about \frac{1}{r}m^{-1}? Or does it not even exist on the account that I read the problem wrong? At first I was like: well; I have something like 1/r inverse meters, so that must mean I have r regular meters.. But that makes no sense. Is there a correct way to view this?
The problem stated that an acceleration has magnitude \mu(r+\frac{a^3}{r^2}), where a is the initial displacement and r is the distance from the origin. What the problem asked was to check what were the dimensions of \mu. When you read it right it's an easy problem, right?
But in the way I had originally read it, it stated: an acceleration has magnitude \mu(\frac{r + a^3}{r^2}). Which makes the thing inside the parenthesis look like \mu(\frac{1}{r}+\frac{a^3}{r^2}).
Well, \frac{a^3}{r^2} would just be something like \frac{(am)^3}{(rm)^2}, where m is some unit of distance and that would simplify to \frac{a^3}{r^2}m. Which is fine.
But what would \frac{1}{rm} mean? Like, a dimensionless something per unit of distance? How should I think about \frac{1}{r}m^{-1}? Or does it not even exist on the account that I read the problem wrong? At first I was like: well; I have something like 1/r inverse meters, so that must mean I have r regular meters.. But that makes no sense. Is there a correct way to view this?
