Scaling of units for equations of motions

[TammyTsang]

Homework Statement

I am currently doing my final year project and I have been trying to scale my variables in my equations of motions, however I'm not sure what it means when the time is scaled also

Relevant Equations

V'=a-gsin(gamma)-V^2/l

In the picture, there is a problem where the t is in units of square root(l/g), and V in square root(gl)

I am wondering
1. What it means when time is in units other than time? Does it mean that when solving I have to take time/squareroot(l/g)
2. How did they get square root(l/g).

Thank you so much

 

[BvU]

Hello Tammy Tsang, !

1. What it means when time is in units other than time? Does it mean that when solving I have to take time/squareroot(l/g)
2. How did they get square root(l/g).

What is described is a simple variable transformation to make the math look simpler. Time is still expressed as time, just not in seconds, or in minutes, or in hours but in a unit $\sqrt{l/g}$ which you may remember from the simple mathematical pendulum (that answers 2.). Note that $\sqrt{l/g}$ has the dimension of time.

So if we write $$t' = t\Bigg/\sqrt{l/g}$$ we get a dimensionless measure t' for the time

I use the ' apostrophe to indicate a different variable, not to indicate differentiation wrt time -- for that I use the dot notation, e.g. $\dot x \equiv {dx\over dt}$. ​

Similarly $$V' = {V\over \sqrt {gl}} $$ so that $${dV'\over dt'} = {dV \over dt} \; { \sqrt {l/g} \over \sqrt{gl} } = {dV \over dt} \Bigg/ g$$

Meaning that, if we divide left and right of the first equation by $g$, what we get is $$
\dot {V'} = {a\over g} - \sin\gamma- {V^2\over gl} = a'- \sin\gamma- V'^{\, 2} $$