Process title?

[Gareth56]

Apparently mg + mv^2/r = mg(1 + v^2/rg) but does anyone know what is the name of the process?

[Shadow]

When I threw it into Google most of the titles had the title circular motion....if that helps.

[timo]

"Factoring out" for the process of going from the left-hand side to the right-hand side.

"Distributivity" for the property telling you that you can do that.

You did not really describe a process - going from the left-hand side of the equation to the right-hand side was just a guess of what you could have meant.

[Gareth56]

Thanks for the reply. What puzzles me is why the mg appears outside the bracket? and there's only a 'g' in one of the terms so how can you distribute 'g'?

[Cap'n Refsmmat]

[math]mg + \frac{mv^2}{r} = mg(1 + \frac{v^2}{rg})[/math]

 

Watch what happens when you distribute the last bit:

 

[math]mg + mg \times \frac{v^2}{rg}[/math]

 

The extra g cancels out, leaving:

 

[math]mg + \frac{mv^2}{r}[/math]

[timo]

The other way round is relatively interesting, too: [math] mg + \frac{mv^2}{r} = mg + \frac{mv^2}{r} \frac{g}{g} = mg\left( 1+ \frac{v^2}{rg} \right) [/math] (to be read as successive steps from the left to the right). The first step I did here is writing 1 in a complicated -but useful for the following steps- manner and using that [math]x = 1x[/math]. Multiplying things with a complicated 1 or adding complicated zeros to terms is something you see very often (at least in physics), hence this post highlighting it.