I know the following involves common factors but I just cannot see how the 1appears in the second equation could some kind soul add an explanation?

 

= mg + m v^2/r .................(1)

 

???????????

 

= mg (1 + v^2/rg) ................(2)

 

It's what happens between (1) & (2) that's bugging me. Maths was so long ago!!!

 

Thanks

Looks like you are dividing through by mg!

 

So - put the whole lot in brackets, then take the mg outside. Leaving the 1 on the LHS. (because 1 x mg is mg)

 

 

To check it out - take whats in the brackets and multiply by mg!

You are left with eq.1. mg x 1 = mg and mg x v^2/rg = mv^2/r

Are you in any doubt that it is true?

 

Just expanding (2) gives (1).

 

It is easiest to go from (1) to (2) in a few steps.

 

I) take out the common factor of m,

 

[math]m (g + \frac{v^{2}}{r})[/math]

 

II) now take out a factor of [math]g[/math], but gang on, there is no factor g in the second part! So, put one in

 

[math]m(g +\frac{v^{2}}{r} \frac{g}{g})[/math]

 

as [math]g/g[/math] = 1

 

III) Now take out the factor to get

 

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

You guys are brilliant. It's so simple if you can "see" these things. 30+ years have dulled my eyesight somewhat

 

Although I'm not too sure how you can just put things in like g/g, clearly it's allowed but why?

 

Also when you say "take out the factor" do you mean take the factor outside the brackets?

 

Thanks again

Edited September 4, 200817 yr by Gareth56

Yes indeed I just mean take factor outside the bracket.

 

Provided we are dealing with things that a notion of divide ( or more generally inverses) exist then we can always have [math] a/a = a a^{-1} = a^{-1} a = 1 [/math], (identity) for "invertible elements".

 

As [math]b= 1b = b1 = 1b1[/math] etc we can insert such a factor [math]a/a[/math] where ever we like. The trick is knowing where to stick it!

I see. I shall always look to where I can stick my "invertible elements", hopefully I won't get arrested!

 

Thanks again ajb and give my regards to Manchester where I was born and educated (in a place on Oxford Road!)

I see. I shall always look to where I can stick my "invertible elements", hopefully I won't get arrested!

 

 

Quite - as ajb said g/g or a/a are equal to 1. So you can multiply anything by 1 anytime you like without changing the equation.

Edited September 4, 200817 yr by DrP
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