Algebraic operation

1. Aug 12, 2015

1. The problem statement, all variables and given/known data

In a rotation of rigid bodies-problem there's a question regarding the speed of a falling block, attached to a solid cylinder which can rotate frictionless around its axis. I have a question regarding the algebraic operations.

2. Relevant equations

So I have this expression: $mgh=1/2mv^2+1/2(1/2MR^2)(v^2/R^2)$.

3. The attempt at a solution

I can operate on the expression to solve for v. So I eventually get that $2mgh=v^2(m+1/2M)$. And from here I just want to divide both sides by $(m+1/2M)$ and take the square root of both sides.

Which would leave me with $v=sqrt((2mgh)/(m+(1/2)M)$

My text book expresses this result as $v=sqrt((2gh)/(1+(M/2m))$

Are those the same results? If so, what operations were made to end up with that result? If not, what did I do wrong?

2. Aug 12, 2015

SteamKing

Staff Emeritus
Divide both sides of $2mgh=v^2(m+1/2M)$ by m and then solve for v.

You should be careful to distinguish (1/2)M from 1/(2M) though.

3. Aug 12, 2015

Thanks. I get that if I divide $(m+1/2M)$ by $m$ I get $(1+M/2m)$. But doesn't dividing the whole right-hand side with $m$ affect $v^2$ and I'll end up with $v^2/m$? I hope you can get what's confusing me. :)

4. Aug 12, 2015

SteamKing

Staff Emeritus
Why should dividing by m affect v2? What you are doing is dividing $(m+1/2M)$ by m.

It's like saying m * y = (1 + k) * x2. If you divide both sides of the equation by the same quantity, what's left must be equal. That's not really algebra even; it's arithmetic.

m * y = (1 + k) * x2 will be the same as y = [(1 + k)/m] * x2, as long as m ≠ 0.

The value of x is not going to be affected by dividing by m.

5. Aug 13, 2015

RUber

Test the relationship:
$\sqrt{ \frac{2mgh}{m+M/2} } = \sqrt{ \frac{(m)2gh}{(m)(1+M/(2m)} } = \sqrt{ \frac{2gh}{(1+M/(2m)} }$
They look the same to me.