# Please could someone help me with an equation

1. Mar 2, 2005

### faramir12345

Could someone help me?

Edit: Thanks all.

Last edited: Mar 2, 2005
2. Mar 2, 2005

### Gamma

You know you need to arrive at $aw=\sqrt{2}v$. Take a good look at the given equations.

Set Equations 1 = 5.

Now use equation 7 to substitute for R.

At this point many terms will cancel off.

Use $A_s = \pi a^2$ and $J=ma^2/3$

At this point more terms will cancel off.

Rearrange to get $aw=\sqrt{2}v$

3. Mar 2, 2005

### faramir12345

re:

I sort of got that before, but I'm not really a maths based person so am useless at rearranging and stuff with equations. This is part of a bigger piece of work, but I need to show how to get that equation.

Sorry if this is asking too much, but could you show each step of the rearrangement for me? :uhh: Would be very grateful...

4. Mar 2, 2005

### Jayboy

I would say that, if showing how you get to the equation is part of the coursework then it's likely to be a useful tool for your course. (Not to mention that basic algebra isn't a bad thing to know generally!)

It's probably best for you in the long run that you have a go and show us where you get in trouble.

Gamma has provided the framework and, as a more explicit tip, notice that $$(a \omega)$$ appears in one of the equations.

5. Mar 2, 2005

### Gamma

edit: I agree with jayboy. You need to take a paper and pen and write down the equations down and do it yourself first and come back to the following solution.

Equation 1 = 5

$$\frac{mv^2}{R} = \frac{1}{4}\rho (v^2 +a^2w^2)C_LA_s$$ -------(1)

From 7 we have, $$R = \frac{4J}{\rho C_L\pi a^4}$$ -------(2)

Sustitute this in (1)

$$\frac{mv^2}{4J}\rho C_L\pi a^4 = \frac{\rho}{4}(v^2 +a^2w^2)C_LA_s$$

cancelling $$\rho C_L$$ in both sides and substituting $$A_s = \pi a^2$$ you get

$$\frac{mv^2}{J} a^2= (v^2 +a^2w^2)$$

Substitute $$J=\frac{ma^2}{3}$$

cancell ma^2 in the left side and rearrange to get

$$2v^2= a^2w^2$$

$aw=\sqrt{2}v$

Last edited: Mar 2, 2005
6. Mar 2, 2005

### faramir12345

re:

oh right!

I have actually got about 6 sides of A4 working from before asking the question.

I kept ending up with 2Vsquared = (aw)2
I didn't know that you could change (aw)2 to a2w2

which would then go to root2v=aw right?

I thought I kept getting 2Vsquared = (aw)2 because I was making a mistake in the rearrangement (which is why I need to see each step), but all along it was because of not expanding the brackets?

:yuck:

Last edited: Mar 2, 2005
7. Mar 2, 2005

### learningphysics

You almost had it right here. You can immediately take square roots of both sides, to get sqrt(2)V = aw.

You don't need to change (aw)2 to a2w2. That's an unnecessary step.

8. Mar 2, 2005

### faramir12345

Thanks very much everyone.