How Do You Solve This Equation?

[faramir12345]

Could someone help me?

Edit: Thanks all.

 

[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

 

[faramir12345]

Thanks for replying.

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? Would be very grateful...

 

[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.

 

[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

 

[faramir12345]

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?

 

[learningphysics]

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

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.

 

[faramir12345]

Thanks very much everyone.