Please could someone help me with an equation

  • #1
Could someone help me?

Edit: Thanks all.
 
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Answers and Replies

  • #2
357
11
You know you need to arrive at [itex]aw=\sqrt{2}v[/itex]. 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 [itex]A_s = \pi a^2 [/itex] and [itex]J=ma^2/3[/itex]

At this point more terms will cancel off.

Rearrange to get [itex]aw=\sqrt{2}v[/itex]
 
  • #3


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? :uhh: Would be very grateful...
 
  • #4
21
0
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 [tex] (a \omega) [/tex] appears in one of the equations.
 
  • #5
357
11
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

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

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

Sustitute this in (1)

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

cancelling [tex]\rho C_L[/tex] in both sides and substituting [tex]A_s = \pi a^2[/tex] you get

[tex]\frac{mv^2}{J} a^2= (v^2 +a^2w^2)[/tex]

Substitute [tex]J=\frac{ma^2}{3}[/tex]

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

[tex]2v^2= a^2w^2[/tex]


[itex]aw=\sqrt{2}v[/itex]
 
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  • #6


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:
 
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  • #7
learningphysics
Homework Helper
4,099
5
faramir12345 said:
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.
 
  • #8
:approve: Thanks very much everyone.
 

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