- #1

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Could someone help me?

Edit: Thanks all.

Edit: Thanks all.

Last edited:

- Thread starter faramir12345
- Start date

- #1

- 4

- 0

Could someone help me?

Edit: Thanks all.

Edit: Thanks all.

Last edited:

- #2

- 357

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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

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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

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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

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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]

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]

Last edited:

- #6

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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:

- #7

learningphysics

Homework Helper

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You almost had it right here. You can immediately take square roots of both sides, to get sqrt(2)V = aw.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 don't need to change (aw)2 to a2w2. That's an unnecessary step.

- #8

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Thanks very much everyone.

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