- #1

- 85

- 0

## Homework Statement

This is an engineering design application, but It contains integration. Sorry if I didnt post it in the right folder.

I think a)'s answer should be (pgw)^2*h^2

h=r

To be honest no Idea what is going on..

Thanks

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- Thread starter gl0ck
- Start date

- #1

- 85

- 0

This is an engineering design application, but It contains integration. Sorry if I didnt post it in the right folder.

I think a)'s answer should be (pgw)^2*h^2

h=r

To be honest no Idea what is going on..

Thanks

- #2

- 20

- 0

## Homework Statement

This is an engineering design application, but It contains integration. Sorry if I didnt post it in the right folder.

I think a)'s answer should be (pgw)^2*h^2

h=r

To be honest no Idea what is going on..

Thanks

What do the values 'p', 'g' and 'w' represent?

If they are just constants, then remove them from the integrand.

Eg.

- #3

- 85

- 0

1a)

(pgwH^2)/2

1b)

Fo=-(pgwH^3)/3

1c)

θ=βsin(ωt)

dθ/dt=ωβcos(ωt)

d^2θ/dt=ω^2βsin(ωt)

Q2 gets something like:

T=-(pgwH^3)/3+Bωβcos(-tan^(-1)(Aω/B))-Aω^2βsin(tan^(-1)(Aω/B)

which seems a bit complicated to be integrated, because we have to find first its derivetive..

Thanks

- #4

- 36,952

- 7,199

I don't think the coefficient is 1/3. Please post your working.1b)

Fo=-(pgwH^3)/3

Check your signs.1c)

θ=βsin(ωt)

dθ/dt=ωβcos(ωt)

d^2θ/dt=ω^2βsin(ωt)

You can simplify cos(arctan(x)) so as not to involve any trig. I think you have a sign wrong, propagated through from 1c. What makes you think you need to integrate this?T=-(pgwH^3)/3+Bωβcos(-tan^(-1)(Aω/B))-Aω^2βsin(tan^(-1)(Aω/B)

which seems a bit complicated to be integrated, because we have to find first its derivetive..

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