Wave paddle application Integration problem

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SUMMARY

The discussion focuses on the integration challenges encountered in an engineering design application involving wave paddle dynamics. Key equations derived include the force equation Fo = -(pgwH^3)/3 and the angular displacement θ = βsin(ωt). Participants emphasize the importance of correctly identifying constants such as 'p', 'g', and 'w', and suggest simplifying the integration process by addressing trigonometric components. The conversation highlights the necessity of verifying signs in equations to ensure accurate results.

PREREQUISITES
  • Understanding of basic calculus and integration techniques
  • Familiarity with engineering mechanics concepts
  • Knowledge of wave dynamics and paddle systems
  • Proficiency in handling trigonometric functions and their derivatives
NEXT STEPS
  • Study advanced integration techniques in engineering applications
  • Explore wave dynamics in fluid mechanics
  • Learn about the role of constants in physical equations
  • Investigate the simplification of trigonometric expressions in calculus
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Engineering students, mechanical engineers, and anyone involved in the design and analysis of wave paddle systems will benefit from this discussion.

gl0ck
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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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gl0ck said:

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.
gif.gif
 
So, I think I've figured out 1a) ,b) ,c)
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
 
gl0ck said:
1b)
Fo=-(pgwH^3)/3
I don't think the coefficient is 1/3. Please post your working.
1c)
θ=βsin(ωt)
dθ/dt=ωβcos(ωt)
d^2θ/dt=ω^2βsin(ωt)
Check your signs.
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..
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?
 

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