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Topic: Vector caculus with applications(catenary)  (Read 427 times)

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Win,odd Dhamnekar

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Vector caculus with applications(catenary)
« on: October 01, 2021, 11:12:48 AM »
This is the math(other science) question.

A particle moves along a catenary s = c tan ψ. The direction of acceleration at any point makes equal angles with the tangent and normal to the path at that point. If the speed at vertex (ψ = 0) be 𝑣0, show that the magnitude of velocity and acceleration at any point are given by 𝑣0eψ and $\frac{\sqrt{2}}{c}$𝑣02e cos2 ψ respectively.

My answer is velocity $\|\frac{ds}{d\psi}\|= \|c(\tan^2(\psi)+1)\|$ and acceleration is |2c tan(ψ)(tan2+1)|

But in the question velocity and acceleration are different. How is that? Where am i wrong?
« Last Edit: October 01, 2021, 11:49:19 AM by Win,odd Dhamnekar »
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mjc123

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Re: Vector caculus with applications(catenary)
« Reply #1 on: October 02, 2021, 11:00:11 AM »
How many times have we been here before? We can't tell how you went wrong unless you show us your working.

Win,odd Dhamnekar

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Re: Vector caculus with applications(catenary)
« Reply #2 on: October 03, 2021, 02:10:14 AM »
You may go through with my following workings:

Any science consists of the following process.
1) See 2) Hear 3) Smell if needed 4) Taste if needed
5) Think 6) Understand 7) Inference take decision [Believe or disbelieve, useful or useless, healthy or unhealthy, cause or effect, favorable or unfavorable, practical or theoretical, practically possible or practically impossible, true or false or  any other required criteria]

mjc123

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Re: Vector caculus with applications(catenary)
« Reply #3 on: October 04, 2021, 04:41:40 PM »
I haven't tried to work this out in detail, but I have a few concerns about what's been written.
1. The equation s = c tanψ doesn't seem to define a unique curve in terms of y vs. x. What are x and y in terms of ψ? I don't know if this matters.
2. Saying velocity is ds/dψ implies dψ/dt is constant. Is this the case?
3. You appear to be deriving the magnitude of the acceleration as the derivative of the magnitude of the velocity. But this is not valid. For example, for circular motion at constant speed, the magnitude of the velocity is unchanging, but the magnitude of the acceleration is non-zero.

Win,odd Dhamnekar

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Re: Vector caculus with applications(catenary)
« Reply #4 on: October 16, 2021, 01:00:39 AM »
Here is the answer to this question provided to me by one great expert mathematician from UK(United Kingdom).

That's it.
Any science consists of the following process.
1) See 2) Hear 3) Smell if needed 4) Taste if needed
5) Think 6) Understand 7) Inference take decision [Believe or disbelieve, useful or useless, healthy or unhealthy, cause or effect, favorable or unfavorable, practical or theoretical, practically possible or practically impossible, true or false or  any other required criteria]