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Velocity and acceeration [ Vector calculus with applications ]


Dhamnekar Win,odd
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is the arc length of a particle P from a fixed point P0(S=0) on curve C.

Velocity is [math] \vec{v}(t)= \frac{d\vec{r}}{ds}\times \frac{ds}{dt}[/math]  where r⃗ (t) is the position of particle P at time t.

 Acceleration is [math] \vec{a}(t) = \frac{d\vec{v}}{dt}= \frac{d^2\vec{r}}{dt^2}[/math]

Edited by Dhamnekar Win,odd
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56 minutes ago, Dhamnekar Win,odd said:

is the arc length of a particle P from a fixed point P0(S=0) on curve C.

Velocity is v⃗ (t)=dr⃗ ds×dsdt   where r⃗ (t) is the position of particle P at time t.

 Acceleration is a⃗ (t)=dv⃗ dt=d2r⃗ dt2

 

So is there a problem making the substitution ?

You have nearly done the question.

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1 hour ago, studiot said:

 

So is there a problem making the substitution ?

You have nearly done the question.

[math] s= c \cdot \tan{(\psi)}, \frac{ds}{dt} = c\cdot (\tan^2{(\psi)} + 1)[/math]. Now how to compute [math] \frac{d\vec{r}}{ds}[/math] to find velocity?      

Edited by Dhamnekar Win,odd
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1 hour ago, exchemist said:

Aha. Thanks very much.

You're welcome. I didn't know that form either. I did remember that the equation of an ellipse in polar coordinates is an ungodly mess if you try to express it in polar coordinates with the foci equidistant from the centre, but looks nice and simple with one focus sitting at the origin.

I just assumed something similar happens for the catenary. The rest was a wikipedestrian approach. ;)

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3 hours ago, exchemist said:

I barely remember anything about catenaries, but I thought they were hyperbolic cosine functions, not tangents. 

There are several ways to handle catenaries, but I often find that splitting the vertical axis into two with one section constant and the other parallel to the horizontala axis, as in the following.

caten1.thumb.jpg.495bfa7e5a66643ef7bb6e5405e2d0a6.jpgcaten2.thumb.jpg.b938fada3618db63aa67e37555f68a56.jpg

 

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On 9/17/2021 at 3:06 AM, exchemist said:

I barely remember anything about catenaries, but I thought they were hyperbolic cosine functions, not tangents. 

y as a function of x, yes, [tex]y= a cosh\left(\frac{x}{a}\right)[/tex].   But this is s, the arclength, as a function of [tex]\phi[/tex], then 

y as a function of x, yes- y= a cosh(x/a).  But this is s, the arclength, as a function of phi, the angle the graph makes with the x-axis.   It is the "Wethwell equation"- Catenary - Wikipedia

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1 hour ago, Country Boy said:

y as a function of x, yes, [tex]y= a cosh\left(\frac{x}{a}\right)[/tex].   But this is s, the arclength, as a function of [tex]\phi[/tex], then 

y as a function of x, yes- y= a cosh(x/a).  But this is s, the arclength, as a function of phi, the angle the graph makes with the x-axis.   It is the "Wethwell equation"- Catenary - Wikipedia

Yes, thanks, that's what @joigus told me.

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  • 2 weeks later...
Posted (edited)

Example: Find the radial and transverse acceleration of a particle moving in a plane curve in Polar coordinates.

image.png.40c0bfc7a5ab078597997a6aabf5e963.png

 

How can we use this example to solve our original catenary problem?

 

We know [math] \tan{\psi}=\frac{s}{a}, s=c\cdot \sinh{\frac{X}{c}},  V_0 = c , e^\psi= \cosh{\psi} + \sinh{{\psi}}[/math]

 I am able to get the magnitude of velocity [math]V = c\cdot e^{\psi=0}=c,V = V_0\cdot e^{\psi}[/math] but i am not getting magnitude of acceleration [math]\|\vec{a}\|= \frac{\sqrt{2}}{c} \cdot c^2 \cdot e^{2\psi} \cdot \cos^2{\psi}[/math] 

 

Edited by Dhamnekar Win,odd
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Posted (edited)

image.png.80c4b71d707fbed2ba693e42377a2106.png

 

image.png.68d25e35b5aabd45be5a80d547d81d35.png

 

Now let us replace ψ by y and solve the aforesaid differential equation.

image.png.50f1ab0e90f9938557990768437fe237.png

image.png.d54b88bf043e1ca07a31dfee353fc46b.png

 

image.png.e5ee8b27e12355eeacc75b949a617dd5.png

image.png.edbf16f7c85c44a9d82b1f4d10d72bda.png

Now we replace y by [math]\psi(t)[/math], we get,

[math]\psi(t)= C_2 - \displaystyle\int {\frac{1}{C_1 + t -2 \int {\psi(t)}dt}dt}[/math]


Now , how to show that velocity and acceleration of a moving particle along a catenary are equal to the given values in the question?

Edited by Dhamnekar Win,odd
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