Dhamnekar Win,odd Posted September 17 Share Posted September 17 How to answer this question? Any math help, hint or even a correct answer will be accepted. Link to comment Share on other sites More sharing options...

studiot Posted September 17 Share Posted September 17 Why is this not in Homework ? Hints What is s ? What is velocity in terms of s? What is acceleration in terms of velocity ? 1 Link to comment Share on other sites More sharing options...

exchemist Posted September 17 Share Posted September 17 I barely remember anything about catenaries, but I thought they were hyperbolic cosine functions, not tangents. Link to comment Share on other sites More sharing options...

Dhamnekar Win,odd Posted September 17 Author Share Posted September 17 (edited) s is the arc length of a particle P from a fixed point P_{0}(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 September 17 by Dhamnekar Win,odd Link to comment Share on other sites More sharing options...

studiot Posted September 17 Share Posted September 17 56 minutes ago, Dhamnekar Win,odd said: s is the arc length of a particle P from a fixed point P_{0}(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. Link to comment Share on other sites More sharing options...

joigus Posted September 17 Share Posted September 17 1 hour ago, exchemist said: I barely remember anything about catenaries, but I thought they were hyperbolic cosine functions, not tangents. https://en.wikipedia.org/wiki/Whewell_equation Homework. No linear algebra or group theory here. 2 Link to comment Share on other sites More sharing options...

exchemist Posted September 17 Share Posted September 17 17 minutes ago, joigus said: https://en.wikipedia.org/wiki/Whewell_equation Homework. No linear algebra or group theory here. Aha. Thanks very much. Link to comment Share on other sites More sharing options...

Dhamnekar Win,odd Posted September 17 Author Share Posted September 17 (edited) 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 September 17 by Dhamnekar Win,odd Link to comment Share on other sites More sharing options...

joigus Posted September 17 Share Posted September 17 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. Link to comment Share on other sites More sharing options...

studiot Posted September 17 Share Posted September 17 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. 1 Link to comment Share on other sites More sharing options...

Phi for All Posted September 17 Share Posted September 17 ! Moderator Note Moved to Homework Help. Link to comment Share on other sites More sharing options...

Country Boy Posted September 19 Share Posted September 19 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 Link to comment Share on other sites More sharing options...

exchemist Posted September 19 Share Posted September 19 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. Link to comment Share on other sites More sharing options...

Dhamnekar Win,odd Posted September 24 Author Share Posted September 24 Link to comment Share on other sites More sharing options...

Dhamnekar Win,odd Posted October 7 Author Share Posted October 7 (edited) Example: Find the radial and transverse acceleration of a particle moving in a plane curve in Polar coordinates. 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 October 7 by Dhamnekar Win,odd Link to comment Share on other sites More sharing options...

Dhamnekar Win,odd Posted October 9 Author Share Posted October 9 (edited) Now let us replace ψ by y and solve the aforesaid differential equation. 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 October 9 by Dhamnekar Win,odd Link to comment Share on other sites More sharing options...

Dhamnekar Win,odd Posted October 16 Author Share Posted October 16 Now, here is the final correct answer provided to me by one great expert mathematician from UK (United Kingdom). That's it. 1 Link to comment Share on other sites More sharing options...

studiot Posted October 16 Share Posted October 16 Well done +1 Link to comment Share on other sites More sharing options...

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