# Velocity and acceeration [ Vector calculus with applications ]

## Recommended Posts

How to answer this question?  Any math help, hint or even a correct answer will be accepted.

##### Share on other sites

Why is this not in Homework ?

Hints

What is s ?

What is velocity in terms of s?

What is acceleration in terms of velocity ?

##### Share on other sites

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

##### Share on other sites

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

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

Acceleration is $\vec{a}(t) = \frac{d\vec{v}}{dt}= \frac{d^2\vec{r}}{dt^2}$

Edited by Dhamnekar Win,odd
##### Share on other sites

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.

##### Share on other sites

1 hour ago, exchemist said:

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

Homework. No linear algebra or group theory here.

##### Share on other sites

17 minutes ago, joigus said:

Homework. No linear algebra or group theory here.

Aha. Thanks very much.

##### Share on other sites

1 hour ago, studiot said:

So is there a problem making the substitution ?

You have nearly done the question.

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

Edited by Dhamnekar Win,odd
##### Share on other sites

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.

##### Share on other sites

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.

##### Share on other sites

!

Moderator Note

Moved to Homework Help.

##### Share on other sites

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, $$y= a cosh\left(\frac{x}{a}\right)$$.   But this is s, the arclength, as a function of $$\phi$$, 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

##### Share on other sites

1 hour ago, Country Boy said:

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

##### Share on other sites

• 2 weeks later...
Posted (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 $\tan{\psi}=\frac{s}{a}, s=c\cdot \sinh{\frac{X}{c}}, V_0 = c , e^\psi= \cosh{\psi} + \sinh{{\psi}}$

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

Edited by Dhamnekar Win,odd
##### Share on other sites

Posted (edited)

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

Now we replace y by $\psi(t)$, we get,

$\psi(t)= C_2 - \displaystyle\int {\frac{1}{C_1 + t -2 \int {\psi(t)}dt}dt}$

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
##### Share on other sites

Now, here is the final correct answer provided to me by one great expert mathematician from UK (United Kingdom).

That's it.

Well done +1

## Create an account

Register a new account