#### Dhamnekar Winod

##### Junior Member

- Joined
- Aug 14, 2018

- Messages
- 181

How to answer this question?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Dhamnekar Winod
- Start date

- Joined
- Aug 14, 2018

- Messages
- 181

How to answer this question?

D

Please check the given equation for catenary. In books that I have, it is a hyperbolic function.

How to answer this question?

If I were to solve this problem, I would:

Sketch catenary, with the particle at an arbitrary position

Determine the directions of velocity and acc celebration

And continue.

Please share your work/thoughts.

Last edited by a moderator:

- Joined
- Aug 14, 2018

- Messages
- 181

Catenary equation can be derived alternatively using the trigonometric function given in the question.Please check the given equation for catenary. In books that I have, it is a hyperbolic function.

If I were to solve this problem, I would:

Sketch catenary, with the particle at an arbitrary position

Determine the directions of velocity and acc celebration

And continue.

Please share your work/thoughts.

D

OK - so use it.Catenary equation can be derived alternatively using the trigonometric function given in the question.

What is the direction of acceleration of the particle?

- Joined
- Aug 14, 2018

- Messages
- 181

Let \(\displaystyle \vec{r}\) be the position vector of point P along the curve \(\displaystyle c\tan{(\psi)}\). Then \(\displaystyle \frac{d\vec{r}}{d\psi}\) is a unit vector along a tangent andOK - so use it.

What is the direction of acceleration of the particle?

velocity v of the particle at any point of the curve is given by \(\displaystyle \vec{v} =\frac{d\vec{r}}{dt}= \frac{d\vec{r}}{d\psi}\frac{d\psi}{dt}=v \hat{T}\) where \(\displaystyle v=\frac{d\psi}{dt} ,\hat{T}=\frac{d\vec{r}}{d\psi}\),

\(\displaystyle \frac{d\vec{r}}{dt}=\frac{c}{\cos^2{(\psi)}}\)

If all the above work is correct, i shall proceed further.

Last edited:

D

Please post a sketch of the catenary - locating the particle and describing \(\displaystyle \psi , \)Let \(\displaystyle \vec{r}\) be the position vector of point P along the curve \(\displaystyle c\tan{(\psi)}\). Then \(\displaystyle \frac{d\vec{r}}{d\psi}\) is a unit vector along a tangent and

velocity v of the particle at any point of the curve is given by \(\displaystyle \vec{v} =\frac{d\vec{r}}{dt}= \frac{d\vec{r}}{d\psi}\frac{d\psi}{dt}=v \hat{T}\) where \(\displaystyle v=\frac{d\psi}{dt} ,\hat{T}=\frac{d\vec{r}}{d\psi}\),

\(\displaystyle \frac{d\vec{r}}{dt}=\frac{c}{\cos^2{(\psi)}}\)

If all the above work is correct, i shall proceed further.

- Joined
- Nov 12, 2017

- Messages
- 14,457

- Joined
- Aug 14, 2018

- Messages
- 181

Let \(\displaystyle \vec{r}\) be the position vector at point P. \(\displaystyle \vec{r}= x(s)\hat{i} + y(s)\hat{j} + z(s)\hat{k}\)

\(\displaystyle \hat{T}\) is a unit tangent vector = \(\displaystyle \frac{d\vec{r}}{ds} = \frac{dx}{ds}\hat{i} +\frac{dy}{ds}\hat{j} + \frac{dz}{ds}\hat{k}=\frac{1}{a\cdot\tan^2{(\psi)}}\times\left[ \frac{dx}{d\psi}\hat{i} + \frac{dy}{d\psi}\hat{j} + \frac{dz}{d\psi}\hat{k} \right]\)

Velocity \(\displaystyle \vec{v}\) of the particle at any point of the curve is given by \(\displaystyle \vec{v}=\frac{d\vec{r}}{d\psi}= \frac{d\vec{r}}{ds}\cdot\frac{ds}{d\psi}\)

\(\displaystyle \frac{ds}{d\psi}= a\cdot (1+\tan^2{(\psi)})\) which is the velocity but velocity given in the question is \(\displaystyle V_0e^{\psi}\)

I know \(\displaystyle e^{\psi} = \cosh{(\psi)} + \sinh{(\psi)}\)

\(\displaystyle \vec{A}= \frac{dv}{d\psi}\hat{T} + \frac{v^2}{\rho}\frac{d\hat{T}}{d\psi} \) since radius of curvature is \(\displaystyle \rho= \frac{ds}{d\psi}\)

\(\displaystyle \hat{N}\) is a unit normal vector which is \(\displaystyle \frac{d\hat{T}}{d\psi}\)

Now the question states the angle between acceleration vector(\(\displaystyle \vec{A}\)) and \(\displaystyle \hat{T} \) is equal to the angle between \(\displaystyle \vec{A}\) and \(\displaystyle \hat{N}\)

How to answer this question now?

- Joined
- Aug 14, 2018

- Messages
- 181