#### Dhamnekar Winod

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How to answer this question?

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- Thread starter Dhamnekar Winod
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How to answer this question?

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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:

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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.

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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?

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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:

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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.

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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?

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