Vector calculus with applications

Question

ANSWER: How can I use the following example and its solution to answer the above catenary problem? What is $V_0?$ Is it 0 or c ?

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

Answer 1

The velocity is $v=\frac{ds}{dt}$ and is directed along the curve.

The acceleration components are $\frac{d^2s}{dt^2}$ along the curve and $\frac{v^2}{\rho}$ in the direction of the normal, where $\rho$, the radius of curvature is given by $\rho=\frac{ds}{d\psi}$.

The question is effectively saying that these two acceleration components are equal.

So $$s=c\tan\psi\implies \frac{ds}{dt}=c\sec^2\psi\frac{d\psi}{dt}$$

At $\psi=0$, the initial velocity is $v_0$, so $$v_0=c\frac{d\psi}{dt}\implies\frac{d\psi}{dt}=\frac{v_0}{c}$$

The tangential component of acceleration is $$\frac{d^2s}{dt^2}=2c\sec^2\psi\tan\psi\left(\frac{d\psi}{dt}\right)^2+c\sec^2\psi\frac{d^2\psi}{dt^2}$$

The normal component of acceleration is $$\frac{v^2}{\rho}=\frac{c^2\sec^4\psi\left(\frac{d\psi}{dt}\right)^2}{c\sec^2\psi}=c\sec^2\psi\left(\frac{d\psi}{dt}\right)^2$$

These components are equal, so setting them equal, cancelling non-zero terms and rearranging gives

$$(2\tan\psi-1)\left(\frac{d\psi}{dt}\right)^2+\frac{d^2\psi}{dt^2}=0$$

Now write $\frac{d\psi}{dt}=\dot{\psi}$, so that $$\frac{d^2\psi}{dt^2}=\dot{\psi}\frac{d\dot{\psi}}{d\psi}$$

Substituting this into the equation above,

$$(2\tan\psi-1)\left(\frac{d\psi}{dt}\right)^2+\dot{\psi}\frac{d\dot{\psi}}{d\psi}=0$$

We can cancel $\dot{\psi}$ since this is not zero: $$(2\tan\psi-1)\dot{\psi}+\frac{d\dot{\psi}}{d\psi}=0$$

This is a separable variable differential equation:

$$\int_{\frac{V_0}{c}}^{\dot{\psi}}\frac{1}{\dot{\psi}}d\dot{\psi}=\int_0^{\psi}(1-2\tan\psi)d\psi$$ $$\implies\ln\left(\frac{\dot{\psi}c}{V_0}\right)=\psi-2\ln\sec\psi$$

$$\implies\frac{\dot{\psi}c}{V_0}=e^{\psi}\cos^2\psi$$

So the velocity is given by

$$\frac{ds}{dt}=c\sec^2\psi\dot{\psi}=V_0e^{\psi}$$ as required.

The radial component of acceleration is now found to be $$c\sec^2\psi\left(\frac{d\psi}{dt}\right)^2=\frac{{V_0}^2}{c}e^{2\psi}\cos^2\psi$$

The tangential component is the same, therefore the magnitude of the resultant acceleration is $$\frac{\sqrt{2}{V_0}^2}{c}e^{2\psi}\cos^2\psi$$