Question 11.151: Determine the radius of curvature of the path described by t...

We have

\mathbf{v}=\frac{d \mathbf{r}}{d t}=  R\left(\cos \omega_{n} t-\omega_{n} t \sin \omega_{n} t\right) \mathbf{i}+c \mathbf{j}+R\left(\sin \omega_{n} t+\omega_{n} t \cos \omega_{n} t\right) \mathbf{k}

and

or

Now

Then \quad\nu=\left[R^{2}\left(1+\omega_{n}^{2} t^{2}\right)+c^{2}\right]^{1 / 2}

and \quad\frac{d \nu}{d t}=\frac{R^{2} \omega_{n}^{2} t}{\left[R^{2}\left(1+\omega_{n}^{2} t^{2}\right)+c^{2}\right]^{1 / 2}}

Now \quad a^{2}=a_{t}^{2}+a_{n}^{2}=\left\lgroup\frac{d \nu}{d t}\right\rgroup^{2}+\left\lgroup\frac{\nu^{2}}{\rho}\right\rgroup^{2}

At t = 0:

\begin{aligned}\frac{d \nu}{d t} & =0 \\\mathbf{a} & =\omega_{n} R(2 \mathbf{k}) \quad \text { or } \quad a=2 \omega_{n} R \\\nu^{2} & =R^{2}+c^{2}\end{aligned}

Then, with \quad \frac{d \nu}{d t}=0

we have \quad a=\frac{\nu^{2}}{\rho}

or 2\omega_{n}R={\frac{R^2+c^2}{\rho}}\quad\quad\quad\quad\quad\quad\quad\quad\rho=\frac{R^{2}+c^{2}}{2 \omega_{n} R}\blacktriangleleft