91Ó°ÊÓ

The velocity $\frac{ds}{dt}.$

The coordinate system primarily utilized in three-dimensional systems is the cylindrical coordinates of the system. The cylindrical coordinate system is used to find the surface area in three-dimensional space.

The unit basis vectors in cylindrical coordinates are given below.

$$\begin{gathered} e_r=i\cos \mathrm{θ}+j\sin \mathrm{θ} \\ {e}_{\mathrm{θ}}=-i\sin \mathrm{θ}+j\cos \mathrm{θ} \\ {e}_z=k \end{gathered}$$

The time derivatives are mentioned below.

$$\begin{gathered} \frac{d{e}_r}{dt}=-\stackrel{˙}{\mathrm{θ}}i\sin \mathrm{θ}+\stackrel{˙}{\mathrm{θ}}j\cos \mathrm{θ} \\ \frac{d{e}_r}{dt}=\stackrel{˙}{\mathrm{θ}}{e}_{\mathrm{θ}} \end{gathered}$$

Find $\frac{d{e}_{\mathrm{θ}}}{dt}$.

$$\begin{gathered} \frac{d{e}_{\mathrm{θ}}}{dt}=-\stackrel{˙}{\mathrm{θ}}i\cos \mathrm{θ}+\stackrel{˙}{\mathrm{θ}j\sin \mathrm{θ}} \\ \frac{d{e}_{\mathrm{θ}}}{dt}=-\stackrel{˙}{\mathrm{θ}}{e}_r \end{gathered}$$

The value role="math" localid="1659270717671" $\frac{d{e}_z}{dt}=0$

Find the value of $\frac{d^{2}s}{d{t}^2}$.

$$\begin{gathered} \frac{d^{2}s}{d{t}^2}=\stackrel{¨}{r}{e}_r+\stackrel{˙}{r}{\stackrel{˙}{e}}_r+\stackrel{˙}{r}\stackrel{˙}{\mathrm{θ}}{e}_{\mathrm{θ}}+r\stackrel{¨}{\mathrm{θ}}{e}_{\mathrm{θ}}+r\stackrel{˙}{\mathrm{θ}}\stackrel{˙}{e_{\mathrm{θ}}}+\stackrel{¨}{z{e}_z} \\ \frac{d^{2}s}{d{t}^2}=\stackrel{¨}{r}{e}_r+\stackrel{˙}{r}\stackrel{˙}{\mathrm{θ}}{e}_{\mathrm{θ}}×2+\stackrel{˙}{r}\stackrel{˙}{\mathrm{θ}}{e}_{\mathrm{θ}}-r{\stackrel{˙}{\mathrm{θ}}}^2{e}_r+\stackrel{¨}{z}{e}_z \\ \frac{d^{2}s}{d{t}^2}=\left(\stackrel{¨}{r}-{\stackrel{˙}{\mathrm{θ}}}^2\right)e_r+\left(2\stackrel{˙}{r}\stackrel{˙}{\mathrm{θ}+r\stackrel{¨}{\mathrm{θ}}}\right)e_{\mathrm{θ}}+\stackrel{¨}{z}{e}_z \end{gathered}$$

Hence, the acceleration is $\left(\stackrel{¨}{r}-{\stackrel{˙}{\mathrm{θ}}}^2\right)e_r+\left(2\stackrel{˙}{r}\stackrel{˙}{\mathrm{θ}+r\stackrel{¨}{\mathrm{θ}}}\right)e_{\mathrm{θ}}+\stackrel{¨}{z}{e}_z$.