A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3.66 m/sand a centripetal acceleration of magnitude$\mathbf{1}\mathbf{.}\mathbf{83}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. Position vectorlocates him relative to the rotation axis.(a) What is the magnitude of$\overrightarrow{\mathbf{r}}$?What is the direction of$\overrightarrow{\mathbf{r}}$when is directed (b) due east and (c) due south?

The centripetal acceleration$\overrightarrow{a}$ with magnitude role="math" localid="1660900234051" $1.83\mathrm{m}/{\mathrm{s}}^2$, with linear velocity 3.66 m/s.

Find the magnitude of position vector of object by using its centripetal acceleration and linear constant speed, and it is important that the direction of$\overrightarrow{\mathbf{a}}$is directed towards the center of rotation.

Formulae:

Acceleration$a=\frac{v^2}{r}$

Let’s take the magnitude of acceleration to find the position of object from axis.

The equation of acceleration is

$$\begin{gathered} a=\frac{v^2}{r} \\ r=\frac{v^2}{a} \\ =\frac{{\left(3.66m/s\right)}^2}{1.83m/s^2} \\ =7.32m \end{gathered}$$

Therefore, the magnitude of the radius of the rotation is 7.32 m.

As we know, the acceleration is directed towards the center of the circle. If the acceleration vector is directed towards east, then the center of the circle is at west from object.

As discussed above, the acceleration vector is directed towards the center of the circle. If the object is at south, then the object is at north from the center of circle.