91Ó°ÊÓ

Radius of merry go round is $R=11m$

Time to complete one round is $t=12s$.

We are asked to determine the speed of an atom in the outer rim. The speed is the change in the distance with time. So, it is given by

$\mathbf{v}\mathbf{=}\frac{\mathbf{d}}{\mathbf{t}}$ …… (1)

When the atom travel through one round, it moves above the circumference of the circle. The circumference is given by,

$2\mathrm{Ï€}R$

Where, $R$is the radius of the circle. So, the distance where the atom travel is,

$d=2\mathrm{Ï€}R$

Plug the expression for $d$into equation (1) to get the new form,

$V=\frac{2\mathrm{π}R}{t}$ …… (2)

Now plug our values for $\mathrm{R}\mathrm{and}\mathrm{t}$ into equation (2) to get the speed of the atom in the outer rim,

$$\begin{gathered} v=\frac{2\mathrm{Ï€}R}{t} \\ =\frac{2\mathrm{Ï€}\left(11m\right)}{12s} \\ =5.76m/s \end{gathered}$$

Therefore, the speed of an atom in the outer rim is $5.76m/s$.