If the radius of a merry-go-round is$\mathbf{5}\mathbf{}\mathbf{m}$, and it takes$\mathbf{14}\mathbf{}\mathbf{s}$to go around once, what is the speed of an atom at the outer rim? What is the direction of the velocity of this atom: toward the center, away from the center, or tangential?

Given that the radius of a merry $r=5\mathrm{m}$ and the time to complete one round is$t=14\mathrm{s}$

Need to determine the speed $\mathit{v}$of an atom in the outer rim.The speed is the change in the distance with time. So, it is given by

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

When the atom travel through one round, it moves above the circumference of the circle. The circumference is given by, $2\pi r$

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

$d=2\pi r$

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

$v=\frac{2\pi r}{t}$ ………………….…… (2)

Now plug the values for $r$and $t$into equation (2) to get the speed of the atom in the outer rim,

$$\begin{gathered} v=\frac{2\pi r}{t} \\ =\frac{2\pi \left(5\mathrm{m}\right)}{14\mathrm{s}} \\ =2.24\mathrm{m}/\mathrm{s} \end{gathered}$$

The motion here is circular, where the atom could repeat the period again. So, the motion or the momentum of the atom is tangential, therefore, the direction of the velocity of this atom is tangential. If it goes away from the center it will not keep the circular motion, the same thing for toward the center.

Therefore, the speed of the atom at outer rim is role="math" localid="1656676376909" $2.24\mathrm{m}/\mathrm{s}$and the direction of the speed of the atom is tangential.