Question

The radius of a merry-go round is $\mathbf{11}\mathbf{m}$, and it takes $\mathbf{12}\mathit{s}$ to go around one. What is the speed of an atom in the outer rim?

Short answer

The speed of an atom in the outer rim is $5.76\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Given information

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

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

Formula used

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)

The speed is the change in the distance with time

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

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

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