The radius of a merry-go-round is$7m$, and it takes$12\mathrm{s}$ to make a complete revolution.

(a) What is the speed of an atom on the outer rim?

(b) What is the direction of the momentum of this atom?

(c) What is the direction of the rate of change of the momentum of this atom?

The given radius is $r=7\mathrm{m}$and the time is $t=12\mathrm{s}$.

The speed v is the change in the distance d with time t. So, it is given by,

$\mathit{v}\mathbf{=}\frac{\mathbf{d}}{\mathbf{t}}$

(a)

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 the above equation to get the new form,

$v\frac{2\pi r}{t}$

Now plug the values for r and t into the above equation to get the speed of the atom in the outer rim,

$$\begin{gathered} v=\frac{2\pi \left(7\mathrm{m}\right)}{12\mathrm{s}} \\ =3.66\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the atom in the outer rim is $3.66\mathrm{m}/\mathrm{s}$.

(b)

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.

(c)

The net force exerted on the atom equals the rate change of the momentum, so, it is calculated by,

$F_{net}=\frac{d\stackrel{\rightharpoonup }{p}}{dt}$

The atom moves on around in the outer rim, therefore, it moves in a circular loop and to keep this circular loop the force direction must be toward the center of the path.

Hence, the direction of the rate change of the momentum $\frac{d\stackrel{\rightharpoonup }{p}}{dt}$is toward the center of the path.