A person of mass 70 kgrides on a Ferris wheel whose radius is 4 m . The person's speed is constant at 0.3 m/s . The person's location is shown by a dot in Figure 5.78 .

(a) What is the magnitude of the rate of change of the momentum of the person at the instant shown?

(b) What is the direction of the rate of change of momentum of the person at the instant shown?

(c) What is the magnitude of the net force acting on the person at the instant shown? Draw the net force vector on the diagram at this instant, with the tail of the vector on the person.

A person of mass $m=70\mathrm{kg}$rides on a Ferris wheel whose radius is $r=4\mathrm{m}$. The person's speed is constant at $v=0.3\mathrm{m}/\mathrm{s}$.

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two parts. The parallel rate of change of momentum ${\mathbf{(}\mathbf{d}\overrightarrow{\mathbf{p}}\mathbf{/}\mathbf{d}\mathbf{t}\mathbf{)}}_{\mathbf{\parallel }}$and the perpendicular rate of change of momentum ${\mathbf{(}\mathbf{d}\overrightarrow{\mathbf{p}}\mathbf{/}\mathbf{dt}\mathbf{)}}_{\mathbf{\perp }}$are the two parts that we are concerned with.

Change in momentum is given by,

localid="1668415797574" $$\begin{gathered} \frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{\mathbf{d}\mathbf{t}}}_{\mathbf{\parallel }}\mathbf{+}{\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{\mathbf{d}\mathbf{t}}}_{\mathbf{\perp }} \\ \mathbf{.}\mathbf{.}\mathbf{.}\left(i\right) \end{gathered}$$

The object's speed is affected by the parallel rate of change of momentum, and because the speed is constant, the parallel rate is zero and equal to the rate of change of the magnitude of the momentum.

localid="1668417357550" $$\begin{gathered} \frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{{\mathbf{dt}}_{\mathbf{\parallel }}}\mathbf{=}\frac{\mathbf{d}\mathbf{\left|}\overrightarrow{\mathbf{p}}\mathbf{\right|}}{\mathbf{d}\mathbf{t}}\hat{\mathbf{p}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \end{gathered}$$

The direction shift generated by the perpendicular rate of change is known as the rate change. The quantity of the perpendicular rate change matches the rate change of the momentum at speeds significantly slower than the speed of light. .

localid="1668417363012" $$\begin{gathered} \frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{{\mathbf{dt}}_{\perp }}\mathbf{=}\mathbf{\left|}\overrightarrow{\mathbf{p}}\mathbf{\right|}\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{\mathbf{dt}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}} \\ \mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{ii}\mathbf{\right)} \end{gathered}$$

It will also be equivalent to the rate of change of momentum$\mathit{d}\overrightarrow{\mathbf{p}}\mathbf{/}\mathit{d}\mathit{t}$.

Substitute $m$, $v$, and $R$ values into equation (i) to get $\mathrm{d}\overrightarrow{\mathrm{p}}/\mathrm{dt}$.

$$\begin{gathered} \frac{\mathrm{d}\overrightarrow{\mathrm{p}}}{\mathrm{dt}}=\frac{{\mathrm{mv}}^2}{\mathrm{R}} \\ =\frac{\left(70\mathrm{kg}\right){\left(0.3\mathrm{m}/\mathrm{s}\right)}^2}{4.0\mathrm{m}} \\ =1.57\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The magnitude of the rate of change of the momentum is$1.57\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2$

The motion is circular, with the participant having the option of repeating the period. As a result, the person's motion and momentum are tangential, and the direction of the change in momentum is toward the path's centre. As a result, the force is directed toward the path's centre.

The direction of the rate of change of momentum is toward the centre of the path.

At speeds significantly slower than the speed of light, the net force exerted on the person equals the rate change of momentum, and the magnitude of the perpendicular rate change is given by

$$\begin{gathered} F_{net}=\frac{d\overrightarrow{p}}{dt} \\ =1.57\mathrm{N} \end{gathered}$$

The magnitude of the net force is 1.57 N