An engineer whose mass is 70kgholds onto the outer rim of a rotating space station whose radius is 14mand which takes 30sto make one complete rotation. What is the magnitude of the force the engineer has to exert in order to hold on? What is the magnitude of the net force acting on the engineer?

An engineer whose mass is $m=70\mathrm{kg}$holds onto the outer rim of a rotating space station whose radius is $R=14\mathrm{m}$and which takes $t=30s$ to make one complete rotation.

A force is a push or pull on an object is because of the interaction of the thing with another object. Every time two things interact, a force is exerted on each of them .The acted force may be of attraction or of repulsion .The two items no longer feel the force after the interaction ends

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two components.

The parallel rate of change of momentum$\frac{d\overrightarrow{p}}{dt}$and the perpendicular rate of change of momentum$\frac{d\overrightarrow{p}}{dt}$are the two elements that we are concerned with.

So, the net force$F_{net}$on the object is given by

$$\begin{gathered} F_{net}=\frac{d\overrightarrow{p}}{dt} \\ =\frac{d\overrightarrow{p}}{dt}+\frac{d\overrightarrow{p}}{dt} \end{gathered}$$

The rock's speed is affected by the parallel rate of change of momentum and given the speed is constant, therefore, the parallel rate is zero, and it equals the size of the momentum rate change.

$\frac{d\overrightarrow{p}}{dt}=0$

The centrifugal force $F_c$equals the rate change, which is the change in direction owing to the perpendicular rate of change.

At speeds significantly slower than the speed of light, the magnitude of the perpendicular rate change is given by

$$\begin{gathered} F_c=\frac{d\overrightarrow{p}}{dt} \\ =\frac{m{v}^2}{R} \end{gathered}$$

The change in distance over time is the speed.

So, it is given by

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

The engineer goes above the diameter of a circle as he travels through one cycle.

The circumference is given by

role="math" localid="1656911269271" $2\pi R$

Whereis the radius of the circle.

So, the distance where the engineer travel is

$d=2\pi R$

Therefore,

$v=\frac{2\pi R}{t}$

Now put the values forandto get the speed of the engineer

$$\begin{gathered} v=\frac{2\pi R}{t} \\ =\frac{2\mathrm{\pi }\left(14\mathrm{m}\right)}{30s} \\ =2.94\mathrm{m}/\mathrm{s} \end{gathered}$$

Now put the values for $m$, $v$and $R$to get$F_c$

$$\begin{gathered} F_c=\frac{m{v}^2}{R} \\ =\frac{\left(70\mathrm{kg}\right)\left(2.49\mathrm{m}/{\mathrm{s}}^2\right)}{14\mathrm{m}} \\ =43.21\mathrm{N} \end{gathered}$$

The magnitude of the force the engineer has to exert in order to hold on is$43.21\mathrm{N}$_.

At speeds far slower than the speed of light, the net force applied on the object equals the rate change of momentum and the magnitude of the perpendicular rate change, which is the centrifugal force$F_c$.

$$\begin{gathered} F_{net}=F_c \\ =\frac{d\overrightarrow{p}}{dt} \\ =43.21\mathrm{N} \end{gathered}$$

Thus, the magnitude of the net force acting on the engineer is $43.21\mathrm{N}$.