91Ó°ÊÓ

The wheel has a radius $R$and rotates at a constant rate, going around once in a time $T$.

A force is a push or pull on an object 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 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${\frac{d\stackrel{→}{p}}{dt}}_{\left|\right|}$and the perpendicular rate of change of momentum${\frac{d\stackrel{→}{p}}{dt}}_{âŠ\yen }$are the two elements that we are concerned with.

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

$$\begin{gathered} F_{net}=\frac{d\stackrel{→}{p}}{dt} \\ ={\frac{d\stackrel{→}{p}}{dt}}_{\left|\right|}+{\frac{d\stackrel{→}{p}}{dt}}_{âŠ\yen } \end{gathered}$$

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.

${\frac{d\stackrel{→}{p}}{dt}}_{\left|\right|}=0$

As a result, the rate change here is the direction change owing to the perpendicular rate of change.

At speeds far slower than the speed of light, the net force applied on the rider equals the rate change of momentum, and the magnitude of the perpendicular rate change is given by,

$$\begin{gathered} F_{net}={\frac{d\stackrel{→}{p}}{dt}}_{âŠ\yen } \\ =\frac{m{v}^2}{R} \end{gathered}$$

Also, the rider is under two forces, the force exerted by the seat where its direction is upward$F_N$, and his weight mg and it is downward as shown in the free-body diagram below, so the net force of both forces is:

_{$F_{net}=F_N-mg$,}

Where the negative sign is due to the opposite direction of the forces.

Therefore,

$$\begin{gathered} F_N-mg=\frac{m{v}^2}{R} \\ {F}_N=\frac{m{v}^2}{R}+mg \end{gathered}$$

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

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

The circumference is given by

$2\mathrm{Ï€}R$

Where, R is the radius of the circle.

So, the distance where the engineer travel is:

$d=2\mathrm{Ï€}R$

Therefore,

$v=\frac{2\mathrm{Ï€}R}{t}$

Now, put the expression for$\mathrm{ν}$to get a relationship

$$\begin{gathered} F_N=\frac{m{v}^2}{R}+mg \\ =\frac{m}{R}{\left[\frac{2\mathrm{Ï€}R}{T}\right]}^2+mg \\ =m\left[\frac{4{\mathrm{Ï€}}^{2}R}{T^2}+g\right] \end{gathered}$$

Thus, the force exerted by the seat is $m\left[\frac{4{\mathrm{Ï€}}^{2}R}{T^2}+g\right]$.