A student of weight${}^{\mathbf{667}\mathbf{}\mathbf{N}}$rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal forceon the student from the seat is${}^{\mathbf{556}\mathbf{}\mathbf{N}}$. (a) Does the student feel “light” or “heavy” there? (b) What is the magnitude of at the lowest point? If the wheel’s speed is doubled, what is the magnitude${}^{{\mathbf{F}}_{\mathbf{N}}}$at the (c) highest and (d) lowest point?

$\mathrm{Weight}\mathrm{of}\mathrm{student}=667\mathrm{N},$

$F_N=556\mathrm{N}$

Ferris wheel ride is in a vertical circular motion. The apparent weight of the rider varies with his position. According to vertical circular motion, a body revolves in vertical circle such that its motion at different points is different, then the motion of the body is said to be vertical circular motion.

Formula is a follow:

$F_{N,bottom}-mg=\frac{m{v}^2}{R}$

where, m is mass, v is velocity, g is acceleration due to gravity, R is radius and${}^{F_{N,bottom}}$ is force at bottom.

The free-body diagrams of the student at the top and bottom of the Ferris wheel are shown below:

Newton’s second law for the radial direction gives,

$mg-F_{N,top}=\frac{m{v}^2}{R}$

The net force toward the centre of the circle is,

$F_{N,bottom}-mg=\frac{m{v}^2}{R}$

The Ferris wheel is “steadily rotating”. So, the value ${}^{F_c=\frac{m{v}^2}{R}}$is the same everywhere.

The apparent weight of the student is given by ${}^{F_{N·}}$

At the top, it is given that, ${}^{F_{N,top}=556\mathrm{N}}$, and${}^{mg=667\mathrm{N}}$. This means that the seat is pushing up with a force that is smaller than the student’s weight, and it can be said that the student experiences a decrease in his “apparent weight” at the highest point. Thus, student feels “light.”

Centripetal force is,

$$\begin{gathered} F_c=\frac{m{v}^2}{R} \\ =mg-F_{N,top} \\ =667\mathrm{N}-556\mathrm{N} \\ =111\mathrm{N} \end{gathered}$$

Thus, the normal force at the bottom is,

$$\begin{gathered} F_{N,bottom}=\frac{m{v}^2}{R}+mg \\ =F_c+mg \\ =111\mathrm{N}+667\mathrm{N} \\ =778\mathrm{N} \end{gathered}$$

Hence, themagnitude of force at the lowest point is ${}^{778\mathrm{N}}$.

If the speed is doubled, the force will be

$$\begin{gathered} {F^{\text{'}}}_c=\frac{m{\left(2v\right)}^2}{R} \\ =4\left(111\mathrm{N}\right) \\ =444\mathrm{N} \end{gathered}$$

Therefore, at the top,

$$\begin{gathered} {F^{\text{'}}}_{N,top}=mg-{F^{\text{'}}}_c \\ =667\mathrm{N}-444 \\ =223\mathrm{N} \end{gathered}$$

Hence, themagnitude of force at the highest point is${}^{233\mathrm{N}}$.

The normal force at the lowest point will be,

$$\begin{gathered} {F^{\text{'}}}_{N,bottom}={F^{\text{'}}}_c+mg \\ =444\mathrm{N}+667\mathrm{N} \\ =1111\mathrm{N} \end{gathered}$$

Hence, themagnitude of force at the lowest point is${}^{1111\mathrm{N}}$.