A person riding a Ferris wheel moves through positions at (1) the top, (2) the bottom, and (3) mid height. If the wheel rotates at a constant rate, rank these three positions according to (a) the magnitude of the person’s centripetal acceleration, (b) the magnitude of the net centripetal force on the person, and (c) the magnitude of the normal force on the person, greatest first.

A person riding a Ferris wheel moving at constant speed goes through positions:

1. The top
2. The bottom
3. The mid-height

To find the acceleration, we have to use the relation between linear velocity and centripetal acceleration acting on the person. This relation also describes the acting centripetal force and the normal force acting direction. This determines the rank of the positions accordingly for the given cases.

Formulae:

The centripetal acceleration acting on a body in a circular motion,$a_c=\frac{v^2}{r}$ (i)

The force according to Newton’s second law, F =ma (ii)

The force due to gravity acting on the body, $F_N=mg$ (iii)

From the equation of (i), we get the relation of centripetal acceleration to the radius of the wheel. Thus, we can say that centripetal acceleration only depends on the linear velocity and radius.

Here both are constant so at any position centripetal acceleration will be the same.

Hence, the rank of these positions is equal to each other.

The net centripetal force on the person is defined by substituting the value of equation (i) in equation (ii) as follows:

$F_c=m\left(\frac{v^2}{R}\right)................\left(a\right)$

Here, also the centripetal force is dependent only on the linear velocity, radius, and mass. All are constant so at any position, the centripetal force will be the same.

Hence, the rank of the positions is equal to each other.

We can find the value of normal force at each position by using equation (ii) for all the net forces acting on the person in the wheel as follows:

$\underset{}{\sum F=ma}$

At the top: The net forces acting on the person can be given using equations (iii) and (a) in the above equation as:

$$\begin{gathered} F+F_C-F_g=0 \\ N=Mg-M\left(\frac{V^2}{R}\right) \end{gathered}$$

At the mid-height: The net forces acting on the person can be given using equations (iii) and (a) in the above equation as: (for,)

$$\begin{gathered} N-M\left(\frac{V^2}{R}\right)=0 \\ N=M\left(\frac{V^2}{R}\right) \end{gathered}$$

At the bottom:The net forces acting on the person can be given using equations (iii) and (a) in the above equation as:

$$\begin{gathered} N-F_g-F_C=0 \\ N=M_g+m\left(\frac{V^2}{R}\right) \end{gathered}$$

So, at the bottom point, the Normal force is at maximum, and at the top, it is at minimum.

Hence, the rank of the positions ${\mathrm{N}}_{\mathbf{Bottom}\mathbf{}}>{\mathrm{N}}_{\mathrm{Midh}\mathrm{eight}}>{\mathrm{N}}_{\mathrm{Top}}$is accordingly.