A Ferris wheel with radius 14.0 m is turning about a horizontal axis through its center (Fig. E3.27). The linear speed of a passenger on the rim is constant and equal to 6.0 m/s. What are the magnitude and direction of the passenger’s acceleration as she lowest passes through (a) the point in her circular motion and (b) the highest point in her circular motion? (c) How much time does it take the Ferris wheel to make one revolution?

The given data can be listed below.

Motion that follows a curved course is known as circular motion. Uniform circular motion is the description of a body traveling along a circular route with a constant (uniform) speed.

The linear velocity of the passenger is constant so the acceleration of the passenger is centripetal and in an upward direction which is given by,

$a=\frac{v^2}{R}$

Here, v is the linear velocity of the passenger, and R is the radius of the wheel.

Substitute all the values in the above,

$\begin{array}{rcl}a& =& \frac{{\left(6m/s\right)}^2}{14m}\\ & =& 2.57m/s^2\end{array}$

Thus, the magnitude of acceleration is $2.57m/s^2$and it is pointed in upward direction.

The centripetal acceleration is constantly pointed in the direction of the center and has the same magnitude at every point along a curved route.

Thus, the acceleration of the passenger is 2.57 m/s² downhill at the maximum point.

The time taken by the Ferris wheel for one revolution is given by,

$T=\frac{2\mathrm{\pi R}}{v}$

Here, v is the linear velocity of the passenger, and R is the radius of the wheel.

Substitute all the values in the above,

$\begin{array}{rcl}T& =& \frac{2\left(3.14\right)\left(14m\right)}{6m/s}\\ & =& 14.66s\end{array}$

Thus, the time taken by the Ferris wheel to complete one revolution is 14.66 s.