Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.100 rev/s. As Jack passes through the highest point of his circular path, the upward force that the chair exerts on him is equal to one-fourth of his weight. What is the radius of the circle in which Jack travels? Treat him as a point mass

The force due to the weight of the object is given by,

F = mg

Here, F is linear force, m is the mass of the object, and g is the acceleration due to gravity.

Here, the rotating speed is 0.100 rev/s.

Convert the given angular velocity into rad/sec ,$\omega =\left(0.100\hspace{0.33em}\frac{rev}{s}\right)\left(2\pi \hspace{0.33em}\frac{{\displaystyle \frac{rad}{s}}}{{\displaystyle \frac{rev}{s}}}\right)=0.628\frac{rad}{s}$

At the highest point, the chair will exert an upward force which will be equal to one-fourth of his weight (mg). The upward force on Jack is the difference between his own weight and the centrifugal force acting on Jack.

The forces on the Jack are given by,

$$\begin{gathered} mg-mr{\omega }^2=0.25mg \\ r{\omega }^2=0.75g \\ r=\frac{0.75g}{{\omega }^2} \end{gathered}$$

Substitute for$9.8m/s^2$, and 0.628 rad/s for $\omega$ in the above equation, and we get,

$$\begin{gathered} \mathrm{r}=\frac{0.75\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{0.628\mathrm{rad}/\mathrm{s}} \\ =18.6\mathrm{m} \end{gathered}$$

Therefore, the radius of the circle is 18.6 m.