8.0 Rotor

Rotor is an amusement park ride, which is essentially a big cylinder, and it starts spinning with sufficient angular velocity and also the floor goes down.

The person leaning to the wall of the cylinder does not fall because the weight is balance by the friction between the person and the cylindrical wall. The normal force provides centripetal force for the rotation.

The case is similar to that of the death well. The only difference is that in death well the wall is stationary and the car is moving but in rotor the wall and the person are relatively at rest.

Let the rotor is rotating with angular velocity $\omega $ as shown in the figure,

$$\begin{equation} \begin{aligned} f = mg\quad ...(i) \\ N = m{\omega ^2}R\quad ...(ii) \\\end{aligned} \end{equation} $$

Minimum angular velocity so that the person does not fall is when maximum friction acts between man and the wall,

$$\begin{equation} \begin{aligned} f = {f_L} \\ f = \mu N\quad ...(iii) \\\end{aligned} \end{equation} $$

From equation $(i)$, $(ii)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} \mu m\omega _{\min }^2R = mg \\ {\omega _{\min }} = \sqrt {\frac{g}{{\mu R}}} \\\end{aligned} \end{equation} $$

Question 9. A car running in a death well of radius $20\ m$ with its maximum speed of $72\ km/h$. Find the minimum coefficient of friction, so that the car does not fall.

Solution: Maximum speed of car, $$\begin{equation} \begin{aligned} {v_{\max }} = 72\;km/h \\ {v_{\max }} = 20\;m/s \\\end{aligned} \end{equation} $$

Radius, $$r = 20\;m$$

Coefficient of friction, $\mu = ?$

For balancing the car we can write, $$\begin{equation} \begin{aligned} v \geqslant \sqrt {\frac{{rg}}{\mu }} \\ \mu \geqslant \frac{{rg}}{{{v^2}}} \\ \mu \geqslant \frac{{20 \times 9.8}}{{400}} \\ \mu \geqslant 0.49 \\\end{aligned} \end{equation} $$