Uniform Circular Motion — Question 8

Question 8

A rotating wall supplies a normal that enables vertical friction

Take $g=10$ m/s$^2$. A 60 kg rider is held against the inside of a vertical cylindrical wall of radius 2 m rotating steadily about its vertical axis. The floor is lowered. Static wall friction has coefficient 0.40; the rider stays at fixed height and rotates with the wall.

1. Find the minimum angular speed $\omega$ that prevents downward sliding.

2. Find the normal and actual friction at $\omega=4$ rad/s.

3. Find the largest allowable rotation period and explain why the normal is not equal to weight.

Question 8 - Solution

A rotating wall supplies a normal that enables vertical friction

(a) Use radial motion to determine available support.

The wall normal supplies radial acceleration, while friction supports weight: $$N=m\omega^2r,\qquad f=mg\leq\mu_sN.$$ $$\boxed{\omega_{\min}=\sqrt{\frac{g}{\mu_sr}}=\sqrt{\frac{10}{.40(2)}}=\frac5{\sqrt2}=3.536\ \mathrm{rad/s}}.$$

(b) Distinguish demand from capacity.

At $\omega=4$ rad/s, $$\boxed{N=60(4)^2(2)=1920\ \mathrm N},\qquad \boxed{f=mg=600\ \mathrm N\text{ upward}}.$$ The capacity is $\mu_sN=.40(1920)=768$ N $>600$ N, so the vertical rest constraint is feasible.

(c) Convert a lower frequency to an upper period.

The angular-speed lower bound becomes a period upper bound: $$\boxed{\mathcal T\leq\frac{2\pi}{\omega_{\min}}=\frac{2\pi\sqrt2}{5}=1.777\ \mathrm s}.$$ The normal is horizontal and cannot balance vertical weight. Its magnitude is determined by radial motion; friction provides the separate vertical support.