Question

In a classic carnival ride, patrons stand against the wall in a cylindrically shaped room. Once the room gets spinning fast enough, the floor drops from the bottom of the room! Friction between the walls of the room and the people on the ride make them the "stick" to the wall so they do not slide down. In one ride, the radius of the cylindrical room is R = 6.1 m and the room spins with a frequency of 23.3 revolutions per minute. What is the speed of a person "stuck" to the wall? What is the normal force of the wall on a rider of m = 48 kg? What is the minimum coefficient of friction needed between the wall and the person? If a new person with mass 96 kg rides the ride, what minimum coefficient of friction between the wall and the person would be needed? Which of the following changes would decrease the coefficient of friction needed for this ride? increasing the riders mass increasing the radius of the ride increasing the speed of the ride increasing the acceleration due to gravity (taking the ride to another planet) To be safe, the engineers making the ride want to be sure the normal force does not exceed 1.9 times each persons weight and therefore adjust the frequency of revolution accordingly. What is the minimum coefficient of friction now needed?
I need the answers for 5 and 6 parts

Answer 1

Answer to part 5:

Since the rider is standing against the wall, the frictional force is acting upwards and is equal to the weight of the rider, i.e,   $F_{fr}= mg$. The normal force acting on the rider is given by : $F_N= \frac{mv^2}{r}$

This acts towards the centre of the cylinder. Now, $F_{fr}=\mu F_N\Rightarrow \mu = \frac{g}{v^2r}$

Therefore, the coefficient of friction will decrease by increasing the radius of the ride and increasing the speed of the ride.

Answer to part 6:

Since $F_N\leq 1.9mg$, so the maximum normal force is :

$(F_N)_{max}= 1.9mg = \frac{mv_{max}^2}{r}\Rightarrow v_{max}^2=1.9gr$

Hence, $\mu_{min}=\frac{g}{v_{max}^2r}=\frac{1}{1.9r^2}=0.014$ (Ans)