Question

An amusement park ride consists of a rotating vertical cylinder with rough canvas walls. The floor is initially about halfway up the cylinder wall as shown. After the rider has entered and the cylinder is rotating sufficiently fast, the floor is dropped down, , yet the rider does not slide down. The rider has mass of 50 kg. The diameter of the cylinder is 6.5 meters. The coefficient of static friction between the rider and wall of the cylinder is 0.65. What is the minimum linear velocity of the passengers so that they remain in place along the wall after the floor drops down? A. 2.4 m/s B. 3.0 m/s C. 3.8 m/s D. 5 m/s E. 7 m/s

It is the diagram at the top

noャ^ max Statie situnta -con' use FN An amusement park ride consists of a rotating vertical cylinder with rough canvas walls. The floor is initially about halfway up the cylinder wall as shown above. After the rider has entered and rotating sufficiently fast, the floor is dropped down, yet the rider does not slide down. The rider of 50 kilograms, the radius R of the cylinder is 5 meters, the angular velocity of the cylinder when rotating is 2 radians per second, and the coefficient of static friction between the rider and the wall of the cylinder is 0.6. has mass On the diagram below. draw and identify the forces on the rider when the system is rotating and the floor has dropped down. a. FM b. Calculate the centripetal force on the rider when the cylinder is rotating and state what provides that force conver radians par secon d foree 5 0 c. Calculate the upward force that keeps the rider from falling when the floor is dropped down and state what provides that force 500 N, foree of rietn proude : Ffric 勹 Frve 500 fric d. At the same rotational speed, would a rider of twice the mass slide down the wall? Explain your answer. [w.ys e. 네 rteba

Answer 1

I have solved parts b to d, also and at the last, the velocity.

M = 50 kg: R = 5 m: w = 2 rad/s: mu_s = 0.6 b) F_c = m v^2/R {Because U = r w} F_l = m/R middot R^2 w^2 = m R w^2 = 50 times 5 times 2^2 F_c = 1000 N c) Limiting friction force f_2 = mu_c N: N = F_c = 1000 f_c = 0.6 times 1000 = > f_c = 600 N Because Limiting friction > applied force (gravitational force) friction, f = f g = m g = 50 times 10 = > f = 500 N d) No, because is that case too the limiting friction force is greater the weight for the rides to just remain , f = f g = > mu N = f g = > mu m V^2/R = m g = > V = squareroot g R/mu = squareroot 10 times 6.5/0.65 = V 7 m/s (E)