Question

001 (part 1 of 2) 10.0 points An amusement park ride consists of a large vertical cylinder that spins about enough that any person inside is held up against the wall when the floor drops away its axis fast 6.87 m What is the minimum angular velocity wmin needed to keep the person from slipping downward? The acceleration due to gravity is 9.8 m/s2, the coefficient of static friction be- tween the person and the wall is 0.77, and the radius of the cylinder is 6.87 m Answer in units of rad/s. 002 (part 2 of 2) 10.0 points Suppose the person, whose mass ing held up against the wall with an angular velocity of 2wmin is m, is be- The magnitude of the frictional force be- tween the person and the wall is 1. F=-mg 2. F-5mg 6, F = mg

Answer 1

F = m u^2/r f lessthanorequalto u N = f_max N = m u^2/R f_max = mu N = mu m^2/R = mg v_max = squareroot g R/mu v_min = squareroot 9.8 times 6.87/0.77 = 9.35 m/sec = v_min/R = 9.35/6.87 = 1.36 rad/sec (b) answer is F = mg option since f lessthanorequalto f_max = mu N one angular velocity has increased and surface the minimum angular velocity required to keep the pressure against angle these it does not depend on functional force