Answer:
a) In order for the riders to stay pinned against the inside of the drum when the floor drops away, the force of friction between the drum and the riders must be greater than or equal to the force of gravity pulling them down.The maximum force of friction that can be exerted on a rider is given by:f_max = μNwhere μ is the coefficient of static friction and N is the normal force acting on the rider.At the minimum speed required for the riders to stay pinned against the inside of the drum, the normal force acting on each rider is equal to their weight, so:N = mgwhere m is the mass of the rider and g is the acceleration due to gravity.The force of gravity pulling the rider down is:F = mgTherefore, the minimum speed required for the riders to stay pinned against the inside of the drum is given by:f_max = FμN = mgv_min = sqrt(rgμ)where r is the radius of the drum, which is half of the diameter, so r = 5m.Plugging in the values, we get:v_min = sqrt(59.80.15) = 2.16 m/sb) The angular velocity of the drum at this speed can be found using the formula:v = ωrwhere v is the linear velocity of the riders at the minimum speed required to stay pinned against the inside of the drum, r is the radius of the drum, and ω is the angular velocity of the drum.Rearranging this equation to solve for ω, we get:ω = v/rPlugging in the values, we get:ω = 2.16/5 = 0.432 rad/sSo the angular velocity of the drum at the minimum speed required for the riders to stay pinned against the inside of the drum is 0.432 rad/s.