Final answer:
The speed of a rider on a carnival ride that rotates with a radius of 2.0m and a frequency of 1.1 times per second is approximately 13.82 m/s. The centripetal acceleration at this speed is approximately 95.3 m/s². The minimum coefficient of friction required to prevent riders from slipping when the floor drops is approximately 9.72.
Step-by-step explanation:
To calculate the speed of a rider on a carnival ride with a 2.0m radius that rotates 1.1 times per second, we use the formula for speed on a circular path, which is circular speed = 2πr × frequency. Plugging in the given values, the speed (v) is 2π × 2.0m × 1.1s-1 which gives approximately 13.82 m/s.
To find the centripetal acceleration (ac), we use the equation ac = v2/r. With v = 13.82 m/s and r = 2.0 m, the centripetal acceleration is (13.82 m/s)2 / 2.0 m, which results in approximately 95.3 m/s2.
The centripetal acceleration is produced by the central force that acts towards the center of the circular path, which in this case is provided by the mechanical structure of the ride that guides the riders along the circular path.
When the floor drops down, riders are held up by the friction force between the walls and the riders. The minimum coefficient of friction (μ) required to prevent slipping can be determined by equating the frictional force (which in this case provides the centripetal force) to the required centripetal force for the motion. Therefore, μ × normal force = mass × centripetal acceleration. Assuming that the normal force is equal to the weight of a person (mg), the equation can be rewritten as: μ × mg = m × ac. Simplifying this gives μ = ac / g, where g is the acceleration due to gravity (approximately 9.8 m/s2). Thus, the coefficient of friction needed is 95.3 m/s2 / 9.8 m/s2, which is approximately 9.72.