Question

A particular Ferris wheel takes riders in a vertical circle of radius \(9.00 \mathrm{~m}\) once every \(12.0 \mathrm{~s}\) a) Calculate the speed of the riders, assuming it to be constant. b) Draw a free-body diagram for a rider at a time when she is at the bottom of the circle. Calculate the normal force exerted by the seat on the rider at that point in the ride. c) Perform the same analysis as in part (b) for a point at the top of the ride.

Short answer

Answer: The speed of the riders on the Ferris wheel is approximately 4.71 m/s. The normal force exerted by the seat on the rider at the bottom of the ride is 12.28m, and at the top of the ride, it is 7.34m.

Step-by-step solution

a) Calculating the speed

We can calculate the speed using the formula: \(v=\frac{2\pi R}{T}\) Given the radius \(R=9.00 \mathrm{~m}\) and the period \(T=12.0 \mathrm{~s}\), we can find the speed as follows: \(v=\frac{2\pi (9.00)}{12.0} \approx 4.71 \mathrm{~m/s}\) The speed of the riders is approximately \(4.71 \mathrm{~m/s}\).

b) Free-body diagram and normal force at the bottom

At the bottom of the Ferris wheel, the forces acting on the rider are: gravitational force (\(mg\)) acting downward and normal force (\(F_N\)) exerted by the seat acting upward. To find the net force, we can use Newton's second law: \(F = ma\) At the bottom of the circle, the net force is \(F_N - mg = ma\). The centripetal acceleration can be found using the formula \(a=\frac{v^2}{R}\), where \(v\) is the speed calculated earlier, and \(R\) is the radius. \(a= \frac{(4.71)^2}{9.00} \approx 2.47 \mathrm{~m/s^2}\) Now, we can find the normal force: \(F_N = mg + ma\) Let's assume the mass of the rider is \(m\). \(F_N = m(g+ a) = m(9.81 + 2.47) = 12.28m\) At the bottom of the ride, the normal force exerted by the seat on the rider is \(12.28m\).

c) Free-body diagram and normal force at the top

At the top of the Ferris wheel, the forces acting on the rider are: gravitational force (\(mg\)) acting downward and normal force (\(F_N\)) exerted by the seat acting downward as well. To find the net force, we can use Newton's second law: \(F = ma\) At the top of the circle, the net force is \(mg - F_N = ma\). We already found the centripetal acceleration earlier, which was \(a = 2.47 \mathrm{~m/s^2}\). Now, we can find the normal force: \(F_N = mg - ma\) \(F_N = m(g - a) = m(9.81 - 2.47) = 7.34m\) At the top of the ride, the normal force exerted by the seat on the rider is \(7.34m\).