A \( 40.0-\mathrm{kg} \) child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of 18.0 m . a) What is the centripetal acceleration of the child? b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride? c) What force does the seat exert on the child at the highest point of the ride? d) What force does the seat exert on the child when the child is halfway between the top and bottom?
Solución de inteligencia artificial de Upstudy
Responder
Solución
Respondido por UpStudy AI y revisado por un tutor profesional


Bonus Knowledge
To solve the problem, we can break it down into parts for clarity. ### a) Centripetal Acceleration The formula for centripetal acceleration (\(a_c\)) is given by: \[ a_c = \frac{v^2}{r} \] First, we need to calculate the speed (\(v\)) of the child. The Ferris wheel completes 4 rotations per minute, which can be converted to radians per second: \[ 4 \text{ rotations/minute} \times \frac{2\pi \text{ radians}}{1 \text{ rotation}} \times \frac{1 \text{ minute}}{60 \text{ seconds}} = \frac{8\pi}{60} \text{ radians/second} \approx 0.4189 \text{ radians/second} \] The radius (\(r\)) of the Ferris wheel is half the diameter: \[ r = \frac{18.0 \text{ m}}{2} = 9.0 \text{ m} \] Now, we can find the centripetal acceleration: \[ a_c = \frac{(9.0 \text{ m} \cdot \frac{8\pi}{60})^2}{9.0 \text{ m}} \approx 0.447 \text{ m/s}^2 \] ### b) Force at the Lowest Point At the lowest point, the forces acting on the child are the force exerted by the seat (\(F_N\)) and the gravitational force (\(F_g = mg\)). The net force provides the centripetal motion: \[ F_N - mg = m a_c \] Where \(m = 40.0 \text{ kg}\) and \(g \approx 9.81 \text{ m/s}^2\): \[ F_N - (40.0 \text{ kg} \cdot 9.81 \text{ m/s}^2) = 40.0 \text{ kg} \cdot 0.447 \text{ m/s}^2 \] Calculating those values gives: \[ F_N - 392.4 \text{ N} = 17.88 \text{ N} \] \[ F_N \approx 410.28 \text{ N} \] So the normal force at the lowest point is approximately **410.3 N upwards**. ### c) Force at the Highest Point At the highest point, the forces also include gravity, but they add to the centripetal motion: \[ F_N + mg = m a_c \] Using the same mass and gravitational force: \[ F_N + 392.4 \text{ N} = 17.88 \text{ N} \] \[ F_N \approx 17.88 \text{ N} - 392.4 \text{ N} \approx -374.52 \text{ N} \] Since the negative value indicates that the normal force is lower than the gravitational force, we conclude that the force exerted by the seat at the highest point is approximately **-374.5 N downwards**. ### d) Force Halfway Between In this position, the vertical forces balance out. The normal force and gravitational force are in play but don’t affect centripetal force directly since the child isn't at a peak or trough: \[ F_N = mg \] Thus, \[ F_N = 40.0 \text{ kg} \cdot 9.81 \text{ m/s}^2 = 392.4 \text{ N} \] So at halfway between the top and bottom, the force exerted by the seat is approximately **392.4 N upwards**. That covers all the parts of the inquiry!
preguntas relacionadas
