A 48.0-kg child takes a ride on a Ferris wheel that rotates four times each minu

Question

A 48.0-kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of 24.0 m.

(a) What is the centripetal acceleration of the child? magnitude m/s2 direction Correct: Your answer is correct.

(b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride? magnitude N direction Correct: Your answer is correct.

(c) What force does the seat exert on the child at the highest point of the ride? magnitude N direction Correct: Your answer is correct.

(d) What force does the seat exert on the child when the child is halfway between the top and bottom? (Assume the Ferris wheel is rotating clockwise and the child is moving upward.) magnitude N direction ° counter-clockwise from the horizontal

Explanation / Answer

m =48 kg ,

w=4*2pi/60, d= 24m , r =12 m

centripetal accleration ac = rw^2 = 12(8*3.14/60) 2

ac =2.103 m/s^2

(a) Fc =mac = 48*2.103 = 100.944 N

(b) At lowest point

From Newton second law net force along verticla direction

Fn- mg = Fc

Fn = mg+Fc = (48*9.8)+100.944

Fn = 571.344 N along upward direction

(b) At highet point

From Newton second law net force along verticla direction

Fn- mg = -Fc

Fn = mg-Fc = (48*9.8)-100.944

Fn = 369.456 N along upward direction

(d) the centripetal force points in purely the x direction, but since the child cannot be accellerating in the y

direction at the time(or he'd stop going in a circle) the resultant force in the y direction must be zero

Fn = Fg

Fny = mg = 470.4 N (y direction)

Fn =Fc

Fnx = 100.944 N (x-direction)

Magnitude Fn = [(100.944)^2+(470.4)^2]^1/2

Fn = 481.11 N

direction tan(theta) = 470.4/100.944

theta= 77.90 with + x-axis

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