A 40.0 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 hughest point of the ride ? (d) What force does the seat exert on the child when the child is halfway between the top and bottom?
Solved 1 Answer
See More Answers for FREE
Enhance your learning with StudyX
Receive support from our dedicated community users and experts
See up to 20 answers per week for free
Experience reliable customer service
Step 1Given Mass of child m=40kgRadius of wheel r=9mSpeed ω=4×2π60=0.42 rad/s Step 2a) acceleration a=rω2=9×0.422=1.58m/s2b)At lowest point N-mg=maN=m(g+a)N=40(9.81+1.58)=455N Force acting on child is 455N upward towards centre.c)At highest point N+mg=maN=m(g-a)N=40(9.81-1,58)N=329N Step 3d) free body diagram is shown below Step 4F=ma=40×1.58=63.2Nmg=40×9.81=392NResultant force acting on child FR=63.22+3922=397Nφ=tan-1(39263.2)=80.80 ...