The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop (in m/s) if the radius of curvature there is 12.0 m and the downward acceleration of the car is 1.50 g

Question

Physics

Boston

10 January, 02:24

The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop (in m/s) if the radius of curvature there is 12.0 m and the downward acceleration of the car is 1.50 g

+1

Answers (1)

Know the Answer?

Jayda Donaldson · Answer 1

v=13.28m/s

Explanation:

centripetal force is the net force that keeps a body in circular motion along a circular path. The direction of a centripetal force is toward the center of curvature, also the direction of centripetal acceleration. According to Newton's second law of motion, the rate of change in momentum is directly proportional to the force applied

f=m (v-u) / t

a = (v-u) / t

f=ma

f=centripetal force

m=mass of the body

DV=change in velocity

t=time

a=centripetal acceleration

a=v^2/r

centripetal acceleration, since centripetal force produces centripetal acceleration

1.5g is the same as 14.7m/s^2

14.7=v^2/12

14.7*12=v^2

176.4=v^2

v=13.28m/s