At amusement parks, there is a popular ride where the floor of a rotating cylind

Question

At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders "plastered" against the wall. Suppose the radius of the room is 3.50 m and the speed of the wall is 11.8 m/s when the floor falls away.

(a) What is the source of the centripetal force acting on the riders?
the gravitational force
the cavitron force between the rider and the wall
the strong force between the rider and the wall
the normal force exerted on the rider by the wall
.

(b) How much centripetal force acts on a 55.0 kg rider?
______ N

(c) What is the minimum coefficient of static friction that must exist between the rider's back and the wall, if the rider is to remain in place when the floor drops away?
__________

Explanation / Answer

a) normal force exerted on the rider by the wall b) Given     mass of rider   m = 55 kg     Radius of room   r = 3.5 m     speed of wall v = 11.8 m/s     centripetal force acts on the rider                      F = m v 2 / r                          = ( 55 ) ( 11.8 ) 2 / 3.5 m                          = 2188 N c) F = mg      coefficient of static friction               = F / mg                    = 2188 / 55 *9.8                     = 4.05                                     

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.