At amusement parks, there is a popular ride where the floor of arotating cylindr

Question

At amusement parks, there is a popular ride where the floor of arotating cylindrical room falls away, leaving the backs of theriders "plastered" against the wall. Suppose the radius of the roomis 3.52 m and the speed of the wall is 11.5 m/s when the floorfalls away. The source of the centripetal force on the riders isthe normal force provided by the wall. (a) How much centripetalforce acts on a 52.9 kg rider? (b) What is the minimum coefficientof static friction that must exist between the rider's back and thewall, if the rider is to remain in place when the floor dropsaway?

Explanation / Answer

The centripetal force is     F = mV^2 /R     m =59.2   , V = 11.5 and R = 3.52m     F = 2224.2 N -------------------------------------------------------------------------------------------------------- the weight should be balanced by frictional force        m g = F         = m g /F            =0.26

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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