On the ride \"Spindletop\" at the amusement park Six Flags Over Texas, people st

Question

On the ride "Spindletop" at the amusement park Six Flags Over Texas, people stood against the inner wall of a hollow vertical cylinder with radius 2.76 m . The cylinder started to rotate, and when it reached a constant rotation rate of 0.760 rev/ s , the floor on which the people were standing dropped about 0.500 m . The people remained pinned against the wall.

What minimum coefficient of static friction is required if the person on the ride is not to slide downward to the new position of the floor?

Does your answer in part A depend on the mass of the passenger? (Note: When the ride is over, the cylinder is slowly brought to rest. As it slows down, people slide down the walls to the floor.)

Yes

No

Explanation / Answer

Circumfrence of the ride:

2*2.76*(pi) = 17.3416 m/s

Centripetal acceleration is V^2/r

a = 17.3416^2 / 2.76 = 108.9604 m/s^2

Friction being based on the normal force:

F = (mu)N

Mass will cancel from both sides of this, leaving:

g <= (mu)*108.9604

Therefore your minimum coefficient of friction is

9.81/108.9604 = 0.09003

Mass does not matter, since it will cancel out.

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

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