If the room radius is 4.5 m, and the rotation frequency is 0.8 revolutions per second when the floor drops out, what is the minimum coefficient of static friction so that the people will not slip down?

Question

Justice Glenn

Physics

31 July, 01:45

If the room radius is 4.5 m, and the rotation frequency is 0.8 revolutions per second when the floor drops out, what is the minimum coefficient of static friction so that the people will not slip down?

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Dakota Galvan · Answer 1

The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration. The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N. a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r. The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec. So a_N = 114 m/sec^2. g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.