An amusement park ride called the Rotor consists of a room in the shape of a ver

Question

An amusement park ride called the Rotor consists of a room in the shape of a vertical cylinder 3.67 m in radius which, once the riders are inside, begins to rotate, forcing them to the wall. When the room reaches the angular speed of 2.8 rad/s, the floor suddenly drops out. It is deemed that this rate of rotation will be sufficient to keep a 400 lbf person pinned against the wall of the cylinder. What is the minimum angular speed required to keep a 200 lbf person pinned against the wall of the cylinder?

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CHEGG,

Please start with (show how you derive from) simple formulas (i.e. ac = (Vt2)/R = wVt = w2R and F = (mV2)/r and [arc length] = [angle in radians]/[t] etc.), and please show ALL your work. Thank you!

Explanation / Answer

m = 400 lbf = 181.4 Kg

Frictional force must balance = weight of person
miu*N = m*g

Now this N mist be coming from centrifugal force
N = m*r*w^2

so,
miu*m*r*w^2 = m*g
miu*r*w^2 = g
w = sqrt (g/miu*r)

As we can see that angular speed is independent of mass.
Same angular speed will be required for all the masses
Answer: 2.8 rad/s

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