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 2.9 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?

Explanation / Answer

To stick, the weight down cannot exceed the maximum possible friction force up. Right at the limit:
W = u * N
where W is weight and N is the normal force. The normal force is the centripetal force, so
N = m * v^2 / r = m * 0 ^2 * r^2

N = (181.436948/9.8) * 2.9^2 * 3.67^2 = 2097.1417

So,

2097.1417 = (90.718474/9.8) * ^2 * 3.67^2

or = 4.10 rad/s

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