Forces in rotational motion An amusement park ride called the Rotor consists of

Question

Forces in rotational motion

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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An object of mass m 182.9 g undergoes constant circular motion with radius 69.9 cm on a flat frictionless table. It is connected by a massless string through a hole in the table to a larger object of mass m2 201.6 g. If, as the result, the hanging mass m2 is stationary, what is the angular speed of the rotating mass m,? Answer Check

Explanation / Answer

1)

lets say that the linear speed of the mass m1 is v, so its angualr speed will be w = v/r

where r is the radius of circular motion = 69.9cm or 0.699m

mass m1 = 182.9gm or 0.1829kg

mass m2 = 201.6g or 0.2016kg

since the mass m2 is stable as it is not moving so we can safely say that it will be lifted up againest the gravitional force by centrifugal force of mass m1

so (m1*v2)/r = m2*g

so v = (m2*g*r/m1)1/2   = 2.747 m/s

so angular velocity of mass m1 will be w = v/r = (2.747/0.699) = 3.929 rad/s

2)

radius r = 3.67m

angular speed w = 2.8 rad/s

lets say that the angular speed required to keep the 200lb man attached to the wall is = w

so here m*w2*r >= m*g

w >= (g/r)1/2 >= (9.8/3.67)1/2 >= 1.63 rad/s

so the angular velocity should be greater than 1.63 rad/s

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