The Rotor is an amusement park ride shown. For this ride, a person enters the Ro
ID: 1861455 • Letter: T
Question
The Rotor is an amusement park ride shown. For this ride, a person enters the Rotor when it is stationary and stands against the wall with their back to the wall. The rotor then begins to rotate about a vertical axis. After reaching cruising speed, the floor drops away and the patron is left suspended against the wall. Assume that the rate of rotation increases according to the expression (alpha) ± = k(theta) until the floor drops (alpha is angular acceleration in rad/s2, theta is angular position in radians). After this point, the Rotor spins at a constant speed until the floor rises again. Determine an expression for the number of revolutions the Rotor should turn before it is safe to drop the floor. The coefficient of friction between the rider and the wall is (mu)s.
Explanation / Answer
Let final angular velocity = w
Centrifugal force = m * w^2 * R
Force balance: N = m*w^2 * R
uN = mg
=> u*m*w^2*R = mg => w = Sqrt[g/(uR)]
alpha = k(theta)
alpha = dw/dt
& w = d(theta)/dt
=> d2(theta) / dt2 = k(theta)
=> w = d(theta)/dt = {k(theta)^2 }/2 + C1
for theta = 0 for w = 0
=> C1 = 0
So w = (k(theta)^2)/2
Sqrt[g/(uR)] = (k/2)(theta)^2
=> theta = [4g/(ukR)]^(1/4) rad
1rad = 1/2pi revolution
=> theta rad = {[4g/(ukR)]^(1/4)}/2pi revolutions
Revolutions = {[4g / ( ukR ) ] ^ (1/4) } / 2pi
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