A partial cross-section of an amusement park ride is shown. While the ride spins

Question

A partial cross-section of an amusement park ride is shown. While the ride spins up to the angular speed omega c , there is a small platform a t F on which the person P stands. Once the ride reaches the desired angular speed, the platform falls away and only friction keeps the person from sliding to the floor of the ride. The wall, against which the person lies, is inclined at the angle theta = 17degree with respect to the vertical. Model the person as a particle that is a distance d = 36 ft from the spin axis A B and let the coefficient of static friction between the person and the wall be mu s = 0.6. Determine the maximum value of omega c before the person slides up the wall and out of the ride.

Explanation / Answer

Just before the point where the man falls out, friction will be downwards along the plane

Hence equating forces==>

md22sin(17) = mgcos(17) + (mgsin(17) + md22cos(17)) ---[1]

substituting all values and canceling m

==> 2 =  (mgcos(17) + mgsin(17))/(md2sin(17) -md2cos(17)) = -ve value which is impossible

If we observe closely.... in eq [1] the right side term is always greater than thge lest of the equation... in fact md22cos(17) > md22sin(17)

Hence, at any speed the person will not slide up and out

If you require a answer.... c = it is infinity

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