
#1
Jan2408, 12:17 PM

P: 1

1. The problem statement, all variables and given/known data
In an oldfashioned amusement park ride, passengers stand inside a 3.0mtall, 5.0mdiameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will “stick” to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.60 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. 2. Relevant equations T = 1/f and so the people dont slide (coefficient of friction) (Fn)= mg ( (coefficient of friction)*m*v^2)/ r = mg 3. The attempt at a solution (coefficient of friction) = (g*r)/ (v^2) T = (2*pi*r)/ square root of ((g*r)/(coefficient of friction)) and f= 1/T I got 0.4068 = f but the program says I am wrong 



#2
Jan2408, 12:25 PM

Mentor
P: 39,601

What did you use for the coefficient of friction? What did you use for the radius?




#3
Jan2408, 12:29 PM

Mentor
P: 40,877

Assuming that the question was to find the minimum frequency to insure that they wouldn't slide, it looks OK to me. How was the question phrased exactly?




#4
Jun1809, 02:02 PM

P: 1

homework question, need help asap, please
The 0.4068 is rotations in a second. You had to multiply that value by 60 to put it into rotations per MINUTE.



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