In an amusement park there is a room where the walls form a vertical cylindrical

Question

In an amusement park there is a room where the walls form a vertical cylindrical surface with its axis perpendicular to the horizontal ground below. People are advised to stand with their back well against the wall. The room is then made to rotate about its axis unit the rotational speed becomes sufficient. At that point the floor is dropped from under the feet but everybody remains as if they were pinned against the wall. Assume the coefficients of friction between any person and the wall are mu_s = 0.6 and mu = 0.4 and that the radius of the cylindrical wall is R = 5 m. Again let g = 10 m/s^2 Derive an algebraic relation for the minimum speed of revolution (omega - v/R) that would allow the people to remain at their positions against the wall (including not to slide down the vertical surface.) Express your result in tombs of the appropriate coefficient of friction mu as well as the radius R and the acceleration of gravity g. [Start by drawing an F. B. D showing the forces a

Explanation / Answer

a) Here,     usN = mg

Also,   N = mw2R

=>   us(mw2R) = mg

=>    us(w2R) = g

=>   w   = sqrt[g/(usR)]

=> minimum speed of revolution = sqrt[g/(usR)]

b)    value of speed =   sqrt[9.8/(0.6 * 5)]

                                  = 1.807 rad/sec

                                  =   0.2877 rev/sec

                                 =   17.262 rev/min

   => period of revolution = 1/0.2877 = 3.475 sec

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