Question

At takeoff a commercial jet has a 55.0 m/s speed. Its tires have a diameter of 0.850 m (a) At how many rpm are the tires rotating? rpm (b) What is the centripetal acceleration at the edge of the tire? m/s2 (c) With what force must a determined 10-15 kg bacterium cling to the rim? (d) Take the ratio of this force to the bacterium's weight. (force from part (c)/bacterium's weight)

Answer 1

Given: speed = 55.0 m/s dia = 0.850m Therefore r = 0.425m Because v = r w = r middot 2 pi/T v = r middot 2 pi N Therefore N = v/r times 2 pi times 60 = 55/2 pi times 0.425 times 60 = 1235.8 Rpm \r\n centripetal acceleration a_c = rw^2 = 0.425 times (v/r)^2 = 7117.6 m/s^2 Force, a determined 10^-15 kg bacterium cling to the rim F_c = m times a = 10^-15 times 7117.6 = 7.12 times 10^-12 N F_c/m times g = 7.12 times 10^-12/9.8 times 10^-15 = 726.53 TildeEqual 727