At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m. (a) At how many rev/min are the tires rotating? (b) What is the centripetal acceleration at the edge of the tire? (c) With what force must a determined 1.00*10-15 kg bacterium cling to the rim? (d) Take the ratio of this force to the bacterium's weight.

(a) Angular speed = 1344.8rev/min

(b) Centripetal acceleration = 8473.4m/s^2

(c) Force = 8.4734*10^-12N

(d) Ratio of force to bacterium's weight is 864.6

Explanation:

(a) Angular speed (w) = v/r

v = 60m/s, d = 0.850m, r = 0.850m/2 = 0.425m

w = 60/0.425 = 141.2rad/s = 141.2*9.524rev/min = 1344.8rev/min

(b) Centripetal acceleration = w^2r = 141.2^2 * 0.425 = 8473.4m/s^2

(c) Force = mass * centripetal acceleration = 1*10^-15 * 8473.4 = 8.4734*10^-12N

(d) Bacterium's weight = mass * acceleration due to gravity = 1*10^-15 * 9.8 = 9.8*10^-15N

Ratio of force to bacterium's weight = 8.4734*10^-12N/9.8*10^-15N = 864.6