Question

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 $1.00 \times 10^{-15}\textrm{ kg}$ bacterium cling to the rim?
(d) Take the ratio of this force to the bacterium's weight.

Final Answer

- $1350 \textrm{ rev/min}$
- $8.47 \times 10^{3}\textrm{ m/s}^2$
- $8.47 \textrm{ -12} \textrm{ N}$
- $864$

Solution Video