Question

At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the tires rotating? Part (b) What is the centripetal acceleration at the edge of the tire in m/s^2?

Answer 1

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r  = 0.89 x 0. 5 = 0.445 m

Angular velocity

         $\omega =\frac{72}{0.445}=161.8rad/s$

Frequency

         $f=\frac{2\pi}{\omega}=\frac{2\times \pi}{161.8}=0.0388rev/s=2.33rev/min$

Revolutions per minute = 2.33

b) Centripetal acceleration

               $a=\frac{v^2}{r}$

  Here linear velocity = 72 m/s

  Radius, r  = 0.445 m

Substituting

   $a=\frac{72^2}{0.445}=11649.44m/s^2$

Centripetal acceleration = 11649.44m/s²