Question

At takeoff, a commercial jet has a 60.0 m/s speed. If its tires have a diameter of 0.850 m, what is the angular speed of the tires during takeoff?

a) 6.39 , {rad/s}
b) 7.51 , {rad/s}
c) 8.63 , {rad/s}
d) 9.75 , {rad/s}

Answer 1

The angular speed of the tires during takeoff is 141.18 rad/s, which is equivalent to 1347.2 revolutions per minute.

Step-by-step explanation:

To find the angular speed of the tires during takeoff, we need to use the relationship between linear speed (v) and angular speed (ω). The formula connecting these quantities is v = rω, where v is the linear speed, r is the radius of the tire, and ω is the angular speed. Given that the linear speed of the jet is 60.0 m/s and the diameter of the tire is 0.850 m, we can calculate the radius of the tire as half of the diameter:

r = diameter / 2 = 0.850 m / 2 = 0.425 m

Now we can rearrange the formula to solve for angular speed:

ω = v / r

ω = 60.0 m/s / 0.425 m

ω = 141.18 rad/s

To convert this value into revolutions per minute (rev/min), we use the conversion factors 1 rev = 2π rad and 1 min = 60 s. The angular speed in rev/min is therefore:

ω = 141.18 rad/s × (1 rev / 2π rad) × (60 s / 1 min)

ω = 1347.2 rev/min