Jet Takeoff: Tire Rotation, Centripetal Acceleration, and Bacterium's

Step 1 of 5

We are required to calculate the angular velocity of the tires, centripetal acceleration, centripetal force on a bacterium and the ratio of the centripetal force on the bacterium to its weight.

Part a

Given data,

Linear speed of the jet, \(v=60.0 \mathrm{m} / \mathrm{s}\)

Diameter of its tire, \(d=0.850 \mathrm{m}\)

Radius, \(r=0.425 \mathrm{m}\)

Angular velocity, \(\omega=\frac{v}{r}\)

\(\begin{array}{l}\omega=\frac{60.0}{0.425 \mathrm{rad} / \mathrm{s}}\\ \omega \approx 141 \mathrm{rad} / \mathrm{s}\end{array}\)

Now, \(1 \mathrm{rev}=2 \pi \text { radians }\)

\(141 \text { radians }=\frac{141}{2 \pi} \text { revolutions }\)

1 min = 60 s

1 s = 1/60 min