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

Textbook Solutions for College Physics

Chapter 6 Problem 20

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 \(1.00 \times 10^{-15}\) kg bacterium cling to the rim?

(d) Take the ratio of this force to the bacterium’s weight.

Solution

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

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Jet Takeoff: Tire Rotation, Centripetal Acceleration, and Bacterium's
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Discover the dynamics of a commercial jet takeoff in this video! We explore tire rotation speed, centripetal acceleration, and a bacterium's grip on the rim. Dive into physics and microscopic forces, all while considering weight ratios.


Title College Physics  1 
Author Paul Peter Urone, Roger Hinrichs
ISBN 9781938168000

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

Chapter 6 textbook questions

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