Problem 6-20: The centripetal acceleration of the commercial jet’s tires, and the force of a determined bacterium in it

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 1.00×10−15 kg bacterium cling to the rim?

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

Solution:

We are given the following quantities: linear speed, v=60.0 m/s v=60.0 \ \text{m/s}, radius is half the diameter, r=0.425 m r=0.425 \ \text{m}.

Part A

We can compute the angular velocity based on the given using the formula, ω=vr \displaystyle \omega = \frac{v}{r}.

ω=vrω=60.0 m/s0.425 mω=141.1765 rad/sec\begin{align*} \omega & = \frac{v}{r} \\ \\ \omega & = \frac{60.0 \ \text{m/s}}{0.425 \ \text{m}} \\ \\ \omega & = 141.1765 \ \text{rad/sec} \end{align*}

Now, we can convert this into the required unit of rev/min.

ω=141.1765 radsec×1 rev2π rad×60 sec1 minω=1348.1363 rev/minω=1.35×103 rev/min  (Answer)\begin{align*} \omega & = 141.1765\ \frac{\text{rad}}{\text{sec}} \times \frac{1\ \text{rev}}{2\pi\ \text{rad}} \times \frac{60\ \text{sec}}{1\ \text{min}} \\ \\ \omega & = 1348.1363 \ \text{rev/min} \\ \\ \omega & = 1.35 \times 10^{3} \ \text{rev/min} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The centripetal acceleration at the edge of the tire can be computed using the formula, ac=rω2 a_{c} = r \omega ^{2}.

ac=rω2ac=(0.425 m)(141.1765 rad/sec)2ac=8470.5918 m/s2ac=8.47×103 m/s2  (Answer)\begin{align*} a_{c} & = r \omega ^2 \\ \\ a_{c} & = \left( 0.425\ \text{m} \right) \left(141.1765\ \text{rad/sec} \right)^2 \\ \\ a_{c} & = 8470.5918 \ \text{m/s}^2 \\ \\ a_{c} & = 8.47 \times 10 ^{3} \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

From the second law of motion, the force is equal to the product of the mass and the acceleration. In this case, we are going to use the formula, Fc=mac F_c = m a_c . We are given the mass to be m=1.00×10−15 kgm=1.00 \times 10 ^{-15}\ \text{kg} , and the centripetal acceleration is solved in Part B.

Fc=macFc=(1×10−15 kg)(8470.5918 m/s2)Fc=8.4705918×10−12 kg m/s2Fc=8.47×10−12 N  (Answer)\begin{align*} F_c & = ma_c \\ \\ F_c & = \left( 1 \times 10^{-15}\ \text{kg}\right) \left(8470.5918 \ \text{m/s}^2\right) \\ \\ F_c & = 8.4705918 \times 10 ^{-12}\ \text{kg m/s}^2 \\ \\ F_c & = 8.47 \times 10^{-12} \ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

The ratio of this force, Fc F_c to the weight of the bacterium is

Fcmg=8.4705819×10−12 N(1×10−15kg)(9.81 m/s2)Fcmg=863.4640Fcmg=863  (Answer)\begin{align*} \frac{F_c}{mg} & = \frac{8.4705819 \times 10 ^{-12}\ \text{N}}{\left( 1 \times 10^{-15} \text{kg} \right)\left(9.81 \ \text{m/s}^2 \right)} \\ \\ \frac{F_c}{mg} & = 863.4640 \\ \\ \frac{F_c}{mg} & = 863 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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