Question

At takeoff a commercial jet has a 70.0 m/s speed. Its tires have a diameter of 0.400 m.

(a)

At how many rpm are the tires rotating?

rpm

(b)

What is the centripetal acceleration at the edge of the tire?

m/s²

(c)

With what force must a determined 10⁻¹⁵ kg bacterium cling to the rim?

N

(d)

Take the ratio of this force to the bacterium's weight.

(force from part (c)/bacterium's weight)

Answer 1

Given linear velocity of jet = 70.0 m/s thus the linear velocity of tires is also = 70.0 m/s

radius of tire = d/2 = 0.400/2 = 0.200 m

therefore angular velocity, w = v/r = 70/0.2 =350 rad/s

therefore revolution per minute of tire rotation = 60*350/2pi = 3343.94 ~ 3344 rpm[Answer part a]

we have centripetal acceleration formula = v²/r = 70²/0.2 = 24500 rad s^-2[Answer part b]

we also have centripetal force = mass * centripetal acceleration

The bacterium with force it cling on rim must be equal to centripetal force to be prevented from getting thrown away.

Therefore for 10-15 kg, force = 10*24500 to 15*24500 = 245000 to 367500 N[Answer part c]

we know weight = m*g

Therefore force/weight = ma / mg = a/g = 24500/10 = 2450 [Answer part d]