Many bacteria exhibit a “run and tumble” motion, swimming in straight lines unti

Question

Many bacteria exhibit a “run and tumble” motion, swimming in straight lines until they stop themselves. They then rotate (“tumble”) for a period of time before moving in a straight line in a new, random direction. This allows them to randomly seek out different directions. Eventually, they run into food. We are going to simplify the situation and analyze it from a physics perspective. Let an E. Coli bacterium be a cylinder. We’ll actually be looking at the motion of two of them. E.coli#1 is 1 micron in diameter and 10 microns in length. E.coli#2 has a diameter of 1 micron and is 2 microns in length. In their “running” state, they move with a speed of 30 microns per second. Their density is 1.15 g/mL (1150 kg/m3 ).

a) What is the mass of E.coli#1 and the mass of E.coli#2?

b) What is the moment of inertia of each? Assume they are rotating about their centers of mass in a “windmill”-like rotation.

c) When they stop and rotate, assume all of the translational energy is transformed into rotational energy. What is the angular velocity of each bacteria during the “tumble” state?

For reference these are the answers but they should only be used for reference. Work must be shown for each part.

a) 9.03x10^-15, 1.8x10^-15

b) 7.1x10^-28, 7.6x10^-26

c) Roughly 10, 50

Explanation / Answer

r1 = 0.5µm=0.5* 10^-6m, r2 = 0.5µm= 0.5*10^-6m, h1=10µm= 10^-5m, L2=2µm= 2*10^-6m, =1150kg/m^3

a) V1= r1^2h1 = 3.14*(0.5*10^-6)^2*10^-5 = 7.85*10^-18 m^3

m1=*V1=1150*7.85*10^-18 = 9.03*10^-15 kg

V2= r2^2h2 = 3.14*(0.5*10^-6)^2*2*10^-5 = 1.57*10^-18 m^3

m2=*V2=1150*1.57*10^-18 = 1.8*10^-15 kg

b) I1= m1/12(3r1^2+h1^2) = (9.03*10^-15)/2*(3(0.5* 10^-6)^2+(10^-5)^2) = 7.6*10^-26 kg*m^2

I2= m2/12(3r2^2+h2^2) = (1.8*10^-15)/2*(3(0.5* 10^-6)^2+(2*10^-6)^2) = 7.1*10^-28 kg*m^2

c) w1=v1/r1 = (30*10^-6)/( 0.5*10^-6) = 60 rad/s

w2=v2/r2 = (30*10^-6)/( 0.5*10^-6) = 60 rad/s

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