
#1
Nov2910, 09:00 PM

P: 115

1. The problem statement, all variables and given/known data
Bacteria use swimming to seek out food. Imagine that the bacterium is in a region of low food concentration. For the bacterium to profit from swimming to a region with more food, it has to reach there before diffusion of food molecules makes the concentrations in the two regions the same. Here we find the smallest distance that a bacterium needs to swim so it can outrun diffusion. (a) ... (b) ... (c) Estimate the number of ATP molecules the bacterium must consume (hydrolyze) per second in order to travel at this speed [speed of bacterium], assuming that all of the energy usages goes into overcoming fluid drag. The amount of energy released from one ATP molecule is approximately 20 kT. Note that the bacterial flagellar motor is actually powered by a proton gradient and this estimate focuses on the ATP equivalents associated with overcoming fluid drag. 2. Relevant equations Speed of bacterium = 30µm/s Diffusion constant of food molecule = 500µm^2 / s Propulsive force: Fp = 2πnLvcos(theta)sin(theta) where n is the viscosity of water, L is the length of the flagellum with L = 10µm, theta is the angle at which a small segment of the flagellum moves with respect to the direction of motion of the bacterium, v is the speed of a section of the flagellum perpendicular to the direction of motion of the bacterium, and v = πDf, where D = 0.5µm and f = 100Hz Speed of bacterium: V = vsin(theta)cos(theta) Also, tan(theta) = (πD)/P, where P = 2µm. 3. The attempt at a solution I can't quite figure what to do with sin(theta)cos(theta).... But really, I can't figure how to move from these equations to energy/time. I thought that maybe F = (gradient)U could be helpful, but when I thought of how to apply that, I wasn't sure what to take F with respect to.... theta seems to be the only value to change in the Fp equation, but even if I found the potential energy, how would I move to something that is energy per time? Thanks! 


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