Energy Needed Per Second to Sustain a Velocity through a Liquid

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SUMMARY

The discussion focuses on calculating the energy required for a bacterium to swim through a liquid, specifically to outrun diffusion. The speed of the bacterium is given as 30µm/s, and the diffusion constant for food molecules is 500µm²/s. The propulsive force equation is defined as Fp = 2πnLvcos(theta)sin(theta), where viscosity, flagellum length, and movement angle are critical variables. The challenge lies in converting these equations into a calculation of ATP consumption per second, with one ATP molecule providing approximately 20 kT of energy.

PREREQUISITES
  • Understanding of fluid dynamics, specifically drag forces in liquids.
  • Familiarity with bacterial locomotion mechanics and flagellar function.
  • Knowledge of basic thermodynamics, particularly energy conversion from ATP.
  • Proficiency in algebra and trigonometry to manipulate equations involving sin(theta) and cos(theta).
NEXT STEPS
  • Research the relationship between power, force, and velocity in physics to apply it to biological systems.
  • Explore the role of viscosity in fluid dynamics, particularly in relation to bacterial movement.
  • Study the mechanics of flagellar propulsion in bacteria to understand the impact of angle and speed on energy consumption.
  • Investigate the conversion of chemical energy from ATP into mechanical work in microorganisms.
USEFUL FOR

Students studying microbiology, biophysics researchers, and anyone interested in the mechanics of bacterial locomotion and energy expenditure in biological systems.

Oijl
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Homework Statement


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.


Homework 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.


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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There is an equation relating power, force and velocity. It should be in any introductory physics book, in the chapter that deals with work, energy, and power.

Hope that helps. Not sure what to do about theta, without further information.
 

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