Resistive Force Proportional to Object Speed Squared

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SUMMARY

The resistive force acting on objects moving at high speeds through air is modeled by the equation $$\sf R=\tfrac12D\rho Av^2$$, where $$\sf D$$ is the drag coefficient, $$\rho$$ is the air density, and $$\sf A$$ is the cross-sectional area. The drag coefficient varies based on the object's shape, with spherical objects typically having a value of about 0.5 and irregular shapes reaching up to 2. The factor of 1/2 in the equation is derived from the Bernoulli equation, which explains why the resistive force is proportional to the square of the speed ($$R \propto v^2$$) rather than linearly ($$R \propto v$$).

PREREQUISITES
  • Understanding of fluid dynamics principles
  • Familiarity with the Bernoulli equation
  • Knowledge of drag coefficients and their significance
  • Basic physics concepts related to motion and forces
NEXT STEPS
  • Study the derivation of the drag equation in detail
  • Explore the application of the Bernoulli equation in fluid dynamics
  • Investigate the impact of different shapes on drag coefficients
  • Learn about computational fluid dynamics (CFD) simulations for drag analysis
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Physics students, aerospace engineers, automotive designers, and anyone interested in understanding the dynamics of high-speed motion through fluids.

Anama Skout
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Straight from my physics textbook:

For objects moving at high speeds through air, such as airplanes, skydivers, cars, and baseballs, the resistive force is reasonably well modeled as proportional to the square of the speed. In these situations, the magnitude of the resistive force can be expressed as $$\sf R=\tfrac12D\rho Av^2\tag{6.7}$$ where ##\sf D## is a dimensionless empirical quantity called the drag coefficient, ##\sf r## is the density of air, and ##\sf A## is the cross-sectional area of the moving object measured in a plane perpendicular to its velocity. The drag coefficient has a value of about ##\sf 0.5## for spherical objects but can have a value as great as ##\sf 2## for irregularly shaped objects.​

I want to know how can one - formally - derive ##\sf eq.(6.7)##? (they didn't show a derivation for that formula) And what's the intuition behind it? For instance why is ##\sf R\propto v^2## and not ##\sf R\propto v##? Where does that ##\sf \frac12## coefficient comes from?
 
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Anama Skout said:
Straight from my physics textbook:

For objects moving at high speeds through air, such as airplanes, skydivers, cars, and baseballs, the resistive force is reasonably well modeled as proportional to the square of the speed. In these situations, the magnitude of the resistive force can be expressed as $$\sf R=\tfrac12D\rho Av^2\tag{6.7}$$ where ##\sf D## is a dimensionless empirical quantity called the drag coefficient, ##\sf r## is the density of air, and ##\sf A## is the cross-sectional area of the moving object measured in a plane perpendicular to its velocity. The drag coefficient has a value of about ##\sf 0.5## for spherical objects but can have a value as great as ##\sf 2## for irregularly shaped objects.​

I want to know how can one - formally - derive ##\sf eq.(6.7)##? (they didn't show a derivation for that formula) And what's the intuition behind it? For instance why is ##\sf R\propto v^2## and not ##\sf R\propto v##? Where does that ##\sf \frac12## coefficient comes from?

Here is the derivation of the drag equation:

http://physics.info/drag/

The factor of 1/2 falls out from the Bernoulli equation.
 
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