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define-viscosity-and-stokes-law

๐Ÿš€ Viscosity is a measure of a fluid's resistance to flow. It describes how thick or thin a fluid is, which affects how easily it can move. For example, honey has a higher viscosity than water, meaning it flows more slowly. Stokes' Law relates to the motion of spheres through a viscous fluid, providing a formula to calculate the drag force experienced by the sphere due to viscosity. This law is particularly useful in understanding the behavior of small particles in fluids, such as in sedimentation processes.

Theory Explanation

Understanding Viscosity

Viscosity is defined as the internal friction in a fluid that resists its flow. It can be thought of as a measure of how 'sticky' a fluid is. The higher the viscosity, the more force is required to move the fluid. Viscosity is affected by temperature; as temperature increases, viscosity typically decreases for liquids.

\[ \eta = \frac{F}{A \cdot \frac{du}{dy}} \]
Stokes' Law

Stokes' Law describes the force of viscosity on a sphere moving through a viscous fluid. The law states that the drag force (F) acting on a sphere is directly proportional to the radius of the sphere (r), the velocity of the sphere (v), and the viscosity of the fluid (ฮท). The formula is given by: F = 6ฯ€ฮทrv, where F is the drag force, ฮท is the viscosity, r is the radius of the sphere, and v is the velocity of the sphere.

\[ F = 6\pi \eta r v \]

Key Points

  • ๐ŸŽฏ Viscosity measures a fluid's resistance to flow.
  • ๐ŸŽฏ Higher viscosity means a thicker fluid that flows more slowly.
  • ๐ŸŽฏ Stokes' Law applies to small spheres moving through a viscous fluid.
  • ๐ŸŽฏ The drag force is proportional to the radius, velocity, and viscosity.
  • ๐ŸŽฏ Temperature affects viscosity, typically decreasing it for liquids as temperature increases.

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Examples:💡

Calculate the drag force on a sphere of radius 0.01 m moving at a velocity of 0.5 m/s through a fluid with a viscosity of 0.89 Pa.s.

Solution:

Step 1: Identify the values: r = 0.01 m, v = 0.5 m/s, ฮท = 0.89 Pa.s.

Step 2: Use Stokes' Law: F = 6ฯ€ฮทrv = 6 ร— ฯ€ ร— 0.89 ร— 0.01 ร— 0.5.

\[ F = 6 \times \pi \times 0.89 \times 0.01 \times 0.5 \]

Step 3: Calculate the drag force: F โ‰ˆ 0.0279 N.

A small ball of radius 0.005 m falls through a viscous liquid with a viscosity of 0.1 Pa.s. If the ball reaches a terminal velocity of 0.2 m/s, calculate the drag force acting on the ball.

Solution:

Step 1: Identify the values: r = 0.005 m, v = 0.2 m/s, ฮท = 0.1 Pa.s.

Step 2: Use Stokes' Law: F = 6ฯ€ฮทrv = 6 ร— ฯ€ ร— 0.1 ร— 0.005 ร— 0.2.

\[ F = 6 \times \pi \times 0.1 \times 0.005 \times 0.2 \]

Step 3: Calculate the drag force: F โ‰ˆ 0.0006 N.

Common Mistakes

  • Mistake: Confusing viscosity with density; students often think thicker fluids are denser, but viscosity is about flow resistance, not mass per volume.

    Correction: Remember that viscosity relates to how easily a fluid flows, while density is about mass per unit volume.

  • Mistake: Forgetting to convert units when applying Stokes' Law; students may use inconsistent units for radius, velocity, or viscosity.

    Correction: Always check that all units are consistent (e.g., SI units) before performing calculations.

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