Force & Viscosity between two parallel plates

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SUMMARY

The discussion focuses on calculating the torque required to rotate a flat circular disc with a radius of 1.0m over a fixed surface, separated by an oil film with a viscosity of 16 times that of water at 20 degrees Celsius. The key equation used is T/A = viscosity(dv/dy), where T represents shear force, A is the area of the plate, v is the velocity, and y is the gap between the plates. The final calculated torque is 2.05 N*m, derived from the parameters provided, including an angular velocity of 0.65 rad/sec and a gap of 0.5mm.

PREREQUISITES
  • Understanding of fluid dynamics, specifically shear force and viscosity.
  • Familiarity with torque calculations in rotational motion.
  • Knowledge of basic geometry for calculating the area of a circle.
  • Ability to convert angular velocity into linear velocity.
NEXT STEPS
  • Explore the relationship between viscosity and shear stress in fluids.
  • Learn about the effects of gap distance on torque in rotating systems.
  • Investigate the application of Newton's law of viscosity in engineering problems.
  • Study the principles of rotational dynamics and their applications in mechanical systems.
USEFUL FOR

Students in mechanical engineering, physicists studying fluid dynamics, and engineers involved in designing rotating machinery or systems requiring lubrication analysis.

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Homework Statement


A flat circular disc of radius 1.0m is rotated at an angular velocity of 0.65 rad/sec over a fixed flat surface. An oil film separates the disk and the surface. If the viscosity of the oil is 16 times that of water (20 degree C) and the space between the disk and the fixed surface is 0.5mm, what is the torque required to rotate the disk?

Homework Equations



(T/A) = viscosity(dv/dy)

Where,
T= Shear Force
A= Area of the plate
v= velocity
y= gap between plates

The Attempt at a Solution



A= pi(0.5)^2= 0.7854 m^2
v= 0.65 rad/sec
y= 5e-4 m
Viscosity = 16(1.002e-3) = 16.032e-3 N*sec/m^2

How do I turn the rad/sec into shear force and thus torque?

Solving I get T = (0.7854)(16.032e-3)(0.65/5e-4) = 16.37 N*Rad/m

I'm at a loss from here.
 
Last edited:
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Nevermind, I've got it figured out. I went about it completely wrong. The answer I have now is 2.05 N*m.
 

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