What Is the Torque Required to Rotate a Disc Between Two Static Discs?

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SUMMARY

The discussion centers on calculating the torque required to rotate a disc situated between two static discs. Richard presents a scenario involving two identical discs with an inner diameter of 0.095m, outer diameter of 0.105m, and a coefficient of friction of 0.4, held together by a force of 10kN. He correctly calculates the torque needed to initiate rotation as 200Nm using the formula (coefficient of friction x radius x force). When a third disc is introduced, the consensus confirms that the torque required to rotate the middle disc is indeed 400Nm, as the resistances from both contact faces add together.

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  • Understanding of static friction and torque calculations
  • Familiarity with basic engineering mechanics
  • Knowledge of disc dimensions and their impact on torque
  • Experience with force application in mechanical systems
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RichyRich85
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Hi everyone

This is my first post on the forum. I work in engineering and use this forum to answer questions from time to time.

I am looking for confirmation of my understanding of static rotation torque (not something I deal with in day to day work).

Imagine 2 identical discs or washers that can rotate. The i/d is 0.095m and o/d is 0.105m. The coefficient of friction is 0.4. They are held together by an invisible force of 10kN. I believe the torque required to start rotation is (coefficient of friction x radius x force) 0.4x0.05×10000=200Nm

Next scenario: A third identical disc is inserted between the other two, and the two outer discs are unable to rotate. The same invisible 10kN force is applied on both of the outer discs. Would the torque required to rotate the middle disc be the same 200Nm, or would it be the sum of the 2 contact faces, I.e. 400Nm. I believe the latter, but just wanted to check.

Thanks in advance.
Richard
 
Engineering news on Phys.org
Yes, the two torque resistances will add.
 
haruspex said:
Yes, the two torque resistances will add.

Thank you haruspex. I appreciate the feedback.
 

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