How to work out frictional torque

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Homework Help Overview

The discussion revolves around calculating frictional torque in a system involving a flywheel and a suspended mass. The mass of 0.5 kg is released and falls a distance of 0.5 m in 1.5 seconds, prompting calculations related to tension in the rope, angular acceleration, and the effects of friction.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various calculations for linear and angular acceleration, tension in the rope, and the relationship between forces acting on the mass and the flywheel.
  • Some question the definitions and types of acceleration needed for their calculations, while others discuss the application of Newton's laws to the system.
  • There is a focus on understanding how to derive frictional torque from net torque and the forces involved.

Discussion Status

The discussion is ongoing, with participants sharing calculations and seeking clarification on concepts. Some have provided guidance on applying Newton's laws, while others are still working through the relationships between different variables in the problem.

Contextual Notes

Participants note potential confusion regarding the definitions of linear versus angular acceleration and the role of gravity in their calculations. There is also mention of the need to consider the kinetic energy of the flywheel in energy conservation discussions.

  • #31


Doc Al said:
Torque = I*alpha, where torque units are N-m and alpha units are rad/s2.

What you are saying is that

N.m = kg.m2(rad/s2)
or(kg.m/s2)(m) = kg.m2(rad/s2)

or am I missing something?
 
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  • #32


motoxYogi said:
What you are saying is that

N.m = kg.m2(rad/s2)
or(kg.m/s2)(m) = kg.m2(rad/s2)

or am I missing something?
Yes. Dimensionally, radians have no significance.
 
  • #33


Doc Al said:
Yes. Dimensionally, radians have no significance.

I apologize for wasting everyones time then. Sorry
 
  • #34


saucysaunders said:
I was confusing myself, by ignoring the fact that the wheel turning actually makes the tension in the rope less. It's all coming together now I think...

Tension in the rope Ft = (0.5* -0.44) + (0.5*9.81) = 4.685N
Torque in the rope = 4.685 * 0.212 = 0.993Nm

So Frictional Torque = 0.993 - 0.2 = 0.793Nm (Lower than previous answer as I said before)


New to the forum thing! Working through the problem I am following it as far as the tension in the rope, but am a little confused by the next line - torque in the rope, which is obviously the tension * the radius of gyration, why are we multiplying these two together? Any help appreciated.
 
Last edited by a moderator:
  • #35


should it not be the radius of the pulley, 300mm? that we are multiplying by 4.685?
 
Last edited:
  • #36


asiwrasse said:
Working through the problem I am following it as far as the tension in the rope, but am a little confused by the next line - torque in the rope, which is obviously the tension * the radius of gyration, why are we multiplying these two together? Any help appreciated.
Oops. That's an error, of course.

asiwrasse said:
should it not be the radius of the pulley, 300mm? that we are multiplying by 4.685?
Yes.
 
  • #37


excellent, no confusion now, thanks for the reply.
 
  • #38


Torque in rope = 4.685 * 0.300 = 1.4055

Frictional Torque therefore = 1.4055 - 0.199935 = 1.21 Nm
 

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