How to work out frictional torque

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The discussion focuses on calculating frictional torque in a system involving a suspended mass and a flywheel. Participants work through various equations to determine linear and angular acceleration, tension in the rope, and net torque. They clarify the distinction between linear and angular acceleration and emphasize the importance of applying Newton's laws correctly to find the frictional torque. Ultimately, the calculations lead to a frictional torque value of approximately 0.793 Nm, with participants confirming the correctness of their approach and calculations. The conversation highlights the complexities involved in understanding rotational dynamics and the interplay of forces.
  • #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.
 
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  • #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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