How to work out frictional torque

[motoxYogi]

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?

 

[Doc Al]

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.

 

[motoxYogi]

Yes. Dimensionally, radians have no significance.

I apologize for wasting everyones time then. Sorry

 

[asiwrasse]

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.

 

[asiwrasse]

should it not be the radius of the pulley, 300mm? that we are multiplying by 4.685?

 

[Doc Al]

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.

should it not be the radius of the pulley, 300mm? that we are multiplying by 4.685?

Yes.

 

[asiwrasse]

excellent, no confusion now, thanks for the reply.

 

[saucysaunders]

Torque in rope = 4.685 * 0.300 = 1.4055

Frictional Torque therefore = 1.4055 - 0.199935 = 1.21 Nm