[hamdani yusuf (OP)]

Angular distance or angular separation is the measure of the angle between the orientation of two straight lines

In other words, angle. Why use two long words when the meaning is conveyed by one short word? 

Worse still, "angular distance" implies some measure of length such as 2r sin (θ/2) or rθ.

In rotational motion caused by a force, the torque is also affected by the angle between the direction of force and the rotating radius, which is complementary to the angle between the direction of force and direction of motion. You need to specify which angle you are referring to.

[hamdani yusuf (OP)]

So if I apply a torque to a bolt that doesn't move, α= 0 so τ= 0? My torque wrench thinks not.

If your torque wrench still shows 0, then it hasn't received any torque yet.
Otherwise, your torque wrench must have rotated somewhat.

[hamdani yusuf (OP)]

And there seems to be some inconsistency here. Or maybe not?

If the bolt doesn't turn, then its net torque is zero. The torque exerted by your torque wrench is canceled out by the torque exerted by the thread.

[hamdani yusuf (OP)]

In practice, lever might be the simplest device employing torque. But they also involve many factors that increase its complexity to analyze, which make them conceptually more complex, such as angle between force and rotational motion, normal force by the floor, also restricted rotational distance. Torque wrench must deal with spring constant, material plasticity and friction.
A better way to analyze torque without additional complexity is by using coaxial pulley.

[alancalverd]

If the bolt doesn't turn, then its net torque is zero. The torque exerted by your torque wrench is canceled out by the torque exerted by the thread.

Quite so. The I disconnect the torque wrench and the bolt magically unscrews itself?

[paul cotter]

That's hat I have always sought, rusty seized bolts that unscrew themselves. So I just torque it, remove the tool and hey presto it unwinds??

[hamdani yusuf (OP)]

If the bolt doesn't turn, then its net torque is zero. The torque exerted by your torque wrench is canceled out by the torque exerted by the thread.

Quite so. The I disconnect the torque wrench and the bolt magically unscrews itself?

You push a box with a force, but it doesn't accelerate. If you stop pushing, will it accelerate to the opposite direction?
Friction seems to have you confused.
Let's not repeat Aristotle's mistake in describing motion.

[hamdani yusuf (OP)]

In practice, lever might be the simplest device employing torque. But they also involve many factors that increase its complexity to analyze, which make them conceptually more complex, such as angle between force and rotational motion, normal force by the floor, also restricted rotational distance. Torque wrench must deal with spring constant, material plasticity and friction.
A better way to analyze torque without additional complexity is by using coaxial pulley.

Let's analyze idealized coaxial pulley as shown in the picture. Consider the mass of the pulley and the strings are negligible. Friction of the bearing as well as the air are negligible too. The strings don't slip with the pulley. Gravitational acceleration is 10 m/s^2, which makes each 1 kg weight A, B, C, and D exert 10 Newton of force to the strings. Inner pulley is 1 meter in radius, while outer pulley is 2 meter. The length of the strings are long enough to allow for multiple turns of the pulley.
Rotational inertia of the system is 2.1^2 + 2.2^2 = 2+8 = 10 kg.m^2/rad^2.

[hamdani yusuf (OP)]

When no additional force is applied, the net torque is zero, thus rotational acceleration is zero. Its rotational velocity is constant, but not necessarily zero.
Newton's first law of motion is basically a special case of the second law. But it's important to mention first to dispell previously widespread misconception that all motion will eventually stop even without external influence.

[hamdani yusuf (OP)]

In practice, lever might be the simplest device employing torque. But they also involve many factors that increase its complexity to analyze, which make them conceptually more complex, such as angle between force and rotational motion, normal force by the floor, also restricted rotational distance. Torque wrench must deal with spring constant, material plasticity and friction.
A better way to analyze torque without additional complexity is by using coaxial pulley.

Let's analyze idealized coaxial pulley as shown in the picture. Consider the mass of the pulley and the strings are negligible. Friction of the bearing as well as the air are negligible too. The strings don't slip with the pulley. Gravitational acceleration is 10 m/s^2, which makes each 1 kg weight A, B, C, and D exert 10 Newton of force to the strings. Inner pulley is 1 meter in radius, while outer pulley is 2 meter. The length of the strings are long enough to allow for multiple turns of the pulley.
Rotational inertia of the system is 2.1^2 + 2.2^2 = 2+8 = 10 kg.m^2/rad^2.

Let's start with the pulley is initially stopped. Another 1 kg weight is added to B. One second later, another 1 kg weight is added to C. How will the system behave  then?

What if the second weight is added to D, instead of C?

[hamdani yusuf (OP)]

Let's start with the pulley is initially stopped. Another 1 kg weight is added to B. One second later, another 1 kg weight is added to C. How will the system behave  then?

When 1 kg is added to B, there's a net 10 N force downward at B.

It causes net torque 10 Nm/rad to turn the pulley counter clockwise.

It also adds the rotational inertia of the system by 1 kg.m^2/rad^2 to become 11 kg.m^2/rad^2.

Thus the rotational acceleration is 10/11 rad/s^2.

1 second later, the rotational velocity is 10/11 rad/s counter clockwise.

Then another 1 kg weight is added to C. This cancels out net torque to the pulley. With 0 net force, its rotational velocity is constant, which is held at 10/11 rad/s counter clockwise.

Note that for this scenario, the added weight at C has the same velocity as C when it's being added.
If that's not the case, we need to calculate rotational momentum of the system before the addition, also the rotational inertia before and after that addition.

[hamdani yusuf (OP)]

Rotational momentum right before additional weight at C is L = I ω = 11*(10/11) = 10 (kg.m^2)/(rad.s)
If the added weight is initially stop, then it doesn't add its rotational momentum to the system.

But if it's initially move at the same velocity as C, it adds rotational momentum by 1*1^2 * (10/11) = 10/11 (kg.m^2)/(rad.s)
It makes the total rotational momentum = 10+(10/11) = 120/11 (kg.m^2)/(rad.s)

Rotational inertia after added weight at C is 11+(1*1^2) = 12 kg.m^2/rad^2

[hamdani yusuf (OP)]

So, if the added weight at C is initially stationary, the angular velocity of the system becomes
ω = L/I = 10/12 rad/s

While if the added weight at C is initially moving at the same velocity as C, the angular velocity of the system becomes
ω = L/I = (120/11)/12 = 10/11 rad/s, the same as our previous calculation.

[hamdani yusuf (OP)]

What if the second weight is added to D, instead of C?

Just like before, the final state of the system depends on the initial velocity of the added weight. With the formulas shown above, you can just plug the numbers in to get the results.

[hamdani yusuf (OP)]

The problem with removing the unit of rotational angle from the denominator in the unit of torque is that it creates inconsistency with other units of rotational quantities, like rotational distance, velocity, acceleration, inertia, momentum, and energy.
IMO, the cost of this removal enormously outweighs its benefit.

[paul cotter]

NONSENSE!

[Bored chemist]

IMO, the cost of this

And apparently, not in the opinion of anyone else at all, ever.

So, it's not changing any time soon.
Face it.

[alancalverd]

The problem with removing the unit of rotational angle from the denominator in the unit of torque

Nobody has removed it. It was never there. And as we have shown, if you add it, it just makes a nonsense of elementary mechanics.   

[hamdani yusuf (OP)]

The problem with removing the unit of rotational angle from the denominator in the unit of torque

Nobody has removed it. It was never there. And as we have shown, if you add it, it just makes a nonsense of elementary mechanics.   

If it was really never there then I'll just have to add it. The most important quality of a standard is its consistency.
Which elementary mechanics are you referring to?

[hamdani yusuf (OP)]

NONSENSE!

Do you have any argumentation?
Is it just your feeling?
Do you understand my example with coaxial pulley above?
Do you have a different answer?