[hamdani yusuf (OP)]

Any professors I would have known would be long dead at this stage: example, I knew professor Wesley Cocker who worked at ICI on the development of polymethyl methacrylate(Perspex) and he was fairly old 55years ago.

You can ask any currently active professor. There should be many in Cambridge alone.

[hamdani yusuf (OP)]

Dimensional analysis says that the volume of a sphere "should" be r^3, but it's not.
In a similar way, the energy change due to a single turn with a torque of 1 n m "should" be 1 joule, but it's not.

That's a limitation of dimensional analysis. It ignores constants.
On the other hand, unit analysis doesn't ignore constants. It can give the right answers if you use the right units and their conversion factors.

[alancalverd]

When the object doesn't move, then the acceleration is zero. So does the rotational acceleration. The net torque must be zero.

But I have hung a weight on the end of a lever, so I am exerting a torque that wasn't there before.

Now let's take my 1N weight and 1m lever  and attach them to  two different spiral springs. Spring A is very stiff and deflects 0.1 rad. Spring B is much more flexible and deflects 1 rad. Hamdani says this is not possible because I have applied a torque of 1 Nm per rad so they must both deflect by 1 rad!

[alancalverd]

Dimensional analysis says that the volume of a sphere "should" be r^3, but it's not.

Beg to differ, my friend. Dimensional analysis gives [volume] = L³ where the brackets indicate "the dimensions of", not "the size of", and the equation applies to the volume of anything of any shape or even no shape.

Where an object is sufficiently symmetrical that it can be completely characterised by a single parameter, dimensional analysis says that its volume will be a multiple of the cube of that parameter, but DA alone doesn't specify the multiplier. In my miss-spent youth as a crystallographer I even knew the multipliers for several regular polyhedra!

[hamdani yusuf (OP)]

When the object doesn't move, then the acceleration is zero. So does the rotational acceleration. The net torque must be zero.

But I have hung a weight on the end of a lever, so I am exerting a torque that wasn't there before.

No. You are exerting a force. When no rotation occurs, you have no rotation radius. Thus no torque.
The length of the lever is not necessarily equivalent to the rotation radius.

The rotation radius must be equal to the arc length of the rotation divided by rotational angle. But if both numerator and denominator are zero, you get 0/0, which is undefined.

[hamdani yusuf (OP)]

Now let's take my 1N weight and 1m lever  and attach them to  two different spiral springs. Spring A is very stiff and deflects 0.1 rad. Spring B is much more flexible and deflects 1 rad. Hamdani says this is not possible because I have applied a torque of 1 Nm per rad so they must both deflect by 1 rad!

Your error is assigning the meter to the length of the lever.
In the unit I proposed, the meter should be assigned to the arc length of the circular trajectory of the applied force.

[hamdani yusuf (OP)]

τ = I α

as {τ} = ML²T^-2 and {I} = ML², α must be in rad/sec²

You can use standard units. But you can also use non-standard units, as long as you are consistent with their usage and the conversion factors.
Let's say the torque is 1 N.m/rotation, and the rotational inertia is 1 kg.m^2/rotation^2. The rotational acceleration is 1 rotation/second^2.
If the torque is applied for 1 second, the rotational velocity of the object will be 1 rotation per second.

In standard unit, the torque above equals 1/(2π) N.m/rad
The rotational inertia is 1/(2π)^2 kg.m^2/rad^2.
The rotational acceleration is 2π rad/second^2.
If the torque is applied for 1 second, the rotational velocity of the object will be 2π rad/second.

I hope this example is simple enough for you to follow my reasoning which concludes that torque is force times distance of rotation divided by rotational angle. This equation works generally, no matter which units of angle that you choose. A complete rotation is a commonly used alternative, especially when dealing with multiple rotation. On the other hand, when dealing with small rotational angle, we can use degree, arc minute, or arc second as the measuring unit.

[alancalverd]

I simply stated what  I did: attach a weight to a lever. That is the applied torque. Then two very different things happened, depending on the spring stiffness.

You now seem to want to use a perfectly well-understood word to describe something completely different - the resulting movement, if any, or maybe the countertorque of the other object attached to the lever.  Problem here is that, even if we physicists allowed you to misappropriate our word, you'd have a problem applying it equally to (a) objects that don't move, (b) springs or other resilient and conservative devices, (c) objects that move with energy loss (eg subject to friction) and (d) objects that are free to rotate and accelerate.

If you want to re-use torque to describe the effect, you will have to invent a new word for the applied cause, and then convince everyone to use it.

[hamdani yusuf (OP)]

Let's say the torque is 1 N.m/degree, and the rotational inertia is 1 kg.m^2/degree^2. The rotational acceleration is 1 degree/second^2.
If the torque is applied for 1 second, the rotational velocity of the object will be 1 degree per second.

The exact same case, but the rotational angle is measured in radian will give following numbers.
torque is 180/π N.m/radian, and the rotational inertia is (180/π)^2 kg.m^2/radian^2. The rotational acceleration is π/180 radian/second^2.
If the torque is applied for 1 second, the rotational velocity of the object will be π/180 radian per second.

[alancalverd]

So now you want to use torque to describe the cause, not the effect? Please make up your mind!

[hamdani yusuf (OP)]

I simply stated what  I did: attach a weight to a lever. That is the applied torque. Then two very different things happened.

No. You are exerting a force. When no rotation occurs, you have no rotation radius. Thus no torque.
The length of the lever is not necessarily equivalent to the rotation radius.

The rotation radius must be equal to the arc length of the rotation divided by rotational angle. But if both numerator and denominator are zero, you get 0/0, which is undefined.

[hamdani yusuf (OP)]

So now you want to use torque to describe the cause, not the effect? Please make up your mind!

You seem to forget Newton's third law of motion.

[alancalverd]

No. You are exerting a force. When no rotation occurs, you have no rotation radius. Thus no torque.
The length of the lever is not necessarily equivalent to the rotation radius.

Poppycock! The torque wrench measures torque whether or not the bolt turns. A stalled motor exerts torque (it's called "stalling torque").

If the system does rotate, the distance from the action point of the force to the center of rotation is both the lever arm and the rotation radius.

[alancalverd]

The rotation radius must be equal to the arc length of the rotation divided by rotational angle

Cart before horse! The rotation angle (in radians) is the arc length divided by the radius. Arc is an effect, not a cause. 

[hamdani yusuf (OP)]

The rotation radius must be equal to the arc length of the rotation divided by rotational angle

Cart before horse! The rotation angle (in radians) is the arc length divided by the radius. Arc is an effect, not a cause. 

The equation works both ways. If A equals B, then B must be equal to A. Suggesting otherwise only creates a contradiction. Why would you?

[hamdani yusuf (OP)]

Poppycock! The torque wrench measures torque whether or not the bolt turns.

You forget that the wrench lever must turn against its head for any non-zero reading. Except you are using a broken torque wrench.

[hamdani yusuf (OP)]

If the system does rotate, the distance from the action point of the force to the center of rotation is both the lever arm and the rotation radius.

Not necessarily. See my example with a work bench.

Can you do it without any change in position of the lever?

In principle, yes. In practice, any real lever will bend a bit (indeed some torque wrenches use the bending to measure the tporque), but the applied torque is independent of the elasticity of the lever: whether you use a rigid bar or a flexible one, torque is just the product of force x distance.

Here's a thought experiment in a workshop. We wanted to release a bolt from a corroded valve. It's clamped on a bench using a vise. A large wrench was used in an attempt to turn the bolt to release it. After a force was applied, the bolt didn't turn. It moved the whole bench instead.
It reminds you the definition of torque in terms of cross product between force and radius of rotation. It's not the radius of the object.

Here's the diagram for simplified version of the case.

The length of the wrench only represents an expected value for the radius of rotation. But the actual rotation in this case is around the bottom of left leg of the bench. The radius that should be used to calculate torque is the radius of actual rotation.
Let me remind you that unexpected results come from false assumptions.

Now, in other case where there is no actual rotation, even so slightly, what is the radius of rotation that you will use to calculate torque?

[hamdani yusuf (OP)]

What will happen if the weight on the right side is increased to 20N?

The beam would rotate clockwise. So what?

What would be the normal forces at each fulcrum?

What if the direction of the force applied to the right end of the lever is upward instead?

To answer this question, you need to calculate the rotational inertia of the system first, because it becomes a dynamic system.

To clarify, the formulas at the right side of the animated picture show that rotational inertia is proportional to the length of the thin rod squared. The poor font choice makes it hard to distinguish between upper case i and lower case l.
But they also show that rotational radius isn't always the same as object length, nor object radius.

[Bored chemist]

Poppycock! The torque wrench measures torque whether or not the bolt turns.

You forget that the wrench lever must turn against its head for any non-zero reading. Except you are using a broken torque wrench.

It's perfectly possible to arrange for that not to happen.

[Bored chemist]

Beg to differ, my friend. Dimensional analysis gives [volume] = L3 where the brackets indicate "the dimensions of", not "the size of", and the equation applies to the volume of anything of any shape or even no shape.

I should have specified that this is the case in a consistent set of units (such as SI).

In principle, you only need odd constants because of things like integrals or if you use inconsistent units like Calories instead of Joules.

My point was that you always end up with odd constants in some cases.
And a couple of hundred years back, they proved that rotational mechanics was one area where you would always need an irrational constant.
Most of us cope with it just fine.
Hamdani seems to want to eliminate it.
I wonder if he will also try to do the same for the reactance of inductors and capacitors.