[Bored chemist]

I see you're ignoring the energy lost to friction,

Friction losses within a clock spring are typically very small.

[hamdani yusuf (OP)]

Thus, we can avoid confusion by...

Thus far, the list of people who are confused by this only seems to have one entry, and that's you.

Which part of my statement made you think that I'm the one who is being confused?

The fact that you think there is confusion.

After the long discussion here, have you found a confusion?
Do you think that the unit of torque is the same as energy?

[Bored chemist]

The units of both are mass length squared per time squared.

If you want to calculate kinetic energy from velocity my multiplying the square of the speed by the mass you have the right dimensions but there's a factor- a half- that you need to multiply by.

If you want to calculate it from a torque and an angle  you have the right dimensions but there's a factor - the angle-  that you need to multiply by.

The 1/2 and the angle are both dimensionless.

[hamdani yusuf (OP)]

The 1/2 and the angle are both dimensionless.

Where does the 1/2 in calculating energy from torque come from?

[Bored chemist]

The 1/2 and the angle are both dimensionless.

Where does the 1/2 in calculating energy from torque come from?

In what circumstances?

[hamdani yusuf (OP)]

The 1/2 and the angle are both dimensionless.

Where does the 1/2 in calculating energy from torque come from?

In what circumstances?

The circumstances you described in reply#42.

[Bored chemist]

Radius is circumference /2 pi

[hamdani yusuf (OP)]

Radius is circumference /2 pi

You were missing the pi in previous post.

[Bored chemist]

Radius is circumference /2 pi

You were missing the pi in previous post.

Not really.
I didn't specify the units in which I measured the angle.
Fundamentally an angle is a ratio.
Whether it's arc to the radius or the arc to the diameter is a matter of convention.
But there's still a constant in there.
And that was the point I was making.
Even with a consistent set of units like the SI, you still end up with constants in your calculations.

[alancalverd]

And here's the flaw in HY's argument. Newton's first law!

Once you have started a circularly symmetric body rolling on a horizontal frictionless surface, no further force is required and no more work is done to keep it rolling. The starting impulse (torque x duration) determines how fast it will roll, depending on its moment of inertia, but an infinitesimal impulse will make it roll indefinitely.

[Halc]

This thread is so full of erroneous comments and outright speculation that I'm moving it to the lighter side.  I mean, 50 posts, and what has been answered?

An error laden quote from Gemini in the OP mentions the Newton-meter, but post 10 (a wiki quote) identifies the Newton-meter (with force being a vector) as the unit of torque.
End of story.  The rest seems to be yet another blog seeking to get off track in as many ways as possible, which is not the intent of the main sections of the forum.

The units of both are mass length squared per time squared.

This is so wrong.  It seems to presume that energy and torque are the same thing, and then applying a relativistic notion of mass-energy equivalence to torque.

Once you have started a circularly symmetric body rolling on a horizontal frictionless surface,

If the surface is frictionless, the wheel will probably just slide and not roll at all. Which way the body rotates (if at all) depends on where relative to the CoM the impulse is applies.

[paul cotter]

As I predicted confusion reigns. I am however completely at a loss to understand any problem with BC's post(#42) where he states the units of torque and of energy? As regards Alan's rolling body: if rotation has been applied and the surface is entirely friction free it will just sit there spinning without any forward/reverse motion.

[hamdani yusuf (OP)]

The "initial kick" refers to releasing the potential energy stored in the tension of the bolt, i.e. the tightening torque.

The "initial kick" refers to static friction. The torque required to keep the bolt/nut rotating is related to kinetic friction.

[hamdani yusuf (OP)]

The units of both are mass length squared per time squared.

If you want to calculate kinetic energy from velocity my multiplying the square of the speed by the mass you have the right dimensions but there's a factor- a half- that you need to multiply by.

If you want to calculate it from a torque and an angle  you have the right dimensions but there's a factor - the angle-  that you need to multiply by.

The 1/2 and the angle are both dimensionless.

Even with a consistent set of units like the SI, you still end up with constants in your calculations.

The 1/2 there in calculating kinetic energy from mass and velocity is a constant,
while the angle in calculating work from torque and the angle of rotation is a variable.

[hamdani yusuf (OP)]

And here's the flaw in HY's argument. Newton's first law!

Once you have started a circularly symmetric body rolling on a horizontal frictionless surface, no further force is required and no more work is done to keep it rolling. The starting impulse (torque x duration) determines how fast it will roll, depending on its moment of inertia, but an infinitesimal impulse will make it roll indefinitely.

Which part of my argument violates Newton's first law?
The formula

still applies in your situation.
(torque x duration) = rotational impulse = change in angular momentum.
To make the body rotate, the duration can't be exactly 0. Thus there must be some change in angular position while the torque is being applied. When the torque is no longer applied, there is no longer change in angular momentum.

[Bored chemist]

The units of both are mass length squared per time squared.

If you want to calculate kinetic energy from velocity my multiplying the square of the speed by the mass you have the right dimensions but there's a factor- a half- that you need to multiply by.

If you want to calculate it from a torque and an angle  you have the right dimensions but there's a factor - the angle-  that you need to multiply by.

The 1/2 and the angle are both dimensionless.

Even with a consistent set of units like the SI, you still end up with constants in your calculations.

The 1/2 there in calculating kinetic energy from mass and velocity is a constant,
while the angle in calculating work from torque and the angle of rotation is a variable.

And pi is a constant.
So what?

[paul cotter]

Argument purely for the sake of argument and not enlightenment.

[hamdani yusuf (OP)]

An error laden quote from Gemini in the OP mentions the Newton-meter,

Can you specify the error that it made?

[hamdani yusuf (OP)]

The units of both are mass length squared per time squared.

If you want to calculate kinetic energy from velocity my multiplying the square of the speed by the mass you have the right dimensions but there's a factor- a half- that you need to multiply by.

If you want to calculate it from a torque and an angle  you have the right dimensions but there's a factor - the angle-  that you need to multiply by.

The 1/2 and the angle are both dimensionless.

Even with a consistent set of units like the SI, you still end up with constants in your calculations.

The 1/2 there in calculating kinetic energy from mass and velocity is a constant,
while the angle in calculating work from torque and the angle of rotation is a variable.

And pi is a constant.
So what?

So you were comparing apple to orange.

[hamdani yusuf (OP)]

As I predicted confusion reigns.

Your post that I quoted below tells one of your source of confusion.

Nonsense. Torque is a force that may or may not produce a rotation. If rotation occurs work is done and then the displacement angle is of significance. In the absence of rotation there is no angle involved and obviously no work.

When a torque produces no angular acceleration, then there must be another torque working in equal magnitude but on opposite direction. It's similar to force and linear acceleration.

We must have learned about it in school.