[hamdani yusuf (OP)]

Now please answer the question: what force is required for the brake pads to prevent the car rolling down the hill? You may use the data I  gave in reply 509 above.

Let's solve it step by step.
The force downhill experienced by the car is F=m.g.sin θ.
To prevent it from accelerating, you need an opposing force with the same magnitude.
In this case, it's produced solely from friction by braking pad. Other forces are considered insignificant.
The tires are assumed to never slip against the road.

[hamdani yusuf (OP)]

The force to stop the downhill rolling is supposed to be exclusively provided by the brake pad through friction.
F = F_brake . μ
F_brake = F / μ = m.g.sin θ / μ
Usually radius of the brake disc is smaller than the tires. Thus the brake is in mechanical disadvantage, and more force is required to balance out.
F_brake = (m.g.sin θ / μ) (R/r).
If the stopping force is distributed equally to four wheels, then the required force on each wheel is
F_brake = (m.g.sin θ / μ) (R/r) / 4.
Applying force more than minimum requirement doesn't change the result, which is the car doesn't move.

[hamdani yusuf (OP)]

Initial velocity = 0 (the question referred to the parking brake, which you only use once the car has stopped moving)
Permitted rolling distance = 0, obviously.

This is the only condition your definition of torque can work, i.e. where angular acceleration is zero.

On the other hand, the proposed new standard units work well in all conditions, whether or not the angular acceleration is zero, whether or not the angular velocity is zero.

[Bored chemist]

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

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....

...

Imagine there's a crack at the middle of the lever of the wrench. When the force is applied at its end, the wrench breaks at the cracked point, and the actual rotational radius is half the expected radius, which makes the torque that you produce half of what you expected.

If the wrench breaks, the torque falls to zero, not half of anything

[Bored chemist]

The force to stop the downhill rolling is supposed to be exclusively provided by the brake pad through friction.

...
F_brake = (m.g.sin θ / μ) (R/r).F_brake = (m.g.sin θ / μ) (R/r) / 4.
Applying force more than minimum requirement doesn't change the result, which is the car doesn't move.

In what direction does that force (applied to the disks) act?

[alancalverd]

So without acknowledging it, Hamdani has calculated the brake pad force by using the entirely conventional  definition of torque that everyone learned at school. The beauty of the definition is that it is equally useful whether or not there is any angular displacement involved.  QED.

[hamdani yusuf (OP)]

If the wrench breaks, the torque falls to zero, not half of anything

The torque at the head of the wrench is zero because the force is disconnected from that point.
But the torque applied to the broken end part of the wrench is not zero, because the force as well as the rotational angle are not zero.

[hamdani yusuf (OP)]

The force to stop the downhill rolling is supposed to be exclusively provided by the brake pad through friction.

...
F_brake = (m.g.sin θ / μ) (R/r).F_brake = (m.g.sin θ / μ) (R/r) / 4.
Applying force more than minimum requirement doesn't change the result, which is the car doesn't move.

In what direction does that force (applied to the disks) act?

Axially.
The conversion of direction of the force from axial to tangential force of friction is covered by the frictional coefficient of the brake.

[hamdani yusuf (OP)]

So without acknowledging it, Hamdani has calculated the brake pad force by using the entirely conventional  definition of torque that everyone learned at school. The beauty of the definition is that it is equally useful whether or not there is any angular displacement involved.  QED.

Not really. In the case where it works, it's thanks to L'Hopital's Rule which tells us that if we have an indeterminate form 0/0 or ∞/∞ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.
torque = Force . rotational distance / rotational angle
rotational radius = rotational distance / rotational angle
torque = Force . rotational radius

Now please answer the question, what is the radius of rotation, when nothing is rotating?

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?

Imagine there's a crack at the middle of the lever of the wrench. When the force is applied at its end, the wrench breaks at the cracked point, and the actual rotational radius is half the expected radius, which makes the torque that you produce half of what you expected.
Every point in the system has a potential torque as the force is applied, according to the distance from the applied force and the angle between the distance vector and the direction of the force. But the actual torque is defined by the actual axis of rotation, which determines the radius of rotation.

[hamdani yusuf (OP)]

Episode 19. Angular Momentum: An old momentum with a new twist.

?The Mechanical Universe,? is a critically-acclaimed series of 52 thirty-minute videos covering the basic topics of an introductory university physics course.

Each program in the series opens and closes with Caltech Professor David Goodstein providing philosophical, historical and often humorous insight into the subject at hand while lecturing to his freshman physics class. The series contains hundreds of computer animation segments, created by Dr. James F. Blinn, as the primary tool of instruction. Dynamic location footage and historical re-creations are also used to stress the fact that science is a human endeavor.

The series was originally produced as a broadcast telecourse in 1985 by Caltech and Intelecom, Inc. with program funding from the Annenberg/CPB Project.


11:17 Visualization of Johannes Kepler?s three laws of planetary motion.
13:20 Visualization of the Kepler?s law of equal areas (swept by the vector connecting a planet to the sun in equal times)
16:23 Explanation of why planets orbiting the sun and vortices are not subject to twist forces.
17:29 Proving that r x F is the derivative of angular momentum L ; since r x F = 0, L is constant, i.e. angular momentum is conserved; r x F is the twisting force, called torque τ , the rate of change of L
19:13 Using the right-hand rule to calculate L ; visualizing why the orbital speed of a planet increases as the distance to the sun r decreases (Kepler?s second law).
22:23 How the conservation of angular momentum shapes the galaxies in the form of a disc

Torque is defined as the rate of change of angular momentum.
For comparison, in linear motion, force is defined as the rate of change of momentum.

[alancalverd]

Torque is defined as the rate of change of angular momentum.

Wrong. In the case of the parking brake, angular acceleration is zero but the holding torque μFR > 0!

[hamdani yusuf (OP)]

Torque is defined as the rate of change of angular momentum.

Wrong. In the case of the parking brake, angular acceleration is zero but the holding torque μFR > 0!

What is the time derivative of angular momentum?
What is R when nothing is rotating?
BTW, 0! = 1, by definition.

[hamdani yusuf (OP)]

It's obvious that I'm not the only one who's not satisfied by the current standard units for some rotational quantities for their inconsistencies with each other. The problem has already been identified at least since 1936, although no satisfying solution has been found.

Let's list down all pros and cons of each option for standard units of rotational quantities, so we can easily understand what's at stake here.

Option 0: keep using current standard units.
Pros:
- Nothing needs to be done. Just business as usual.
- Currently existing textbooks can still be used.
- Requires less characters because some unit of angle can be omitted.
Cons:
- It leads inevitably to ghostly appearances and disappearances of the radian in the dimensional analysis of physical equations.
- A perennial problem in the teaching of mechanics, where radian appears on one side of an equation, but not on the other side.
- The typical advice of ignoring radians during dimensional analysis and adding or removing radians in units according to convention and contextual knowledge is "pedagogically unsatisfying".

The new standard can also have the same benefit of brevity as option 0. Like the unit for power, which we usually state in Watt instead of Newton meter per second, we can introduce a new unit equals to Newton meter per radian.
What would it be? Here are some options.
Wenn
Woo
Wyy
Werr
You might see a pattern here.

[hamdani yusuf (OP)]

People seem to forget that radian is not the only valid unit for rotational angle. Likewise, radian per second is not the only valid unit for rotational velocity, and radian per second square is not the only valid unit for rotational acceleration. They don't seem to realize that the other units of rotational angle are also valid for derived rotational quantities like rotational inertia, angular momentum, and torque.

[hamdani yusuf (OP)]

The new standard can also have the same benefit of brevity as option 0. Like the unit for power, which we usually state in Watt instead of Newton meter per second, we can introduce a new unit equals to Newton meter per radian.
What would it be? Here are some options.
Wenn
Woo
Wyy
Werr
You might see a pattern here.

Alternatively, we can use someone's name with significant contributions to the understanding of torque. According to various AI models, they are:
Archimedes
da Vinci
Newton
Leibniz
Euler
d'Alembert
Lagrange
Hamilton
Poisson
Thomson
Thompson

[alancalverd]

What is the time derivative of angular momentum?

irrelevant in the case of the parking brake since the angular momentum is zero.

What is R when nothing is rotating?

My apologies, obviously the brake force acts over r, the effective radius of the brake disc, to provide the holding torque. r does not change with angular velocity.

[hamdani yusuf (OP)]

What is the time derivative of angular momentum?

irrelevant in the case of the parking brake since the angular momentum is zero.

What is R when nothing is rotating?

My apologies, obviously the brake force acts over r, the effective radius of the brake disc, to provide the holding torque. r does not change with angular velocity.

Time derivative of angular momentum is torque, or rotational force.
For comparison, in linear motion, force is defined as the time derivative of momentum.

Radius of rotation can change with angular velocity if the spokes are made from elastic materials.

It seems like we also need to distinguish between individual torque and action-reaction pair of torque. They are analogous to tension and compression in linear motion.

Tension is the pulling or stretching force transmitted axially along an object such as a string, rope, chain, rod, truss member, or other object, so as to stretch or pull apart the object. In terms of force, it is the opposite of compression. Tension might also be described as the action-reaction pair of forces acting at each end of an object.

https://en.m.wikipedia.org/wiki/Tension_(physics)

[alancalverd]

Radius of rotation can change with angular velocity if the spokes are made from elastic materials.

(a)  what would be the value of a deformable brake disc? 

(b) there is no angular velocity in the question I asked.

Once again,  you have failed to answer the  simplest of questions by the application of your proposed new quantity.

[hamdani yusuf (OP)]

Once again,  you have failed to answer the  simplest of questions by the application of your proposed new quantity.

This one?

What is the time derivative of angular momentum?
What is R when nothing is rotating?
BTW, 0! = 1, by definition.

[hamdani yusuf (OP)]

When angular momentum is not changing, the net torque is zero, obviously.
Changing angular velocity doesn't necessarily involve torque. Because angular velocity of a system can change without changing angular momentum, like a spinning ice skater while redistributing her body mass.