[alancalverd]

Please, please, use your definition of torque to address the simple question: what force is required on the parking brake pads to prevent the car from rolling down a hill?

I guess around a million cars will have been made since I asked that question, so umpteen engineers who you assert do not understand torque, seem to have worked it out satisfactorily, and all you have done is introduce irrelevancies.

Please don't accuse me of defining or using the term "rotational radius" - it's entirely yours.

[paul cotter]

"Rotational radius" is a fictitious concept that no engineer uses.

[hamdani yusuf (OP)]

"Rotational radius" is a fictitious concept that no engineer uses.

What radius do you use instead?

[hamdani yusuf (OP)]

Please, please, use your definition of torque to address the simple question: what force is required on the parking brake pads to prevent the car from rolling down a hill?

I guess around a million cars will have been made since I asked that question, so umpteen engineers who you assert do not understand torque, seem to have worked it out satisfactorily, and all you have done is introduce irrelevancies.

Please don't accuse me of defining or using the term "rotational radius" - it's entirely yours.

I've already gave the equations. If you want the number, give your numbers.

[hamdani yusuf (OP)]

This road is floating freely in outer space. What's the torque produced by each forces?

F = 1 Newton
L = 1 meter
Mass of the rod = 1 kg

[paul cotter]

"What radius do you use instead?" The distance from where the force is applied to the fulcrum point.

[alancalverd]

I've already gave the equations.

And nowhere did they involve angular displacement.

What's the torque produced by each forces?

Zero, as there is no fulcrum point. But if you reverse or remove one of the forces it is obvious that the rod will rotate. The interesting question is where the center of rotation would be!

[hamdani yusuf (OP)]

"What radius do you use instead?" The distance from where the force is applied to the fulcrum point.

Where is the fulcrum in this diagram?

If the fulcrums are flexible, and become shorter when compressed?

[hamdani yusuf (OP)]

I've already gave the equations.

And nowhere did they involve angular displacement.

What's the torque produced by each forces?

Zero, as there is no fulcrum point. But if you reverse or remove one of the forces it is obvious that the rod will rotate. The interesting question is where the center of rotation would be!

What's important is ∂d/∂θ.
If the right force is moved slightly closer to the center of the rod, what the torque will be?
If the right force is slightly reduced, what the torque will be?
Is there any change in the number of fulcrum?

[hamdani yusuf (OP)]

What's the unit of angular momentum?

Angular Momentum

In this video Paul Andersen explains rotating object have angular momentum.  The angular momentum of a point object is the product of the distant from the center of rotation and the linear momentum.  The angular momentum of an extended object is a product of the rotational inertia and the angular velocity.  The change in angular momentum is equal to the product of the net torque and the change in time.

Grok's answer to "how are Keppler's laws related to conservation of angular momentum?"

Kepler's laws of planetary motion are deeply connected to the conservation of angular momentum, a fundamental principle in physics stating that the angular momentum of a system remains constant unless acted upon by an external torque. Here's how each of Kepler's three laws relates to this concept:
First Law (Law of Ellipses): Planets move in elliptical orbits with the Sun at one focus. This law describes the shape of the orbit but doesn't directly address angular momentum. However, the elliptical path ensures that the planet's distance from the Sun varies, which ties into the second law. The absence of external torques (assuming a two-body system with no significant external forces) means angular momentum is conserved throughout the orbit, allowing the planet to follow a stable elliptical path.

Second Law (Law of Equal Areas): A line segment joining a planet to the Sun sweeps out equal areas in equal intervals of time. This is a direct consequence of angular momentum conservation. Angular momentum for a planet is given by L=mr^2ω, where ( m ) is the planet's mass, ( r ) is the distance from the Sun, and ω is the angular velocity. Since no external torques act on the planet-Sun system, ( L ) is constant. When the planet is closer to the Sun (smaller ( r )), it moves faster (higher ω) to keep r^2ω constant, sweeping out the same area per unit time as when it is farther away and moving slower. The area swept per unit time is proportional to r^2ω, which remains constant due to conserved angular momentum.

Third Law (Law of Periods): The square of a planet?s orbital period is proportional to the cube of the semi-major axis of its orbit (T^2 ∝ a^3). This law relates to angular momentum indirectly through the dynamics of the orbit. For a circular orbit (a simplified case), angular momentum is L=mrv, and velocity v=√(GM/r)
 (from gravitational force balancing centripetal force). Substituting, L=m√GMr. For elliptical orbits, the semi-major axis ( a ) plays a similar role. The conservation of angular momentum ensures the orbit remains stable, and the third law emerges from the balance of gravitational forces and orbital dynamics, with angular momentum constraining the relationship between period and distance.

In essence, conservation of angular momentum underpins Kepler?s second law explicitly and supports the first and third laws by ensuring the stability and predictability of planetary orbits in the absence of external torques. The gravitational force between the planet and Sun is radial (along the line connecting them), producing no torque, so angular momentum remains constant, shaping the elliptical orbits and their timing as described by Kepler.

[paul cotter]

More obfuscation, as usual.

[alancalverd]

In view of Hamdani's consistent refusal to use his new definition of torque to solve a simple question of statics, and his apparent inability to identify a fulcrum in his own drawings, I can only conclude that his redefinition serves no purpose other than to confuse him.

That being said, if a really clever engineer such as Hamdani can't understand the fault in his own proposal, it has no place in the education of the young.

So I won't contribute any further.

[hamdani yusuf (OP)]

More obfuscation, as usual.

If you are still confused about my proposed new standard units of rotational quantities, you can compare their advantages and disadvantages to current standars one by one.

https://www.thenakedscientists.com/forum/index.php?topic=87006.msg741787#msg741787

[hamdani yusuf (OP)]

In view of Hamdani's consistent refusal to use his new definition of torque to solve a simple question of statics, and his apparent inability to identify a fulcrum in his own drawings, I can only conclude that his redefinition serves no purpose other than to confuse him.

That being said, if a really clever engineer such as Hamdani can't understand the fault in his own proposal, it has no place in the education of the young.

So I won't contribute any further.

I did use my new definition. Rotational radius equals derivative of rotational displacement with respect to rotational angle.

If your definition of rotational radius doesn't equal to ∂d/∂θ, you're defining it wrong.

[hamdani yusuf (OP)]

I've already gave the equations.

And nowhere did they involve angular displacement.

What's the torque produced by each forces?

Zero, as there is no fulcrum point. But if you reverse or remove one of the forces it is obvious that the rod will rotate. The interesting question is where the center of rotation would be!

What's important is ∂d/∂θ.
If the right force is moved slightly closer to the center of the rod, what the torque will be?
If the right force is slightly reduced, what the torque will be?
Is there any change in the number of fulcrum?

Since you have difficulty in answering these questions, I'll answer them myself, so everyone else who read this thread can learn.
The torque will no longer be zero.
Still there will be no fulcrum.

[Eternal Student]

Hi.

.... so everyone else who read this thread can learn...   

    I genuinely don't like upsetting people.  I've written a few forum posts myself that weren't well recieved and I've just had to move on.   The forum is useful for discussion and sometimes I / we  just won't like the replies.  Sometimes I've had to recognise that my own ideas had some flaws and sometimes I've just worked through the problems again and become more convinced they may be right while the replies were wrong.  The thing is, forum moderators and other regular forum users are just human in the same way that you are just human.   Most of us are just doing our best.

    It may be you're absolutely correct,  I haven't read all of it so I don't know.   It's just that with this many pages, no-one else is ever going to want to read it.

   Sorry.  I really am sorry.   Typically a forum is only useful for the original poster and a handful of people to learn and especially to discuss something.   This forum does not work as a platform to teach the entire world and I doubt that the original aims, or terms of use, for this forum would want people to use it in that way.

       Extract from the Forum Acceptable Usage Policy,   available at   https://www.thenakedscientists.com/forum/index.php?topic=8535.0
      The site is not for evangelising your own pet theory.  It is perfectly acceptable that you should post your own theory up for discussion, but if all you want to do is promote your own idea and are not inviting critical debate about it, then that will not be acceptable.

      Once again, I am genuinely sorry that the forum isn't providing the service or facillities you are seeking.   That may be a failing on   our / their   part but it just is what it is.   It may be that you have got all that you can get out of this forum.   If you're sure your idea is a golden one, then you could always find another way to promote it and bring it to public attention.   For example, take it to a university and propose these ideas as something you would like to research and hopefully publish.   It may cost you money in course fees and it will take you some time but if you're driven enough to let the world know about your idea then be positive, "pro-active", "fully committed" and find a way to make something like that happen.

   Good Luck and, as always, please be assured of my best wishes towards you.

[paul cotter]

Hi ES. We all make mistakes and I certainly have had to be corrected on many occasions but I cannot recall any of yours apart from typos. Hamdani is seemingly incapable of error and I have never met anyone with such dogged intransigence to learning. When proven wrong he digresses and obfuscates. I have repeatedly said that I am finished dealing with such exasperation but I am drawn back when I see blatant error that may mislead the general reader.

[alancalverd]

 

Rotational radius equals derivative of rotational displacement with respect to rotational angle.

In the case of a parking brake, there is no rotation, so δs/δθ is undefined.

[Bored chemist]

"What's the unit of Torque?"
Same as it was in August when this thread started...

[paul cotter]

Same as it ever was, as the song says.