Understanding Torque: Applying the Right Hand Rule in Real Life

[Aeronautic Freek]

What does right hand rule torque mean in real life?

engine crankshaft produce torque,if apply right hand rule to crankshat that will mean that torque is act in crankshaft axiis of roatation,which mean axials berings will have huge load on it,,that is not the case in reality,axail bering don't have load on it at all when engine is working.

 

[Dale]

if apply right hand rule to crankshat that will mean that torque is act in crankshaft axiis of roatation,

Yes.

which mean axials berings will have huge load on it

Why do you believe that?

It seems like you may be confusing a torque and a force. A torque is a twist around an axis, not a push along the axis.

 

[Aeronautic Freek]

It seems like you may be confusing a torque and a force. A torque is a twist around an axis, not a push along the axis.

Yes i confused with that..
i understand torque in sense of ROTATION :clockwise or anti clockwise... but i can't understand concept of torque in sense of direction!
torque=rotation, force has direction...
so what direction of torque along some axis really mean,in physcal sense(not math)?!

crankshaft has torque clockwise or anticlockwise but not out of cranckshaft axis of rotation!

 

[A.T.]

crankshaft has torque clockwise or anticlockwise

Clockwise or anticlockwise are 2D concepts. In 3D they depend on the perspective (projection) and are therefore ambiguous.

 

[berkeman]

i understand torque in sense of ROTATION :clockwise or anti clockwise... but i can't understand concept of torque in sense of direction!
torque=rotation, force has direction...
so what direction of torque along some axis really mean,in physcal sense(not math)?!

Have you learned about the use of vectors in general? How they involve a magnitude and direction (versus a scalar quantity that only has a magnitude)? Using vectors in math and physics equations usually simplifies the calculations (makes them more compact and intuitive), because they contain more information versus scalars.

For torque, one intuitively useful thing about the vector equation $\vec \tau = \vec r \times \vec F$ is that if the force vector and radius vector (radius out from the axis of the rotational torque) are at a right angle to each other and to the axis of rotation, that gives you the most torque possible for that force and lever arm length. If the force vector and the radius vector and the axis of rotation are not at right angles, the torque is diminished.

In the Hyperphysics note below, do you see how the force and radius vectors and the axis of rotation are at right angles to each other? That maximizes the usefulness of that force for generating the most torque that you can for the values of the force and radius length. If the force were pointing more backward a bit instead of at a right angle to the shaft, less torque would be transferred to the shaft. The equation that they show $\tau = rF sin(\theta)$ is the scalar version of the vector equation that I listed above.

Does that help at all?

http://hyperphysics.phy-astr.gsu.edu/hbase/tord.html

 

[Dale]

Yes i confused with that..
i understand torque in sense of ROTATION

Do you understand that rotation has an axis?

so what direction of torque along some axis really mean,in physcal sense(not math)?!

When you exert a torque on a screwdriver to put a screw in a board, the torque points in the direction the screwdriver points, the axis. Notice that the torque direction, the axis, is perpendicular to the direction of the forces applied by your hand on the handle of the screwdriver

crankshaft has torque clockwise or anticlockwise but not out of cranckshaft axis of rotation!

The torque points in the same direction as the axis.

 

[robphy]

In my opinion, the fundamental concept in the torque is the
planar-area on the plane containing $\vec r$ and $\vec F$,
whose area is equal to $|\vec r\times\vec F|$ (the area of a parallelogram formed by $\vec r$ and $\vec F$),
and whose orientation follows the sense of rotation implied by $\vec F$'s tail at the tip of $\vec r$.

The right-hand-rule is a prescription to assign an arrow (a [pseudo or axial] vector)
to this oriented planar area. That arrow's line can be thought of as an axis of rotation.

The torque [its physics and the sense of rotation] really occurs on that plane.

 

[A.T.]

In my opinion, the fundamental concept in the torque is the planar-area ...

I don't think its fundamental, but allows to visualize it nicely.

 

[robphy]

I don't think its fundamental, but allows to visualize it nicely.

A torque (a cross-product of two ordinary polar vectors) isn't an ordinary vector, it's a pseudovector... mathematically, it's better described as the antisymmetric part of a 3-d matrix.

The right-hand-rule seems so "3 dimensional".
In a 4-d space, what is perpendicular to the plane containing $\vec r$ and $\vec F$? Not a single direction, but a whole 2-dimensional vector space.UPDATE:
When I teach torque, especially balancing torques on a see-saw,
I do emphasize the oriented areas. I don't need to talk about the right-hand-rule.
(Another way to represent torques is with a https://en.wikipedia.org/wiki/Bivector )

 

[Dale]

I don't need to talk about the right-hand-rule.

How do you specify the orientation without the right hand rule?

 

[robphy]

How do you specify the orientation without the right hand rule?

From above,

whose orientation follows the sense of rotation implied by $\vec F$'s tail at the tip of $\vec r$

(I should have said "by putting $\vec F$'s tail...")

Using an image from https://en.wikipedia.org/wiki/Bivector ,

To visualize torque, $\vec r=\vec a$ and $\vec F=\vec b$.If I align my right hand fingers curled with the specific orientation on that oriented area,
my right-hand thumb would point along "the [pseudo-]vector perpendicular to the oriented area"
... if I want such a pseudo-vector.

The geometrical and physical content of the torque has already been established by the oriented area,
before using any right-hand rule.

 

[Dale]

whose orientation follows the sense of rotation implied by F→'s tail at the tip of r→.

I guess I am not seeing how that is different from a right hand rule. The + side of the area is the side where my right thumb would be.

 

[robphy]

I guess I am not seeing how that is different from a right hand rule. The + side of the area is the side where my right thumb would be.

I am specifically not restricting myself to thinking in three dimensions.
If space were 4-d dimensional, the oriented area would still contain the geometrical and physical content.
A right-hand-rule (to determine a pseudo-vector) would not produce a useful quantity associated with the oriented area [since there is a whole 2-D vector space orthogonal to that plane].

The point is that
the direction of a "pseudo-vector" (in 3-D space)
is a proxy (with mathematical but not really physical content (although one may try to come up with one)[hence, the possible misconceptions])
for the oriented-area associated with $\vec r$ and $\vec F$.

(I would argue that the magnetic-field as a vector field or as a pseudo-vector field
may not be the best representation of the physics. A bivector field or two-form field is better.)

 

[vanhees71]

How do you specify the orientation without the right hand rule?

That's the magic of the Hodge dual. You can describe equivalently the torque as an axial vector or as an alternating 2-form. In Cartesian components you have
$$(^{\dagger}\tau)_{jk}=r_j F_k-r_k F_j$$,
which is the Hodge dual of an axial vector with components
$$\tau_i=\epsilon_{ijk} r_j F_k=\frac{1}{2} \epsilon_{ijk} (^{\dagger} \tau)_{jk}.$$
Of course you also have
$$(^{\dagger} \tau)_{jk}=\epsilon_{jki} \tau_i.$$
I'm only wondering whether it isn't more an obstacle to learn about tensors rather than sticking as much as possible to vectors in introductory lectures, but that's another question.

 

[robphy]

I'm only wondering whether it isn't more an obstacle to learn about tensors rather than sticking as much as possible to vectors in introductory lectures, but that's another question.

In my opinion, the order of presentation and emphasis should be

1. oriented-area (signed area)
2. [axial] vector (possibly with a comment that this works only in 3-D)
3. antisymmetric tensor (bivector or two-form)

Depending on the level of the discussion,
(in my opinion, fundamental concept of) the oriented-area may be sufficient
and one doesn't need to introduce the additional constructions.
That is, one might not need to further introduce the right-hand-rule or the Hodge-dual.By the way, when I do introduce the axial-vector (from the cross-product of two vectors),
I emphasize that one cannot add an axial-vector to an ordinary vector. (Similarly,
one cannot add a magnetic field vector to an electric field vector...
indeed, only in three dimensions,
does the magnetic field vector have the same number of components as the electric field.)

 

[vanhees71]

Yes, I usually to 1. and 2. in the introductory lectures (mechanics and electrodynamics); 3. of course comes in when discussing infinitesimal rotations, but then also one can introduce the infinitesimal rotations via thecross product. In this way I avoid the more advanced topic of the Hodge dual, which of course is needed at the more advanced level, particularly in relativity.

Of course the "super selection rule" that you cannot add a polar and an axial vector only holds in theories, where a space-reflection symmetry operation exists. The weak interaction breaks this rule "maximally" due to the famous "vector-minus-axialvector" current.

 

[Aeronautic Freek]

Do you understand that rotation has an axis?

When you exert a torque on a screwdriver to put a screw in a board, the torque points in the direction the screwdriver points, the axis. Notice that the torque direction, the axis, is perpendicular to the direction of the forces applied by your hand on the handle of the screwdriver

The torque points in the same direction as the axis.

yes i know that roatatio has axis...

if professor do that (at 23:00min) staying on skateboard,will skate goes forwad,can he use gyro for linear motion?

 

[Lnewqban]

if professor do that (at 23:00min) staying on skateboard,will skate goes forwad,can he use gyro for linear motion?

When you steer your bicycle, the gyroscopic effect does not accelerate it forward or slow it down, the effect only helps you steering and leaning your bike.

Copied from
https://en.wikipedia.org/wiki/Bicycle_and_motorcycle_dynamics#Gyroscopic_effects

"The role of the gyroscopic effect in most bike designs is to help steer the front wheel into the direction of a lean. This phenomenon is called precession and the rate at which an object precesses is inversely proportional to its rate of spin.
...
Another contribution of gyroscopic effects is a roll moment generated by the front wheel during countersteering. For example, steering left causes a moment to the right. The moment is small compared to the moment generated by the out-tracking front wheel, but begins as soon as the rider applies torque to the handlebars and so can be helpful in motorcycle racing."

Note that the chair in this animation has wheels, which allow the chair to slightly move to compensate for the movement of the body of the subject; nevertheless, the chair does not slide (in any direction) during the precession of the wheel:

 

[Aeronautic Freek]

Note that the chair in this animation has wheels, which allow the chair to slightly move to compensate for the movement of the body of the subject; nevertheless, the chair does not slide (in any direction) during the precession of the wheel:

View attachment 265789

yes same will happened if men hold fan in hands,chair will roatate but not slide in any direction..
but if he stay on skateboard and hold fan in hands ,it will go forward
will it go forward with gyro if he point gyro in correct position?

 

[Lnewqban]

yes same will happened if men hold fan in hands,chair will roatate but not slide in any direction..
but if he stay on skateboard and hold fan in hands ,it will go forward
will it go forward with gyro if he point gyro in correct position?

Never forward movement, always rotation, as long as the spinning object has that degree of freedom (there is no restriction to rotate about that axis).

All rotating bodies “prefer” staying rotating on same plane.
If you or any torque force a rotating body to change from one plane of rotation to another, there is a “in protest” reaction, which is the reactive torque that we know as torque-induced gyroscopic precession.

Copied from
https://en.m.wikipedia.org/wiki/Precession#Torque-induced

“Torque-induced precession (gyroscopic precession) is the phenomenon in which the axis of a spinning object (e.g., a gyroscope) describes a cone in space when an external torque is applied to it. The phenomenon is commonly seen in a spinning toy top, but all rotating objects can undergo precession. If the speed of the rotation and the magnitude of the external torque are constant, the spin axis will move at right angles to the direction that would intuitively result from the external torque. In the case of a toy top, its weight is acting downwards from its center of mass and the normal force (reaction) of the ground is pushing up on it at the point of contact with the support. These two opposite forces produce a torque which causes the top to precess.”

 

[Aeronautic Freek]

Never forward movement, always rotation, as long as the spinning object has that degree of freedom (there is no restriction to rotate about that axis).

ok so we can not use gyro as thrust/propulsion...

 

[Lnewqban]

ok so we can not use gyro as thrust/propulsion...

You could use a mechanism to convert the rotation into linear forward movement at the expense of slowing down the rotating object, which will reduce the gyroscopic effect.

The energy used to overcome the friction of that rotating man-chair (shown in above post) comes from the kinetic energy that the wheel initially had.
After a while both, the chair and the whell, will expend that energy, coming to a stop.

I recommend you reading these two good articles:
https://www.physicsforums.com/insights/what-is-the-moment-of-inertia-a-10-minute-introduction/

https://geekswipe.net/science/physics/how-gyroscopes-work-intuitive-explanation/

Also this very good one about the “mysterious” concept of vectors, which originates the “right hand rule”:
https://www.physicsforums.com/insights/frequently-made-errors-vectors-elementary-use/

 

[jbriggs444]

You could use a mechanism to convert the rotation into linear forward movement at the expense of slowing down the rotating object, which will reduce the gyroscopic effect.

Only if you use an external force to do so.

Absent an external force, linear momentum is conserved. You cannot generate thrust from any purely internal arrangement of torques and rotations.

 

[Aeronautic Freek]

Only if you use an external force to do so.

Absent an external force, linear momentum is conserved. You cannot generate thrust from any purely internal arrangement of torques and rotations.

if you put shaft instead string and connect with gears to the wheels than you make propulsion from gyro..
but you need external force to rotate gyro..

you mean on this?

 

[jbriggs444]

if you put shaft instead string and connect with gears to the wheels than you make propulsion from gyro..
but you need external force to rotate gyro..

you mean on this?

I was simply pointing out that that every possible mechanism for achieving linear propulsion depends on an external linear force.

 

[Dale]

if you put shaft instead string and connect with gears to the wheels than you make propulsion from gyro..
but you need external force to rotate gyro..

you mean on this?

The wheels will provide an external linear force at the contact point with the ground.