Understanding Torque: Why is it Outward/Inward?

[oneplusone]

Hello,

on the topic of torque, my textbook says it's a vector quantity with direction either outward, or inward (perpendicular to the page). Can someone explain WHY it is outward/inwards?

I saw this animation, but it doesn't really provide intuition.

 

[pbuk]

Can someone explain WHY it is outward/inwards?

Because vectors defined in this way are a good model of the way torques behave in practice. For instance the resultant torque from two torques described by vectors τ₁ and τ₂ is described by the vector sum τ = τ₁ + τ₂.

I am not sure how intuitive this can be: I have a mental image of the right hand rule but I don't think I can "see" the addition of two torques in the same way as I can "see" the parallelogram law of forces, but perhaps there are those who can.

 

[PhanthomJay]

Torque is vector with magnitude and direction.. It's direction is a clockwise or counterclockwise rotation in a plane defined by the force (F) and position (r) vectors. It is convenient to define this plane by a vector perpendicular to the plane. In this manner, torques can be vectorially added using the rules of vector addition. If it helps, the direction of the torque vector is the axis about which the torque rotates.

 

[oneplusone]

@PhanthomJay,

WHy is it convenient to define it perpendicular to the plane? I don't see how torque can be OUTSIDE the plane when its direction is clockwise/cc.

THanks.

 

[dipole]

@PhanthomJay,

WHy is it convenient to define it perpendicular to the plane? I don't see how torque can be OUTSIDE the plane when its direction is clockwise/cc.

THanks.

Well what direction should it be? If the object is rotating in a plane, what direction in that plane do you think the torque should point in? Can you think of a definition that would make sense in any situation?

 

[oneplusone]

I understand the definition is a measure of how much a force acting on an object causes that object to rotate, but why would the rotation be perpendicular to the plane? for example, a force causes an object to move, and is directed in the direction of the movement of the object. But for torque, what exactly is it, physically?
Sorry if I'm being vague-im just starting to learn physics, and don't have a teacher/mentor.

 

[bhillyard]

A torque produces a rotation about an axis. The axis would be perpendicular to the plan of rotation. So the axis describes the rotation with a clockwise or anticlockwise direction.

 

[oneplusone]

But when my textbook says, the torque is 'perpendicular to the page, and points inward to the book'
what does this mean?

 

[dipole]

Well, if its perpendicular to the page it can either point "out" of the page, toward the reader, or "in" to the page away from the reader. If it points out, then the rotation is anti-clockwise, if it points in it's clockwise.

The reason torque is defined in the way it is is because the direction of the torque corresponds to the axis of rotation. This is really the only direction that has any use for a rotating body, because the point in the body which remains fixed is the one which passes through the axis of rotation. The definition also makes sense in other ways, think of how torque is defined in terms of a force:

\vec{\tau} = \vec{r}\times\vec{F}

Now imagine the force is acting on some thin rod, at a point \vec{r} from the axis of rotation. If the force were pointing exactly along the direction of the rod, would you expect the rod to rotate? The answer is no, and the torque produced by a force which is along the same direction of the vector \vec{r} is zero by the definition of cross products.

 

[oneplusone]

Okay, so there's no actual force which pushes the object 'outward' or 'inward' correct? It is just defined that way to show if it's clockwise or counterclockwise?

Thanks again, i think i get it now

 

[dipole]

The force acting on the object is always perpendicular to the torque, so no there's no motion in the direction of the torque, but rotation about that direction.

 

[oneplusone]

But when you say in and out of the page, why does 'out' make it anti clockwise and 'in' makes it clockwise?
If it's just an axis...

 

[PhanthomJay]

That is correct the torque lies inside the plane. But the orientation of the plane can be defined by a vector perpendicular to it. If a plane lies within the x-y axis , the z axis defines its orientation. It is just a tool . Supposing you have a clockwise torque T_z about a point in the x-y plane, and then another counterclockwise torque T_x about that point in the y-z plane, and you wish to find the resultant (sum) of those two torques. The magnitude of the resultant of those two torques would be the sq rt of (T_z^2 + T_x^2), where T_z points into the page in the z direction, and T_x points along the positive x axis, using the right hand rule. The direction of the resultant torque in the x-z plane could be found by using the usual trig for vector direction . That direction defines the plane perpendicular to it , in which the resultant torque lies. Now as difficult at that may seem, imagine trying to get the resultant of the two torques without taking advantage of this convenient tool.

 

[PhanthomJay]

But when you say in and out of the page, why does 'out' make it anti clockwise and 'in' makes it clockwise?
If it's just an axis...

that is the convention using the right hand rule. Look at the palm of your hand and stick out your thumb in the same plane of your palm. Now curl the fingers of your right hand in the direction of the rotation of the torque. If the torque in he xy plane is counterclockwise, your thumb points out along the positive z axis. If the torque is clockwise, your thumb points into the page along the negative z axis , and you might sprain your wrist to determine this.

 

[e.chaniotakis]

Perhaps there is another way to depict it but you need to know some physics first.
Assume that you have a spinning wheel rotating at constant angular velocity (ω).
Now, if you start acting on it with a tangential force, then it starts speeding up. Its angular velocity increases with time, thus it has angular acceleration.
When you act on the wheel with the force, you provide the system with energy.
The energy per unit time (power) is the dot product of the torque times angular velocity :
P = τ.ω = τωcosθ
θ is the angle between torque and angular velocity.

Assuming that you are familiar with the right hand rule for rotation (ω points in or out of the wheel, perpendicular to it), then if τ and ω were perpendicular, cosθ=0 and thus no power is delivered to the system even though you consume power to accelerate it.

Therefore the angle between torque and ω must NOT be 90 degrees.
Assuming that you act with a tangential force that forces the wheel to spin on one direction (there is no precession), the math described above dictates that torque is parallel to ω and thus perpendicular to the plane of rotation.

 

[sophiecentaur]

Okay, so there's no actual force which pushes the object 'outward' or 'inward' correct? It is just defined that way to show if it's clockwise or counterclockwise?

Thanks again, i think i get it now

Torque is 'a new idea' and needs to be thought of differently from other. familiar mechanics concepts. Confusion about its 'direction' is understandable but I could ask you where else could it be pointing? If a turning only is involved then there would be no preferred direction in the plane of the forces (a wheel doesn't care where the force is 'coming from' if the shaft is being twisted). The only direction which involved no 'preference' is actually normal to the plane.
I think this is yet another example, in Physics, where it's a big help to go along with the Maths and let it deliver the answer, rather than to seek to hard for a more 'physical' feel - which will often lead you into false ally-ways.
The Maths that is used to describe Vectors is not always intuitive - take the 'Cross Product', for instance - and, imo, you just have to use it enough until it becomes second nature. You will have had a similar problem with things like Algebra, in the past but you may have forgotten the pain of the transition into accepting it as a way of solving problems. ("X is an unknown quantity? What the . . . . .?")