Torque

Well, anything that we use in physics regularly we define that quantity.

Si when it was found that the Force when applied tangentially to any body cause rotation of it and it did also depend on the perpendicular distance of that force from the axis of rotation. So they defined that quantity to be Torque.

For example, you may ask why we call unit vectors i j or k. no matter u call it C H E . But they is decided by an international agreement

For detailed study see http://en.wikipedia.org/wiki/Torque.

whatever i can gather is that when i start proving the moment of inertia i come across r cross f which has been designated as torque .so may be it has been defined as the rule of the game.

for ex you cannot prove F=ma
(by the way how many of you doubt this formula. i seriously do because i think
F=maⁿ where n varies from one part of universe to another.although what i am speaking may be useless stuff but i think that is why Newtonian mechanics fails in front of relativistic mechanics.the difference comes because of the fourth dimension time.)

for ex you cannot prove F=ma
Very wrong.
This is Newton's second law and we have a very good proof of it.
F=maⁿ
Can you prove this?

Actually, this question has two parts in it:

(a) What is the motivation for bringing in a product that involves r and F to define a quantity called torque, and

(b) Why is that written in the precise fashion r X F

This post will deal with the first part

Think about opening a door. Your experience tells you that you need to expend less effort by applying force as far away from the hinge as possible. How does this happen? Why does the point of application of force make a difference?

Notice that for the same angular displacement, a point further away from the hinge moves through a greater displacement than a point closer to the hinge. That means the same force does greater work when it is applied further away from the hinge. This provides us with a starting point.

Let us say it is applied at a distance R from the hinge and at the time interval dt it moves through an angle dθ. Let the force be applied at an angle φ

The work done is F \sin \phi \times R d \theta

Now, this example has been chosen so so that the effect of the force on the body is pure rotation. The work done causes only a change in KE. Hence, now an element of mass dm at a distance r undergoes a change in kinetic energy of

\frac{1}{2} dm \times r^2 (\omega + d \omega)^2 - \frac{1}{2} dm \times r^2 \omega ^2 = dm \times r^2 \omega d \omega

Now if you take this for the entire body, you will get

d \omega \times \int r^2 dm.

This quantity \int r^2 dm is denoted by I.

So the change in KE is just I d \omega

So now we have F \sin \phi \times R d \theta = I \omega d \omega

Then, F \sin \phi \times R \frac{d \theta}{dt} = I \omega \frac{d \omega}{dt}

and hence F \sin \phi \times R \omega = I \omega \alpha \Rightarrow F \sin \phi \times R = I \alpha

Denoting F \sin \phi \times R by \tau we have

\tau = I \alpha.

@Prophet Sir,
You are infact using the conservation of angular momentum to derive torque = I(alpha). How do you derive it solely?

@swordfish

what do you mean by the word 'solely' here?

Here is another proof for
Torque=I(alpha)

We know that
F_t=ma_t where F_t is the component of force that is tangent to circular path and a_t is the tangential acceleration(for a rotating body).

Now,Torque=F_tr=ma_tr where r is the positon vector of the particle.
Using(a_t=(alpha)r),we can write

Torque=m[(alpha)r]r=(mr²)(alpha)
Torque=I(alpha) where mr² is moment of inertia denoted by I.

I completely agree with you Sir. We have to accept certain axioms without any reason.

Good.

Thank you all;especially you,Prophet (SIR) for your detailed explaination.