Question

establish the relation between torque and angular momentum .

Answer 1

Hey .
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When a torque is applied to an object it begins to rotate with an accelerationinversely proportional to its moment of inertia. This relation can be thought of as Newton's Second Law for rotation. The moment of inertia is the rotational mass and the torque is rotational force. Angular motion obeys Newton's First Law.
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Answer 2

Here is the proof that I think you're looking for. As Ali remarks in his answer, the results holds true for a rigid body undergoing rotation with constant angular velocity.

Let r⃗ ir→i denote the position of some particle in a rigid body. Suppose this rigid body is undergoing rotation with angular velocity ω⃗ ω→, then

r⃗ ˙i=ω⃗ ×r⃗ ir→˙i=ω→×r→i

See the appendix for a proof of this. By taking the derivative of both sides with respect to time and multiplying both sides by mimi, the mass of particle ii, we obtain

p⃗ ˙i=ω×p⃗ ip→˙i=ω×p→i

Now we simply note that if F⃗ iF→idenotes the net force on particle ii, then Newton's Second Law gives F⃗ i=p⃗ i˙F→i=p→i˙ so that
τ⃗ i=r⃗ i×F⃗ i=r⃗ i×p⃗ i˙=r⃗ i×(ω⃗ ×p⃗ i)=−p⃗ i×(r⃗ i×ω⃗ )−ω⃗ ×(p⃗ i×r⃗ i)=p⃗ i×(ω⃗ ×r⃗ i)+ω⃗ ×(r⃗ i×p⃗ i)=p⃗ i×r⃗ ˙i+ω⃗ ×L⃗ i=ω⃗ ×L⃗ iτ→i=r→i×F→i=r→i×p→i˙=r→i×(ω→×p→i)=−p→i×(r→i×ω→)−ω→×(p→i×r→i)=p→i×(ω→×r→i)+ω→×(r→i×p→i)=p→i×r→˙i+ω→×L→i=ω→×L→i

This is basically the identity you were looking for. In the fourth equality, I used the so-called Jacobi identity. Now, by taking the sum over ii, the result can readily be seen to also hold for the net torque ττ on the body and the total angular momentum L⃗ L→ of the body;

τ⃗ =ω⃗ ×L⃗ τ→=ω→×L→


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