show the relation between angular momentum and moment if inertia by calculus method
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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 i
r→˙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 i
p→˙i=ω×p→i
Now we simply note that if F⃗ iF→i denotes 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 i
τ→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→
Appendix. The motion of a rigid body undergoing rotation is generated by rotations. In other words, there is some time-dependent rotation R(t)R(t) for which
r⃗ (t)=R(t)r⃗ (0)
r→(t)=R(t)r→(0)
It follows that
r⃗ ˙(t)=R˙(t)r⃗ (0)=R˙(t)R(t)Tr⃗ (t)=ω⃗ (t)×r⃗ (t)
r→˙(t)=R˙(t)r→(0)=R˙(t)R(t)Tr→(t)=ω→(t)×r→(t)
In the last step, I used the fact that R(t)R(t) is an orthogonal matrix for each tt which implies that R˙RTR˙RT is antisymmetric. It follows that there exists some vector ω⃗ ω→, which we call the angular velocity of the body, for which R˙RTA⃗ =ω⃗ ×A⃗ R˙RTA→=ω→×A→ for any A⃗ A→.
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 i
r→˙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 i
p→˙i=ω×p→i
Now we simply note that if F⃗ iF→i denotes 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 i
τ→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→
Appendix. The motion of a rigid body undergoing rotation is generated by rotations. In other words, there is some time-dependent rotation R(t)R(t) for which
r⃗ (t)=R(t)r⃗ (0)
r→(t)=R(t)r→(0)
It follows that
r⃗ ˙(t)=R˙(t)r⃗ (0)=R˙(t)R(t)Tr⃗ (t)=ω⃗ (t)×r⃗ (t)
r→˙(t)=R˙(t)r→(0)=R˙(t)R(t)Tr→(t)=ω→(t)×r→(t)
In the last step, I used the fact that R(t)R(t) is an orthogonal matrix for each tt which implies that R˙RTR˙RT is antisymmetric. It follows that there exists some vector ω⃗ ω→, which we call the angular velocity of the body, for which R˙RTA⃗ =ω⃗ ×A⃗ R˙RTA→=ω→×A→ for any A⃗ A→.
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