Derive the equation of angular momentum in the case of rotational motion about a fixed axis.

Explanation

Angular momentum of a particle of rigid body is in the direction of a fixed Z-axis
`therefore vec(l_(1))=vec(r_(1))xxvec(p_(1))`
If the rigid body consisting of n particle, then the vector sum of angular momentum of all particle will be obtain in the direction of Z-axis.
`therefore vec(L_(Z))=vec(l_(1))+vec(l_(2))+vec(l_(3))+....vec(l_(n))`
`=(vec(r_(1))xxvec(p_(1)))+(vec(r_(2))xxvec(p_(2)))+(vec(r_(3))xxvec(p_(3)))+....(vec(r_(n))xxvec(p_(n)))`
`=underset(i=1)overset(n)sumvec(r_(i))xxvec(p_(i))" where "i=1,2,3,...,n`
`=underset(i=1)overset(n)sumvec(r_(i))xxm_(i)vec(v)_(i)`
`=underset(i=1)overset(n)summ_(i)r_(i)^(2)omegahatk" " [because v_(i)=r_(i)omega]`
`=underset(i=1)overset(n)sumI_(i)omegahatk" where "underset(i=1)overset(n)summ_(i)r_(i)^(2)=I`
but for all particle `SigmaI_(i)=I`,
`vec(L_(Z))=Iomegahatk` where `hatk` is a unit vector in Z-axis. but angular momentum `vec(L_(1))` perpendicular to the axis in rigid body, then total angular momentum
`vecL=vec(L_(Z))+vec(L_(_|_))`
but `vec(L_(_|_))=0`
`therefore vecL=vec(L_(Z))`
`vecL=Iomegahatk`
`therefore L=Iomega` similar to `p=mv` in linear motion.