Deduce the relation between angular momentum and angular velocity.

Explanation

Let us consider the point mass m is at a distance r from the axis of rotation. Its linear momentum at any instant is tangential to the circular path. Then the angular momentum `vecL` is perpendicular to `vecr` and `vecp`. Hence it is directed along the axis of rotation. The angle `theta` between `vecr` and `vecp` in this case is `90^(@)`. The magnitude of the angular momentum L could be written as
`LK=r mv sin 90^(@)=r mv`
where v is the linear velocity. The relation between linear velocity v and angular velocity `omega` in a circular motion is `v=r omega`.
Hence
`L=r m r omega`
`L=(mr^(2))omega`
THe directions of L and `omega` are along the axis of rotation. the abvoe expression can be written in the vector notation as
`vec=L(=mr^(2))vec(omega)`
`vec(L)=(Sigma m_(i)r_(i)^(2)vec(omega)`
`vecL=Ivec(omega)`