Question

derive relation b/w moment of inertia and angular momentum​

Answer 1

We know that,

$L=pr\\\\L=mvr\\\\L=m(r\omega)r\\\\L=mr^2\omega$

Let the body consist of a number of particles where angular velocity is constant but linear momentum is not constant for every particles due to change in mass and difference in separation from the axis of rotation. Then the angular momentum of the i-th particle will be,

$L_i=m_i(r_i)^2\omega$

Then the total linear momentum of the body is,

$\displaystyle L=\sum m_i(r_i)^2\omega\\\\L=\omega\sum m_i(r_i)^2$

But, moment of inertia, $\displaystyle I=\sum m_i(r_i)^2.$

Then,

$\large\boxed {L=I\omega}$