Question

A disc of mass $M$ and radius $R$ is rolling with angular speed $\omega$ on a horizontal plane. The magnitude of angular momentum of the disc about the origin O is $x$ times $M{R}^2\omega$. Then the value of $x$ is

Answer 1

The angular momentum about origin O is
$L_0=L_c+I_c\omega$
$L_c$ is Angular momentum of centre of mass about Origin
$I_c$ is moment of inertia of disc about centre of mass $=\frac{M{R}^2}{2}$

$\therefore$ Magnitude $L_0=I_c\omega +M{v}_{c}R$
But as the body is rolling $V_c=R\omega$
$\Rightarrow L_0=\frac{M{R}^2}{2}\omega +M{R}^2\omega$
$\Rightarrow L_0=\frac{3M{R}^2}{2}\omega$