Question

A disc of mass $M$ and radius $R$ is rolling with angular speed to on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origin $O$ is

• A. $(\frac{1}{2})M{R}^2\omega$
• B. $M{R}^2\omega$
• C. $(\frac{3}{2})M{R}^2\omega$
• D. $2M{R}^2\omega$

Answer 1

Answer: (C)

$L_0=L_{CM}+M(r\times v)$ (i)
We may write Angular momentum about $O$= Angular momentum about
$CM+$ Angular momentum of $CM$ about origin
$\therefore L_0=I\omega +MRv$
$=\frac{1}{2}M{R}^2\omega +MR(R\omega )=\frac{3}{2}M{R}^2\omega$
NOTE that in this case [ Figure (a) ] both the terms in Eq. (i)
i.e $L_{CM}$ and $(r\times v)$ have the same direction $A$. That is why we have used $L_0=/L\omega +MRv$ . We will use $L_0=l\omega -MRv=)MRv$ if they are in opposite direction as shown in figure (b).