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}) MR^2 \omega $
• B. $MR^2 \omega $
• C. $( \frac{3}{2}) MR^2 \omega $
• D. $ 2MR^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} MR^2 \omega +MR (R \omega ) = \frac{3}{2} MR^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 $L_0 = /L \omega + MR v$ . We will use $L_0 = l \omega - MRv =)MRv$ if they are in opposite direction as shown in figure (b).