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. $\bigg( \frac{1}{2} \bigg) MR^2 \omega$
• B. $MR^2 \omega$
• C. $\bigg( \frac{3}{2} \bigg) MR^2 \omega$
• D. $2MR^2 \omega$

Answer 1

Answer: (C)

$\bigg( \frac{3}{2} \bigg) MR^2 \omega$

Answer 2

$L_0 = L_{CM} +M (r \times v)$ \hspace27mm ..........(i)
We may write
Angular momentum about O = Angular momentum about
CM + Angular momentum o f 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 E (i)
i.e $L_{CM} \, \, and \, \, (r \times v )$have the same direction A. That is why we haveused $L_0 = /L \omega + MR v$ . We will use $L_0 = l \omega - MRv$ =/o)~MRv if they are in opposite direction as shown in figure (b).