Question

A disc of mass 'M' and radius 'R' is rolling with an angular speed of $\omega$ rad/s on a horizontal plane as shown in the figure. The magnitude of angular momentum of the disc about the origin O is

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

Answer 1

The correct option is C $\left(\frac{3}{2}\right)M{R}^2\omega$

The angular momentum of a body $\overrightarrow{L}$ may be expressed as the sum of two parts,
(a) one arising from the motion of the centre of mass of the body $\left({\overrightarrow{L}}_{orbital}\right)$
(b) the other from the motion of the body with respect to its centre of mass $\left({\overrightarrow{L}}_{spin}\right)$
i.e., ${\overrightarrow{L}}_{total}={\overrightarrow{L}}_{C.M}+{\overrightarrow{r}}_{C.M}\times \overrightarrow{p}$
$\Rightarrow {\overrightarrow{L}}_{total}={\overrightarrow{L}}_{C.M.}+M(\overrightarrow{C.M}\times {\overrightarrow{v}}_{C.M})$
For this problem
$L_{C.M}=I\omega =\frac{1}{2}M{R}^2\omega$ and
$M({\overrightarrow{r}}_{C.M}\times {\overrightarrow{v}}_{C.M})=MR{v}_{CM}=MR\left(R\omega \right)$
$\Rightarrow M({\overrightarrow{r}}_{C.M}\times {\overrightarrow{v}}_{C.M.})=M{R}^2\omega$
$\Rightarrow L_{total}=\frac{1}{2}M{R}^2\omega +M{R}^2\omega =\frac{3}{2}M{R}^2\omega$