Question

A disc of mass $M$ and radius $R$ is rolling without slipping with angular speed $\omega$ on a horizontal plane as shown in figure. The magnitude of angular momentum of the disc about the $x-$axis 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

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


For pure rolling, $v=R\omega$
where, $v=$ linear velocity of $\text{COM}$ of disc.
$\omega =$ angular velocity of disc.
$\Rightarrow$Angular momentum about $x-$axis will include the angular momentum due to translation as well as angular momentum due to rotational motion.
$\therefore \overrightarrow{L}={\overrightarrow{L}}_{Trans}+{\overrightarrow{L}}_{Rot}$
Angular momentum about $x-$axis is given by $\left(L\right)$
$\Rightarrow L=M{v}_{\text{CM}}R+I_{\text{CM}}\omega$

$\because I_{\text{CM}}=\frac{M{R}^2}{2}$ and $v_{\text{CM}}=R\omega$
So, $L=M\left(R\omega \right)R+(\frac{M{R}^2}{2}\times \omega )$
$+ve$ sign taken since both angular momentums have a clockwise sense about $x-$axis.
$\therefore L=\frac{3}{2}M{R}^2\omega$