Question

A disc of mass M and radius R is rolling with angular speed $\omega$ on a horizontal plane. 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

The correct option is C

$\frac{3}{2}M{R}^2\omega$

Let OC = R and let $V_c$ be the velocity of the centre of mass of the disc. The linear momentum of the center of mass is $P_c=M{v}_c.$ If Lc is angular momentum of the disc about C, then the angular momentum about origin O is $L_0=L_c+R_c\times P_c$

$\therefore$ Magnitude $L_0=l_c\omega +R_c\times M{v}_{c}sin\theta =\frac{1}{2}M{R}^2\omega +M{R}_c{v}_{c}sin\theta$ $(\because I_c=\frac{1}{2}M{R}^2)=\frac{1}{2}M{R}^2\omega +MR\times R\omega (\because R_{c}sin\theta =Rand{v}_c=R\omega )=\frac{3}{2}M{R}^2\omega$