A uniform disc is rolling on a horizontal sur | vedclass

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(D) For a disc of mass $m$ and radius $R$ rolling with angular velocity $\omega$ and linear velocity $v = R\omega$:The angular momentum about the cent

Vedclass · Accepted Answer

All of the above.

Vedclass · Answer

(D) For a disc of mass $m$ and radius $R$ rolling with angular velocity $\omega$ and linear velocity $v = R\omega$:The angular momentum about the center of mass $O$ is $L_O = I_O \omega = (\frac{1}{2}mR^2)\omega$.The angular momentum about any point $P$ is given by $\vec{L}_P = \vec{L}_{cm} + \vec{r}_{cm/P} 	imes m\vec{v}_{cm}$.For point $B$ (point of contact): $\vec{L}_B = I_O \omega + mR^2 \omega = \frac{1}{2}mR^2 \omega + mR^2 \omega = \frac{3}{2}mR^2 \omega$.For point $A$ (topmost point): $\vec{L}_A = I_O \omega - mR^2 \omega = \frac{1}{2}mR^2 \omega - mR^2 \omega = -\frac{1}{2}mR^2 \omega$.The magnitude $|L_B| = \frac{3}{2}mR^2 \omega$ and $|L_A| = \frac{1}{2}mR^2 \omega$. Thus,$|L_B| = 3|L_A|$.Since $L_A$ is negative,it is anticlockwise,and $L_B$ is positive,it is clockwise. Therefore,all statements are correct.

$A$ uniform disc is rolling on a horizontal surface. At a certain instant,$B$ is the point of contact and $A$ is the topmost point of the disc,where $R$ is the radius of the disc.

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