Question

Find the ratio of magnitude of angular momentum of a rolling disc about origin to that about the point $P$ on the disc, as shown in the figure.

• A. $2:1$
• B. $1:1$
• C. $1:3$
• D. $4:3$

Answer 1

The correct option is A $2:1$

Angular momentum about $c$ is $I_w$ but $I=\frac{1}{2}M{R}^2\Rightarrow I_c=\frac{M{R}^{2}w}{2}$

but angular momentum of $c$ w.r.t $O$

is $mVR$

so $I_o=I_c+mVR=I_c+m\left(wR\right)R$

$\Rightarrow I_o=\frac{m{R}^{2}w}{2}+m{R}^{2}w=\frac{3}{2}m{R}^{2}w$

angular momentary of $c$ w.r.t $P$ is

$m(V_c-V_p)$ separation

$=m\frac{V}{2}.\frac{R}{2}=\frac{mVR}{4}$

$V_p=$ velocity of $P$ w.r.t $O$ is $V_P=\frac{R}{2}w$

$=\frac{V}{2}$

$\Rightarrow$ angular momentum w.r.t $P$ will be $I_p=I_c+\frac{mVR}{4}$

$I_p=\frac{m{R}^{2}w}{2}+\frac{m{R}^{2}w}{4}$

velocity of $C$ w.r.t $O$ is

$V_c=V$

$\Rightarrow$ velocity of $C$ w.r.t $P$ will be $V_C-V_P=V-\frac{V}{2}=\frac{V}{2}$

$\frac{I_o}{I-p}=\frac{2}{1}$