Question

A rod of mass m and length 2R is fixed along the diameter of a ring of same mass m and radius R as shown in figure. The combined body is rolling without slipping along x – axis. The angular momentum about z – axis is

• A. $\left(\frac{10}{3}mvR\right)\hat{k}$
• B. $\left(\frac{5}{3}mvR\right)\hat{k}$
• C. $(-\frac{10}{3}mvR)\hat{k}$
• D. $(-\frac{5}{3}mvR)\hat{k}$

Answer 1

The correct option is C $(-\frac{10}{3}mvR)\hat{k}$

$\overline{L}={\overline{L}}_{COM}+m({\overline{r}}_0\times {\overline{v}}_0)$
$L_{COM}=(\frac{m.4R^2}{12}+m{R}^2).\frac{4}{3}mvR$
$m({\overline{r}}_0\times {\overline{v}}_0)=mvR+mRv=2mvR$
$\therefore L=\frac{4}{3}mvR+2mvR=\frac{10}{3}mvR\otimes or{I}_2=(-\frac{10}{3}mvR)\hat{k}$