Question

A rod of mass $1\text{kg}$ and length $2\text{m}$ is performing combined translational and rotational motion as shown in figure. Find the magnitude of total angular momentum about the origin.

• A. $10{\text{kgm}}^2/\text{s}$
• B. $1{\text{kgm}}^2/\text{s}$
• C. $11{\text{kgm}}^2/\text{s}$
• D. $20{\text{kgm}}^2/\text{s}$

Answer 1

The correct option is C $11{\text{kgm}}^2/\text{s}$

For combined rotation and translation motion,
Total angular momentum $\overrightarrow{L}={\overrightarrow{L}}_{\text{translational}}+{\overrightarrow{L}}_{\text{rotational}}$
$\overrightarrow{L}=M({\overrightarrow{r}}_0\times {\overrightarrow{v}}_0)+{\overrightarrow{L}}_{\text{COM}}$

Both angular momentum due to translation and rotation, have clockwise sense of rotation about origin. Hence both will add up
$\therefore L=M\left(r{v}_0\sin \theta \right)+I_{\text{COM}}.\omega$
i.e $L=M{v}_0{r}_{\perp }+\frac{M{L}^2}{12}.\omega ...\left(i\right)$
Here $r_{\perp }$ is the perpendicular distance of velocity vector from origin.
$r_{\perp }=\frac{l}{2}=\frac{2}{2}=1\text{m}$
Putting in Eq $\left(i\right)$, the magnitude of total angular momentum is,
$L=(1\times 10\times 1)+\left(\frac{1\times 2^2}{12}\right)\times 3$
$\therefore L=11{\text{kgm}}^2\text{/s}$