Question

Two identical rods each of mass M and length 'L' are performing general plane motion in horizontal plane as shown in figure. If v is the velocity of circular motion of both rods and 'w' is the angular speed about vertical axis, then angular momentum of rod 1 in the reference frame of centre of mass of rod 2 at given instant will be –

• A. $\left( Mv\frac{3L}{2} + \frac{ML^{2}}{12}\omega \right)(–\widehat{k})$
• B. Mv. $\frac{3L}{2}(–\widehat{k})$
• C. $\left( \frac{ML^{2}}{12}\omega \right)(–\widehat{k})$
• D. None

Answer 1

Answer: (C)

$\left( \frac{ML^{2}}{12}\omega \right)(–\widehat{k})$

Answer 2

$\overrightarrow{L} = {\overrightarrow{L}}_{cm} + {\overrightarrow{r}}_{cm} \times M{\overrightarrow{v}}_{rel}$