# The question is based on Rotational Mechanics. Find the correct option (s)?

Apr 6, 2017

See below.

#### Explanation:

We have

$l = \sqrt{24} a$
${l}_{g} = l + \frac{1}{5} l = \frac{6}{5} l$
${\vec{r}}_{g} = \left(0 , {l}_{g} / \cos \alpha , 0\right)$

where $\left(u , v , w\right)$ must be understood as

$u \hat{i} + v \hat{j} + w \hat{k}$

Here

$\alpha = \arctan \left(\frac{1}{\sqrt{24}}\right)$
$\vec{\omega} = \left(0 , \cos \alpha , \sin \alpha\right) \omega$
${\vec{v}}_{g} = {\vec{r}}_{g} \times \vec{\omega} = \frac{17}{2} {a}^{2} m \omega \left(0 , \cos \alpha , \sin \alpha\right)$

then

$\left\mid {\vec{v}}_{g} \right\mid = \frac{17}{2} {a}^{2} m \omega$

and

(D) $\Omega = \frac{\left\mid {\vec{v}}_{g} \right\mid}{\left\mid {\vec{r}}_{g} \right\mid} = \frac{1}{5} a \omega$

or

$\vec{\Omega} = \left(0 , 0 , 1\right) \Omega$

Now

${\vec{L}}_{O} = {J}_{\omega} \vec{\omega} + {J}_{\Omega} \vec{\Omega}$

with

${J}_{\omega} = \frac{m {a}^{2}}{2} + \frac{4 m {\left(2 a\right)}^{2}}{2}$

We have also

${\vec{L}}_{g} = {J}_{\omega} \vec{\omega}$ and

(B) $\left\lVert {\vec{L}}_{g} \right\rVert = \frac{17}{2} {a}^{2} m \omega$

and

${J}_{\Omega} = \frac{m {a}^{2}}{4} + m {l}_{0}^{2} + \frac{\left(4 m\right) {\left(2 a\right)}^{2}}{4} + 4 m {\left(2 {l}_{0}\right)}^{2}$

with ${l}_{0} = l \cos \alpha$

so

(A) $\left\langle{\vec{L}}_{O} , \hat{k}\right\rangle = {J}_{\Omega} \Omega + {J}_{\omega} \left\langle\vec{\omega} , \hat{k}\right\rangle = {J}_{\Omega} \Omega + {J}_{\omega} \omega \sin \alpha$

and

$\left\lVert {\vec{L}}_{O} \right\rVert = \sqrt{{\left\langle{\vec{L}}_{O} , \hat{k}\right\rangle}^{2} + {\left({J}_{\omega} \omega \cos \alpha\right)}^{2}}$

or

(C) $\left\lVert {\vec{L}}_{O} \right\rVert = \sqrt{{\left({J}_{\Omega} \Omega\right)}^{2} + 2 {J}_{\Omega} {J}_{\omega} \Omega \omega \sin \alpha + {\left({J}_{\omega} \omega\right)}^{2}}$

The final numeric results are left to the reader as an exercise.