Acceleration of the centre of mass of the rolling body is given by $$a = \frac{{g\sin \theta }}{{1 + \left( {\frac{I}{{M{R^2}}}} \right)}}$$
Moment of inertia of the ring about an axis perpendicular to the plane of the ring and passing through its centre is given by $$I = M{R^2}$$
$$\eqalign{ & \therefore a = \frac{{g\sin \theta }}{{1 + \frac{{M{R^2}}}{{M{R^2}}}}} \cr & = \frac{{g\sin {{30}^ \circ }}}{{1 + 1}} \cr & = \frac{g}{4} \cr} $$

Moment of inertia of the whole system about the axis of rotation will be equal to the sum of the moments of inertia of all the particles.

$$\eqalign{ & \therefore I = {I_1} + {I_2} + {I_3} + {I_4} \cr & = 0 + 0 + 27 + 16 \cr & = 43\,kg\,{m^2} \cr} $$

We know that density   $$\left( d \right) = \frac{{{\text{mass}}\left( M \right)}}{{{\text{volume}}\left( V \right)}}$$
$$\therefore M = d \times V = d \times \left( {\pi {R^2} \times t} \right)$$
The moment of inertia of a disc is given by $$I = \frac{1}{2}M{R^2}$$

$$\eqalign{ & \therefore \,\,I = \frac{1}{2}\left( {d \times \pi {R^2} \times t} \right){R^2} = \frac{{\pi d}}{2}t \times {R^4} \cr & \therefore \,\,\frac{{{I_X}}}{{{I_Y}}} = \frac{{{t_X}R_X^4}}{{{t_Y}R_Y^4}} \cr & = \frac{{t \times {R^4}}}{{\frac{t}{4} \times {{\left( {4R} \right)}^4}}} \cr & = \frac{1}{{64}} \cr & {\text{So, }}{I_Y} = 64{I_X} \cr} $$

$$r = \sqrt 2 \frac{a}{2}\,\,\,\,\,\,or,\,{r^2} = \frac{{{a^2}}}{2}$$
Net torque about $$O$$ is zero.
Therefore, angular momentum $$\left( L \right)$$  about $$O$$  will be conserved, or $${L_i} = {L_f}$$
$$\eqalign{ & MV\left( {\frac{a}{2}} \right) = {I_0}\,\omega = \left( {{I_{cm}} + M{r^2}} \right)\omega \cr & \omega = \left\{ {\frac{{M{a^2}}}{6} + M\left( {\frac{{{a^2}}}{2}} \right)} \right\}\omega = \frac{2}{3}M{a^2}\omega = \frac{{3v}}{{4a}} \cr} $$

The system is made up of five bodies (three circles and two straight lines) of uniform mass distribution. Therefore we assume the system to be made up of five point masses where the mass of each body is considered at its geometrical centre.

The y-coordinate of the centre of mass is
$$\eqalign{ & {y_{cm}} = \frac{{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3} + {m_4}{y_4} + {m_5}{y_5}}}{{{m_1} + {m_2} + {m_3} + {m_4} + {m_5}}} \cr & \therefore {y_{cm}} = \frac{{6m \times 0 + m \times 0 + m \times a + m \times a + m\left( { - a} \right)}}{{6m + m + m + m + m}} \cr & = \frac{{ma}}{{10m}} \cr & = \frac{a}{{10}} \cr} $$

We know that $$\overrightarrow {{\tau _c}} = \frac{{d\overrightarrow {{L_c}} }}{{dt}}$$
Where $$\overrightarrow {{\tau _c}} = $$  torque about the center of mass of the body and $$\overrightarrow {{L_c}} = $$  Angular momentum about the center of mass of the body. Central forces act along the center of mass. Therefore torque about center of mass is zero.
When $$\overrightarrow {{\tau _c}} = 0$$     then $$\overrightarrow {{L_c}} = {\text{constant}}$$

We know that $$\left| {\vec \tau } \right| = \left| {\frac{{d\vec L}}{{dt}}} \right|$$     where $$L = I\omega $$
$$\therefore \,\,\tau = \frac{d}{{dt}}\left( {I\omega } \right) = \omega \frac{{dI}}{{dt}}\,.....(i)$$
From the situation it is clear that the moment of inertia for (rod + insect) system is increasing.

Let at any instant of time $$'t ’,$$ the insect is at a distance $$x$$  from $$O.$$   At this instant, the moment of inertia of the system is
$$\eqalign{ & I = \frac{1}{3}M{L^2} + m{x^2}\,\,.....(ii) \cr & {\text{From }}\,(i)\,\,\& \,\,(ii) \cr & \tau = \omega \frac{d}{{dt}}\left[ {\frac{1}{3}M{L^2} + m{x^2}\,} \right] = \omega m\frac{d}{{dt}}\left( {{x^2}} \right) \cr & = 2\omega mx\frac{{dx}}{{dt}} \cr & = 2\omega mxv\,\,\,\,\,\left[ {\because x = vt} \right] \cr & \therefore \tau \propto t\,\,\,\,\,\left( {{\text{till }}t = T} \right) \cr} $$
When the insect stops moving, $${\vec L}$$  does not change and therefore $$\tau $$  becomes constant.

As velocity of an automobile vehicle,
$$v = 54\,km/h = 54 \times \frac{5}{{18}} = 15\,m/s$$
Angular velocity of a vehicle, $$v = {\omega _0}r$$
$$ \Rightarrow {\omega _0} = \frac{V}{R} = \frac{{15}}{{0.45}} = \frac{{100}}{3}rad/s$$
So, angular acceleration of an automobile,
$$\alpha = \frac{{\Delta \omega }}{t} = \frac{{{\omega _f} - {\omega _0}}}{t} = \frac{{0 - \frac{{100}}{3}}}{{15}} = \frac{{ - 100}}{{45}}rad/{s^2}$$
Thus, average torque transmitted by its brakes to wheel
$$\tau = I\alpha \Rightarrow 3 \times \frac{{100}}{{45}} = 6.66\,kg{m^2}{s^{ - 2}}$$

Vertical line from hinge $$A$$ must pass through $$C.M.$$  of rod system.
$$\eqalign{ & \tan \theta = \frac{{OP}}{{AP}} = \frac{{\frac{\ell }{2}}}{{\frac{{2\ell }}{3}}} \cr & \tan \theta = \frac{3}{4} \Rightarrow \theta = {\tan ^{ - 1}}\left( {\frac{3}{4}} \right) \cr} $$

The net force acting on a particle undergoing uniform circular motion is centripetal force which always passes through the centre of the circle. The torque due to this force about the centre is zero, therefore, $${\vec L}$$  is conserved about $$O.$$