Rotational Motion

Each bodies is sliding along the frictionless inclined plane and there is no rolling, therefore the acceleration of all the bodies is same $\left( {g\,\sin \,\theta } \right).$

Angular momentum $(L)$

$=$ ( linear momentum ) $ \times $ ( perpendicular distance of the line of action of momentum from the axis of rotation)

$ = mv \times r$

$ = mv \times 0$

$=0$

[ Here $r=0$ because the particle is moving through the line PQ and r is the perpendicular distance from line PQ of the particle ]

When two small spheres of mass $m$ are attached gently, the external torque, about the axis of rotation, is zero.

So, ${{d\overrightarrow L } \over {dt}} = \overline z $ = 0

$\overrightarrow L $ = conserved

So the angular momentum about the axis of rotation is conserved.

$\therefore$ ${I_1}{\omega _1} = {I_2}{\omega _2}$

$ \Rightarrow {\omega _2} = {{{I_1}} \over {{I_2}}}{\omega _1}$

Here Moment of inertia of Disc ${I_1} = {1 \over 2}M{R^2}$ and

After adding two sphere Moment of Inertia of disc and two sphere,

${I_2} = {1 \over 2}M{R^2} + $$2\left( {{1 \over 2}m{R^2} + {1 \over 2}m{R^2}} \right)$

$\therefore$ ${\omega _2} = {{{1 \over 2}M{R^2}} \over {{1 \over 2}MR + 2m{R^2}}} \times {\omega _1} = {M \over {M + 4m}}{\omega _1}$

To determine the moment of inertia of a quarter section of a uniform circular disc of radius $R$ and mass $M$ about an axis perpendicular to its plane and passing through the center of the original disc, we start with the moment of inertia of the full disc.

The moment of inertia of a full uniform circular disc about an axis perpendicular to its plane and passing through its center is given by:

$ I_{\text{full}} = \frac{1}{2} M_{\text{full}} R^2 $

Since the original disc is uniformly distributed, the mass of the full disc $M_{\text{full}}$ can be split evenly into four equal parts, and mass $M$ corresponds to the quarter section. Hence, we have:

$ M = \frac{M_{\text{full}}}{4} $

We also know that the moment of inertia of a quarter section of the disc about the same axis would be one-fourth of that of the entire disc because the mass and area are uniformly distributed. Therefore, the moment of inertia of the quarter section $I_{\text{quarter}}$ is:

$ I_{\text{quarter}} = \frac{1}{4} I_{\text{full}} $

Substituting $ I_{\text{full}} $ from above:

$ I_{\text{quarter}} = \frac{1}{4} \left( \frac{1}{2} M_{\text{full}} R^2 \right) $

Now, replacing $ M_{\text{full}} $ with $ 4M $ (since $ M = \frac{M_{\text{full}}}{4} $), we get:

$ I_{\text{quarter}} = \frac{1}{4} \left( \frac{1}{2} (4M) R^2 \right) $

This simplifies to:

$ I_{\text{quarter}} = \frac{1}{4} \left( 2 M R^2 \right) $

$ I_{\text{quarter}} = \frac{1}{2} M R^2 $

Therefore, the moment of inertia of the quarter section of the disc about the given axis is:

$ \boxed{ \frac{1}{2} MR^2 } $

So, the correct answer is Option A: $ \frac{1}{2} MR^2 $

Mass of the ring $M=\rho L$. Let $R$ be the radius of the ring, then

$ L=2 \pi R \quad \text { or } \quad R=\frac{L}{2 \pi} $

The moment of inertia about an axis passing through the centre $O$ and perpendicular to plane of the loop is $I_z=m R^2$. Using symmetry and perpendicular axis theorem, $I_x+I_y=I_z$, we get

$ I_x=I_y=\frac{1}{2} I_z=\frac{1}{2} m R^2, $

where $I_x$ is the moment of inertia about an axis passing through $O$ and lying in the plane of the loop. The centre of mass $O$ of the loop lies at a perpendicular distance $d=R$ from the axis $\mathrm{XX}^{\prime}$. The parallel axis theorem gives

$ I_{X X^{\prime}}=\frac{1}{2} M R^2+M R^2=\frac{3}{2} M R^2 $

Substituting values of $M$ and $R$

$ I_{X X^{\prime}}=\frac{3}{2}(\rho L)\left(\frac{L^2}{4 \pi^2}\right)=\frac{3 \rho L^3}{8 \pi^2} $