Rotational Motion

Let the mass of each particle is m.

Then force experienced by each particle, $F = m{\omega ^2}R$

$\therefore$ ${{{F_1}} \over {{F_2}}} = {{m{\omega ^2}{R_1}} \over {m{\omega ^2}{R_2}}}$

$ \Rightarrow $ ${{{F_1}} \over {{F_2}}} = {{{R_1}} \over {{R_2}}}$

Let mass of the semi circular disc = M

Now assume a disc which is combination of two semi circular parts. Let $I$ be the moment of inertia of the uniform semicircular disc. So $2I$ will be the moment of inertia of the full circular disc and 2M will be the mass.

$ \Rightarrow 2I = {{2M{r^2}} \over 2}$

$ \Rightarrow I = {{M{r^2}} \over 2}$

For solid sphere the moment of inertia of $A$ about its diameter

${I_A} = {2 \over 5}M{R^2}.$

The moment of inertia of a hollow sphere $B$ about its diameter

${I_B} = {2 \over 3}M{R^2}.$

$\therefore$ ${I_A} < {I_B}$

Solid sphere is rotating in free space that means no external torque is operating on the sphere.

Angular momentum will remain the same since external torque is zero.

We know that density $\left( d \right) = {{mass\left( M \right)} \over {volume\left( V \right)}}$

$\therefore$ Mass of disc $M = d \times V = d \times \left( {\pi {R^2} \times t} \right).$

The moment of inertia of any disc is $I = {1 \over 2}M{R^2}$

$\therefore$ $I = {1 \over 2}\left( {d \times \pi {R^2} \times t} \right){R^2} = {{\pi d} \over 2}t \times {R^4}$

$\therefore$ ${{{I_X}} \over {{I_Y}}} = {{{t_X}R_X^4} \over {{t_Y}R_Y^4}}$

$ = {{t \times {R^4}} \over {{t \over 4} \times {{\left( {4R} \right)}^4}}}$

$ = {1 \over {64}}$

We know Rotational Kinetic Energy$={1 \over 2}I{\omega ^2},$

Angular Momentum $L = I\omega \Rightarrow I = {L \over \omega }$

$\therefore$ Initial $K.E. = {1 \over 2}{L \over \omega } \times {\omega ^2} = {1 \over 2}L\omega $

Final $K.E'$ = ${{K.E} \over 2}$ = ${1 \over 2}{L'} \times 2\omega $

$\therefore$ ${{K.E} \over {K.E'}} = {{L \times \omega } \over {L' \times \omega '}} $

$\Rightarrow {{K.E} \over {{{K.E} \over 2}}} = {{L \times \omega } \over {L' \times 2\omega }}$

$\therefore$ $L' = {L \over 4}$

As we know $\overrightarrow \tau = \overrightarrow r \times \overrightarrow F $

So the angle between $\overrightarrow \tau $ and $\overrightarrow r $ is ${90^ \circ }$ and the angle between $\overrightarrow t $ and $\overrightarrow F $ is also ${90^ \circ }.$

We also know that the dot product of two vectors which have an angle of ${90^ \circ }$ between them is zero.

$\therefore$ $\overrightarrow {r.} \vec \tau = 0{\mkern 1mu} $ and $\overrightarrow {F.} \overrightarrow \tau = 0\,\,$

Therefore $(d)$ is the correct option.

Moment of Inertia of a circular wire about an axis $nn'$ passing through the centre of the circle and perpendicular to the plane of the circle $ = M{R^2}$

As shown in the figure, $X$-axis and $Y$-axis lies in the plane of the ring. Then by perpendicular axis theorem

${I_X} + {I_Y} = {I_Z}$

$ \Rightarrow 2{I_X} = M{R^2}\,$ $\left[ \, \right.$ as ${I_X} - {I_Y}$ (by symmetry) and ${I_Z} = M{R^2}$ $\left. \, \right]$

$\therefore$ ${I_X} = {1 \over 2}M{R^2}$

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)$

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