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}$

A bullet of mass $m$ strikes the wooden log of mass M and length L and sticks to it.

Angular momentum of the system, about point O, remains conserved. Initial angular momentum = Final angular momentum.

$\therefore$ $mv \times L = {I_0}\omega $ ..... (i)

${I_0} = {I_{\log }} + {I_b}$

${I_0} = {{M{L^2}} \over 3} + m{L^2}$

${I_0} = {L^2}\left( {{M \over 3} + m} \right) = {L^2}\left( {{{M + 3m} \over 3}} \right)$

$\therefore$ $mvL = {L^2}\left( {{{M + 3m} \over 3}} \right)\omega $

$\omega = {{3mv} \over {L(M + 3m)}}$

Acceleration of a body rolling down the inclined plane

$a = {{g\sin \theta } \over {1 + {I \over {m{R^2}}}}}$

For cylinder, $\quad \mathrm{I}=\frac{1}{2} m \mathrm{R}^{2}$

$a = {{g\sin \theta } \over {1 + {{{1 \over 2}m{R^2}} \over {m{R^2}}}}} = {{g\sin \theta } \over {1 + {1 \over 2}}}$

$a = {{g\sin \theta } \over {{3 \over 2}}}$

$a = {2 \over 2}g\sin \theta $

Let the block of mass $m$ hangs from the point $P$ by a string attached to the hinges of the two ladders.

Newton's second law gives the tension $T$ in the string as $T=m g$. By symmetry, string tension pulls down each ladder by a force

$ T / 2=m g / 2 . $

By Newton's third law, the reaction forces acting on the two ladders at the hinge point $P$ are equal and opposite. These are shown by $R_1$ and $R_2$ in the figure. By symmetry, the normal reaction $N$ at $A$ and $B$ are equal, the friction forces $f$ at $A$ and $B$ are equal in magnitude but opposite in direction. Another force acting on both the ladders is their weight $M g$ which pass through their centre of mass. In equilibrium, the resultant forces on the two ladders are zero i.e.,

$ \begin{aligned} & N+R_2-M g-m g / 2=0, ........(1)\\ & f=R_1, .........(2)\\ & N-R_2-M g-m g / 2=0 .........(3) \end{aligned} $

The equations (1) and (3) give $R_2=0$ (as expected!) and $N=(M+m / 2) g$. In equilibrium, the net torque about any point is zero. Thus, the torque about the point $P$ for the left ladder is

$ M g(L / 2) \cos \theta+f L \sin \theta-N L \cos \theta=0 $ ..........(4)

Substitute $N=(M+m / 2) g$ in equation (4) and simplify to get

$ f=\left(\frac{M+m}{2}\right) g \cot \theta . $

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)$

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

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} $