Rotational Motion

Let the large stationary disk have its centre at the origin 0 . The two smaller disks Disk 1 and Disk 2 each have a radius $\mathrm{r}=\frac{\mathrm{R}}{50}$.

When the two smaller disks are in contact with each other while resting on the circumference of the large disk, the distance between the centre of the large disk and the centre of either smaller disk is :

$ \mathrm{R}_{\mathrm{c}}=\mathrm{R}+\mathrm{r} $

Since the two smaller disks are in contact, the distance between their centres is exactly 2 r .

Let $\Delta \theta$ be the angular separation between their centres at centre of the large disk $O$.

Using the properties of an isosceles triangle formed by O and the two small disk centres :

$ \sin \alpha=\frac{r}{R+r} $

Here, $\alpha=\frac{\Delta \theta}{4} \Rightarrow \sin \left(\frac{\Delta \theta}{4}\right)=\frac{\mathrm{r}}{\mathrm{R}+\mathrm{r}}$

Given the small-angle approximation $\sin (\Delta \theta) \approx \Delta \theta$, we can also approximate

$ \sin \left(\frac{\Delta \theta}{4}\right) \approx \frac{\Delta \theta}{4} $

Therefore $\frac{\Delta \theta}{4}=\frac{\mathrm{r}}{\mathrm{R}+\mathrm{r}}$

$ \Delta \theta=\frac{4 r}{R+r} $

Substituting $\mathrm{R}=50 \mathrm{r}$ into this equation:

$ \Delta \theta=\frac{4 \mathrm{r}}{50 \mathrm{r}+\mathrm{r}}=\frac{4}{51} \text { radians } $

Let the small disk rotate about its own centre with an angular velocity $\omega^{\prime}$. The point of contact with the stationary large disk must have a linear velocity of zero to satisfy the no-slip condition.

From rotation about its own centre relative to the contact point $\mathrm{v}_{\mathrm{cm}}=\omega^{\prime} \cdot \mathrm{r}$

From orbital path around the large disk's centre $\mathrm{v}_{\mathrm{cm}}=\Omega \cdot(\mathrm{R}+\mathrm{r})$

$\Omega (R + r) = \omega' \cdot r$

$\Rightarrow $ $\Omega = \omega' \left(\frac{r}{R + r}\right)$

Substituting $R = 50r$:

$\Omega = \omega' \left(\frac{r}{51r}\right) = \frac{\omega'}{51}$

It is given that the two smaller disks spin with constant angular velocities $\omega$ and $2\omega$ in opposite directions along the circumference.

For Disk 1 ($\omega'_1 = \omega$):

$\Omega_1 = \frac{\omega}{51}$

For Disk 2 ($\omega'_2 = 2\omega$):

$\Omega_2 = \frac{2\omega}{51}$

Since they travel in opposite directions around the large disk, their relative orbital angular velocity

$\Omega_{\mathrm{rel}} = \Omega_1 + \Omega_2 = \frac{\omega}{51} + \frac{2\omega}{51} = \frac{3\omega}{51}$

Initially, the two disks are in contact with an angular separation between their centres equal to

$\Delta \theta = \frac{2}{51}$

For them to come into contact again, the total angular path covered by the two centres relative to each other must equal to $\Phi = 2\pi - \Delta \theta$

Since each disk has a radius $r$, when they meet again, their centres will once again be separated by $\Delta \theta$. Thus, the net angular displacement between the two centres around the large disk is :

$\theta_{\mathrm{rel}} = 2\pi - 2\Delta \theta$

Substituting the value of $\Delta \theta = \frac{2}{51}$:

$\theta_{\mathrm{rel}} = 2\pi - \frac{4}{51}$

The time $\tau$ taken to cover this relative angular displacement at a relative angular speed $\Omega_{\mathrm{rel}}$ is :

$\tau = \frac{\theta_{\mathrm{rel}}}{\Omega_{\mathrm{rel}}}$

$\tau = \frac{2\pi - \frac{4}{51}}{\frac{3\omega}{51}}$

$\tau = 51 \times \left(2\pi - \frac{4}{51}\right) / 3\omega$

Therefore, the correct option is (C).

When the uniform solid cylinder rolls along the horizontal table and strikes the sharp corner edge, its smooth rolling motion changes. The corner point P acts as a temporary pivot.

The cylinder begins to pivot or rotate purely about this fixed corner point P . As long as it stays in contact with this corner, the centre of mass ( $C$ ) executes circular motion of radius $R$ around $P$.

The moment of inertia of a uniform solid cylinder of mass M and radius R about its central longitudinal axis is :

$ \mathrm{I}_{\mathrm{cm}}=\frac{1}{2} \mathrm{MR}^2 $

According to the parallel axis theorem, the moment of inertia about an axis at a distance $\mathrm{d}=\mathrm{R}$ from the centre is :

$ \mathrm{I}_{\mathrm{p}}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2=\frac{1}{2} \mathrm{MR}^2+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2 $

The centre of mass immediately after the cylinder pivots onto the corner gravity is a non-impulsive force, so its effect over the infinitesimally short duration is negligible.

The reaction forces at the corner P are impulsive, but they pass directly through P . Therefore, the net torque about the pivot P during the collision is zero, which means angular momentum about P is conserved.

The cylinder is rolling without slipping with a centre of mass velocity $\mathrm{v}_0$ and angular velocity $\omega_0=\mathrm{v}_0 \mathrm{R}$.

The total angular momentum about the level of the corner P which is at a height R below the centre of mass path is:

$L_{\text{initial}} = Mv_0R + I_{\text{cm}}\omega_0$

$L_{\text{initial}} = Mv_0R + \left(\frac{1}{2}MR^2\right)\left(\frac{v_0}{R}\right) = \frac{3}{2}Mv_0R$

After impact the cylinder is rotating about the pivot P with a new angular velocity $\omega_1$. Its angular momentum about P is:

$L_{\text{final}} = I_p\,\omega_1 = \frac{3}{2}MR^2\omega_1$

Using conservation of angular momentum

$L_{\text{initial}} = L_{\text{final}}$

$\frac{3}{2}Mv_0R = \frac{3}{2}MR^2\omega_1 \rightarrow \omega_1 = \frac{v_0}{R}$

As the cylinder tilts forward around the corner, the angle $\theta$ with the vertical increases from $0$. The centre of mass descends vertically. When the line PC has tilted by an angle $\theta$.

Initial height of centre of mass relative to pivot $= R$

Height of centre of mass at angle $\theta$ $= R\cos\theta$

Loss in potential energy

$= Mg(R - R\cos\theta) = MgR(1 - \cos\theta)$

By conservation of total mechanical energy between the initial vertical position of PC ($\theta = 0$, angular speed $\omega_1$) and a tilted position $\theta$ (angular speed $\omega$):

$\Delta K = -\Delta U$

$\frac{1}{2}I_p\omega^2 - \frac{1}{2}I_p\omega_1^2 = MgR(1 - \cos\theta)$

Substituting $I_p = \frac{3}{2}MR^2$ and $\omega_1 = \frac{v_0}{R}$:

$\frac{1}{2}\left(\frac{3}{2}MR^2\right)\omega^2 - \frac{1}{2}\left(\frac{3}{2}MR^2\right)\left(\frac{v_0}{R}\right)^2 = MgR(1 - \cos\theta)$

$\frac{3}{4}MR^2\omega^2 - \frac{3}{4}Mv_0^2 = MgR(1 - \cos\theta)$

Since the linear velocity of the centre of mass at any instant during this circular path is $v = \omega R$,

So, $R^2\omega^2 = v^2$:

$\frac{3}{4}Mv^2 = \frac{3}{4}Mv_0^2 + MgR(1 - \cos\theta)$

$v^2=v_0^2+\frac{4}{3} \operatorname{gR}(1-\cos \theta) \ldots (i)$

The centre of mass travels along a circular arc of radius R around the pivot P .

For motion along the radial direction pointing toward the centre of rotation P .

The component of gravity pulling toward the pivot is $\mathrm{Mg} \cos \theta$

The normal contact force exerted by the corner pulling outward away from the pivot is N

The net inward centripetal force is $Mg\\cos\theta - N = \frac{Mv^2}{R}$

The cylinder loses contact with the corner when the normal contact force drops to zero $N = 0$:

$Mg\cos\theta = \frac{Mv^2}{R} \rightarrow v^2 = gR\cos\theta \ldots \text{(ii)}$

So, from energy conservation $v^2 = v_0^2 + \frac{4}{3}gR - \frac{4}{3}gR\cos\theta \ldots \text{(i)}$

From the radial force equilibrium $\cos\theta = \frac{v^2}{gR} \rightarrow \frac{4}{3}gR\cos\theta = \frac{4}{3}v^2 \ldots \text{(i)}$

$v^2 = v_0^2 + \frac{4}{3}gR - \frac{4}{3}v^2$

$v^2 + \frac{4}{3}v^2 = v_0^2 + \frac{4}{3}gR$

$\frac{7}{3}v^2 = v_0^2 + \frac{4}{3}gR$

The initial velocity is $v_0 = \sqrt{\frac{gR}{3}}$, $\rightarrow v_0^2 = \frac{gR}{3}$.

$\frac{7}{3}v^2 = \frac{gR}{3} + \frac{4}{3}gR$

$\frac{7}{3}v^2 = \frac{5}{3}gR$

$7v^2 = 5gR$

$v^2 = \frac{5gR}{7}$

$v = \sqrt{\frac{5gR}{7}}$

Therefore, the correct option is (B).

The moment of inertia of single rod about the axis passing through its center and making angle $\theta$ along the length of rod is :

$ \mathrm{I}_{\text {center }}=\frac{1}{12} \mathrm{ml}^2 \sin ^2 \theta $

If the axis instead passes through one end of the rod at an angle $\theta$,

$ \mathrm{I}_{\mathrm{end}}=\frac{1}{3} \mathrm{ml}^2 \sin ^2 \theta $

Structure (P) consists of two uniform rods, each of mass m and length l, joined at a right angle ($90^\circ$) at point C.

The axis OCO' passes through the junction C and bisects the angle, making an angle $\theta = 45^\circ$ with both rods.

Since the axis passes through the end of both rods,

$I_1 = \frac{1}{3}ml^2\sin^2(45^\circ) = \frac{1}{3}ml^2\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{6}ml^2$

$I_2 = \frac{1}{3}ml^2\sin^2(45^\circ) = \frac{1}{3}ml^2\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{6}ml^2$

The total moment of inertia of structure (P) is the sum of the two parts:

$I_P = I_1 + I_2 = \frac{1}{6}ml^2 + \frac{1}{6}ml^2 = \frac{1}{3}ml^2$

$P \rightarrow 5$

Rods CA and CB have one end attached at vertex C, which lies directly on the axis of rotation. Each rod is inclined at an angle $\theta = 60^\circ$ to the axis.

$I_{CA} = \frac{1}{3} m l^2 \sin^2(60^\circ) = \frac{1}{3} m l^2 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{3} m l^2 \left(\frac{3}{4}\right) = \frac{1}{4} m l^2$

$I_{CB} = \frac{1}{3} m l^2 \sin^2(60^\circ) = \frac{1}{3} m l^2 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{3} m l^2 \left(\frac{3}{4}\right) = \frac{1}{4} m l^2$

Base rod AB is horizontal and completely parallel to the axis OCO'.

This distance is equal to the altitude (height h) of the equilateral triangle.

$h = l \sin(60^\circ) = \frac{\sqrt{3}}{2} l$

Since the entire mass m of rod AB is distributed uniformly at a constant distance h from the axis, its moment of inertia is:

$I_{AB} = m h^2 = m \left(\frac{\sqrt{3}}{2} l\right)^2 = \frac{3}{4} m l^2$

So, the total moment of inertia for structure Q

$I_Q = I_{CA} + I_{CB} + I_{AB}$

$I_Q = \frac{1}{4} m l^2 + \frac{1}{4} m l^2 + \frac{3}{4} m l^2 = \frac{5}{4} m l^2$

$Q \rightarrow 1$

The rods $A B$ and $B C$ meet at vertex $B$.

The axis passing through A and C , it forms a right-angled triangle ABC where the diagonal AC is the axis. The angle made by axis with every rod is $45^{\circ}$. As all four sides are symmetric about axis of rotation therefore, the total moment of inertia of the square loop about its diagonal axis is:

$ \mathrm{I}=\mathrm{I}_{\mathrm{AB}}+\mathrm{I}_{\mathrm{BC}}+\mathrm{I}_{\mathrm{CD}}+\mathrm{I}_{\mathrm{DA}}=4 \times \frac{1}{3} \mathrm{ml}^2 \sin ^2 45^{\circ}=4 \times \frac{1}{3} \mathrm{ml}^2 \times \frac{1}{2}=\frac{2}{3} \mathrm{ml}^2 $

R→ 4

Rods CA and CB have one end attached at vertex C, which lies directly on the axis of rotation. Each rod is inclined at an angle $\theta = 30^{\circ}$ to the axis.

$I_{CA} = \frac{1}{3}ml^2 \sin^2(30^{\circ}) = \frac{1}{3}ml^2\left(\frac{1}{2}\right)^2 = \frac{1}{3}ml^2\left(\frac{1}{4}\right) = \frac{1}{12}ml^2$

$I_{CB} = \frac{1}{3}ml^2 \sin^2(30^{\circ}) = \frac{1}{3}ml^2\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{3}ml^2\left(\frac{1}{4}\right) = \frac{1}{12}ml^2$

So, the total moment of inertia for structure S

$I_S = I_{CA} + I_{CB}$

$I_S = \frac{1}{12}ml^2 + \frac{1}{12}ml^2 = \frac{2}{12}ml^2 = \frac{1}{6}ml^2$

$S \rightarrow 2$

So, the correct matches are $\mathrm{P} \rightarrow 5, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 4, \mathrm{~S} \rightarrow 2$.

Therefore, the correct option is (A).

Let's disc rolls to $\phi$ angular displacement about its centre:

As the disc rolls without slipping. we can write,

$ x=r \phi=(R-r) \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Now, Let's apply $\tau=I \alpha$ about contact

$ \begin{aligned} & \text { point } \\ & \tau=m g \sin \theta \cdot r+k x r=\left(\frac{m r^2}{2}+m r^2\right) \alpha^{\prime} \\ & \Rightarrow m g \theta \cdot r+k(R-r) \theta r=\frac{3 m r^2}{2} \alpha^{\prime} \\ & \Rightarrow[m g+k(R-r)] \theta r=\frac{3}{2} m r^{\prime} \alpha^{\prime} \\ & \Rightarrow[m g+k(R-r)] \theta=\frac{3}{2} m r \alpha^{\prime}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

here, $\alpha^{\prime}=\frac{d^2 \phi}{d t^2}$

from (1), $\phi=\left(\frac{R-r}{r}\right) 0$

$ \begin{aligned} & \Rightarrow \quad \alpha^{\prime}=\frac{d^2 \phi}{d t^2}=\left(\frac{R-r}{r}\right) \frac{d^2 \theta}{d t^2} \\ & \Rightarrow \quad \alpha^{\prime \prime}=\left(\frac{R-r}{r}\right) \alpha\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

Now, from (2) and (3),

$ \begin{aligned} & {[m g+k(R-r)] \theta=\frac{3}{2} m r\left(\frac{R-r}{r}\right) \alpha} \\ & \Rightarrow[m g+k(R-r)] \theta=\frac{3}{2} m(R-r) \alpha \end{aligned} $

Now, we know, for SHM,

$ \begin{gathered} \alpha=\omega^2 \theta \\ \Rightarrow \quad\left[\frac{m g+k(R-r)}{\frac{3}{2} m(R-r)}\right] \phi=\omega^2 \phi \\ \Rightarrow \frac{2}{3} \frac{g}{(R-r)}+\frac{2}{3} \frac{k}{m}=\omega^2 \\ \Rightarrow \omega=\sqrt{\frac{2}{3}\left[\frac{g}{(R-r)}+\frac{k}{m}\right]} \end{gathered} $

A bar, having a mass of $ M = 1 $ kg and a length of $ L = 0.2 $ m, is hinged at a point P. A particle with a mass of $ m = 0.1 $ kg moves in a direction perpendicular to the bar at a velocity of $ u = 5 $ m/s. This particle collides elastically with the bar at a point that is $ \frac{L}{2} $ (0.1 m) away from the pivot point P. As a result of this collision, the bar begins to rotate with an angular velocity $ \omega $, and the particle recoils with a new velocity $ v $.

(I)

${v_{net}} = \sqrt 2 \omega R = \sqrt 2 $

I $\to$ S

(II) ${v_A}(t = 0.13) = {{5\pi } \over {\sqrt 2 }}\cos 45\widehat i + \left[ {{{5\pi } \over {\sqrt 2 }} \times {1 \over {\sqrt 2 }} - g \times 0.1} \right]\widehat j$

$ = {{5\pi } \over 2}\widehat i + \left( {{{5\pi } \over 2} - 1} \right)\widehat j$

${v_B}(t = 0.1\,\sec ) = {{ - 5\pi } \over 2}\widehat i + \left( {{{5\pi } \over 2}} \right)\widehat j$

After $t = 0.1$, relative velocities should not change.

${v_{rel}}(t = 0.1\,\sec ) = \left| {5\pi \widehat i - \widehat j} \right|$

$ = \sqrt {25{\pi ^2} + 1} $

II $\to$ T

(III) $x = {x_A} - {x_B}$

$ = {x_0}\sin t - {x_0}\sin \left( {t + {\pi \over 2}} \right)$

$ = \sqrt 2 {x_0}\sin \left( {t - {\pi \over 4}} \right)$

${v_{rel}} = {{dx} \over {dt}} = \sqrt 2 {x_0}\cos \left( {t - {\pi \over 4}} \right)$

$ = \sqrt 2 \cos \left( {{\pi \over 3} - {\pi \over 4}} \right)$

$ = \sqrt 2 \times {{\sqrt 3 + 1} \over {2\sqrt 2 }} = {{\sqrt 3 + 1} \over 2}$

III $\to$ P

(IV) ${v_{rel}} = \sqrt {{3^2} + {1^2}} = \sqrt {10} $

IV $\to$ R

Given, force applied on body

$\overrightarrow F = (\alpha t\widehat i + \beta \widehat j)$

Here, $\alpha$ = 1.0 N s^$-$1 and $\beta$ = 1.0 N. Therefore,

$\overrightarrow F = t\widehat i + \widehat j$

By Newton's second law of motion, F = ma. Given m = 1.0 kg. Therefore,

$\overrightarrow F = \overrightarrow a \Rightarrow \overrightarrow a = (t\widehat i + \widehat j)$

We know $\overrightarrow a = {{d\overrightarrow v } \over {dt}} \Rightarrow \overrightarrow v = \int {\overrightarrow a dt} $

$ \Rightarrow \overrightarrow v = \int {(t\widehat i + \widehat j)dt = {{{t^2}} \over 2}\widehat i + t\widehat j} $ ..... (1)

Also, $\overrightarrow v = {{\overrightarrow {dr} } \over {dt}} \Rightarrow \overrightarrow r = \int {\overrightarrow v dt} $

$ \Rightarrow \overrightarrow r = \int {\left( {{{{t^2}} \over 2}\widehat i + t\widehat j} \right)dt = {{{t^3}} \over 6}\widehat i + {{{t^2}} \over 2}\widehat j} $ ..... (2)

Now, $\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {{t^3}/6} & {{t^2}/2} & 0 \cr t & 1 & 0 \cr } } \right|$

$\overrightarrow \tau = \widehat i(0) - \widehat j(0) + \widehat k\left( {{{{t^3}} \over 6} - {{{t^3}} \over 2}} \right) = {{ - 1} \over 3}{t^3}\widehat k$ N m

At t = 1.0 s, $\overrightarrow \tau = {{ - 1} \over 3}\widehat k$ N m

Hence, option (A) is correct.

Now, from Eq. (1), $\overrightarrow v = {{{t^2}} \over 2}\widehat i + t\widehat j$

At t = 1.0 s, $\overrightarrow v = \left( {{1 \over 2}\widehat i + \widehat j} \right)$ m s^$-$1 = $ = {1 \over 2}(\widehat i + 2\widehat j)$ m s^$-$1

Hence, option (C) is correct.

Let n be the number of sides of a regular polygon. By symmetry, its centre of mass O will be equidistant from each vertex i.e., it lies at the centre of the circumscribed circle. Let r be the radius of circumscribed circle and h be the perpendicular distance of O from any side (see figure). The angle subtended by any side on the centre O is 2$\pi$/n and $\angle$PON = $\pi$/n.

When polygon rolls about the vertex P (without slipping or sliding), the point O moves in a circle of radius r centred at P. The point O reaches the maximum height (point O' in the figure) when PO' is perpendicular to PQ. Thus, the maximum increase in height of the locus of the centre of mass O is given by

$\Delta = r - h = {h \over {\cos (\pi /n)}} - h = h\left( {{1 \over {\cos (\pi /n)}} - 1} \right)$.

We discuss the options as follows:

$\bullet$ For option (B) : From law of conservation of momentum, we have

mv = MV ...... (1)

where v is the velocity of point mass m and V is velocity of block of mass M.

Now, by using law of conservation of energy, after the point mass is released, we have the following:

Loss in P.E. of mass m = Gain in K.E. of both mass M and the mass m.

$ \Rightarrow mgh = {1 \over 2}m{v^2} + {1 \over 2}M{V^2}$

Substituting $V = {{mu} \over M}$ [from Eq. (1)], we get

$mgh = {1 \over 2}m{v^2} + {1 \over 2}M{\left( {{{mv} \over M}} \right)^2}$

$mgh = {1 \over 2}m{v^2} + {1 \over 2}M{{{m^2}{v^2}} \over {{M^2}}}$

$mgR = {1 \over 2}m{v^2} + {1 \over 2}{{{m^2}{v^2}} \over M}$

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

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

Rearranging this equation, we get

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

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

$ \Rightarrow v = \sqrt {{{2gR} \over {\left( {1 + {m \over M}} \right)}}} $

Hence, option (B) is correct.

$\bullet$ For option (C) : Now, from Eq. (1), we have

$V = {m \over M}v = {m \over M}\sqrt {{{2gR} \over {\left( {1 + {m \over M}} \right)}}} $

$V = \sqrt {{{{m^2}2gR} \over {{M^2}\left( {1 + {m \over M}} \right)}}} = \sqrt {{{{m^2}2gR} \over {{M^2}\left( {{{M + m} \over M}} \right)}}} $

$V = \sqrt {{{{m^2}2gR} \over {M(m + M)}}} \Rightarrow V = m\sqrt {{{2gR} \over {M(m + M)}}} $

Since, there is no external force acting on the system, the centre of mass does not change.

Now, if the change in position of mass m is $\Delta$x and the change in position of mass M is $\Delta$X, then

m$\Delta$x + M$\Delta$x = 0 (since centre of mass will not change)

Now, we know that the change in position of point mass m is

$\Delta$x = R $-$ x

and the change in position of block of mass M is

$\Delta$X = $-$x

Therefore,

m(R $-$ x) $-$ Mx = 0

m(R $-$ x) = Mx

mR $-$ mx = Mx

mR = Mx + mx

mR = (M + m)x

$\Rightarrow$ x = ${{mR} \over {M + m}}$

The change in position of block of mass M is

$\Delta$x = $-$x = ${{ - mR} \over {M + m}}$

Thus, x-component of the displacement of the centre of mass of the block is

$M = {{ - mR} \over {M + m}}$

Hence, option (C) is also correct.

The given particle of mass m experiences a centrifugal force:

F = ma = mr$\omega$²

$\Rightarrow$ a = r$\omega$²

$\Rightarrow$ ${{dv} \over {dt}} = r{\omega ^2}$

or, $\left( {{{dv} \over {dr}}} \right)\left( {{{dr} \over {dt}}} \right) = r{\omega ^2}$ ($\because$ ${{{dr} \over {dt}}}=v$)

$ \Rightarrow vdv = r{\omega ^2}dr$

$\int\limits_0^v {vdv = {\omega ^2}\int\limits_{R/2}^r {rdr} } $ ($\because$ at t = 0, $r = {R \over 2}$)

$ \Rightarrow {{{v^2}} \over 2} = {\omega ^2}\left( {{{{r^2}} \over 2}} \right)_{R/2}^r$

$ \Rightarrow v = \omega \sqrt {{r^2} - {{{R^2}} \over 4}} $

Now, $v = {{dr} \over {dt}}$

Therefore, ${{dr} \over {dt}} = \omega \sqrt {{r^2} - {{{R^2}} \over 4}} $

or, $\int\limits_{R/2}^r {{{dr} \over {\sqrt {{r^2} - {{{R^2}} \over 4}} }} = \omega \int\limits_0^t {dt} } $ ...... (1)

To solve this integral, we substitute $r = {R \over 2}\sec \theta $; therefore,

$dr = {R \over 2}\sec \theta \tan \theta d\theta $

$ \Rightarrow \int\limits_{{\pi \over 2}}^{{{\sec }^{ - 1}}(2r/R)} {{{(r/2)\sec \theta \tan \theta d\theta } \over {\sqrt {{{{R^2}} \over 4}{{\tan }^2}\theta } }} = \omega t} $

$ \Rightarrow \omega t = \int\limits_{\pi /2}^{{{\sec }^{ - 1}}(2r/R)} {\sec \theta d\theta } $

$ = {\left[ {\ln \left| {\sec \theta + \tan \theta } \right|} \right]^{{{\sec }^{ - 1}}(2r/R)}}$

$ = \left[ {\ln \left| {\sec \theta + \sqrt {{{\sec }^2}\theta - 1} } \right|} \right]_{\pi /2}^{{{\sec }^{ - 1}}(2r/R)}$

$ \Rightarrow \omega t = \ln \left[ {{{2r} \over R} + {{\sqrt {4{r^2} - {R^2}} } \over R}} \right]$

Let ${{2r} \over R} = x$. Therefore,

${e^\omega } = x + \sqrt {{x^2} - 1} $

$ \Rightarrow ({e^{\omega t}} - x)^2 = {x^2} - 1$

$ \Rightarrow {e^{2\omega t}} - 2x{e^{\omega t}} + x^2 = x^2 - 1$

$ \Rightarrow {e^{2\omega t}} - 2x{e^{\omega t}} + 1 = 0$

$ \Rightarrow x = {{{e^{2\omega t}} + 1} \over {2{e^{\omega t}}}} = {{{e^{\omega t}} + {e^{ - \omega t}}} \over 2}$

On replacing x by ${{2r} \over R}$, we get

${{2r} \over R} = {{{e^{\omega t}} + {e^{ - \omega t}}} \over 2}$

or, $r = {R \over 4}[{e^{\omega t}} + {e^{ - \omega t}}]$

Hence, option (c) is correct.

The rotational quantity of velocity given by

${v_{rot}} = {{dr} \over {dt}} = {R \over 4}[\omega {e^{\omega t}} - \omega {e^{ - \omega t}}]$

${v_{rot}} = {{\omega R} \over 4}[{e^{\omega t}} - {e^{ - \omega t}}]$ ...... (1)

The force experienced by the block in the inertial frame is

${\overrightarrow F _{in}} = - m{\omega ^2}r\widehat i - mg\widehat k$ ..... (2)

Therefore, the force experienced by the block in the rotating frame is

${\overrightarrow F _{rot}} = {\overrightarrow F _{in}} + 2m({\overrightarrow v _{rot}} \times \omega \widehat k) + m(\omega \widehat k \times r\widehat i) \times \omega \widehat k$

${\overrightarrow F _{rot}} = ( - m{\omega ^2}r\widehat i - mg\widehat k) + 2m{{\omega R} \over r}[{e^{\omega t}} - {e^{ - \omega t}}]\widehat i \times \omega \widehat k + m{\omega ^2}r\widehat i$

$ = - {{m{\omega ^2}R} \over 2}[{e^{\omega t}} - {e^{ - \omega t}}]\widehat j - mg\widehat k$

Therefore, the net reaction of the disc on the block is

${\overrightarrow F _{reaction}} - {\overrightarrow F _{rot}} = {{m{\omega ^2}R} \over 2}[{e^{\omega t}} - {e^{\omega t}}]\widehat j + mg\widehat k$

Hence, option (B) is correct.

Ring has mass M and radius R. Initial angular speed of ring = $\omega$. Two point masses, each of mass are at rest at O. Initial angular momentum of ring and point masses system,

${L_i} = {I_R}\omega + {I_m}\omega + {I_m}\omega $

$ = M{R^2}\omega + 0 + 0 = M{R^2}\omega $

After some time, situation is changed as shown in the figure.

Angular speed of the system, $\omega ' = {8 \over 9}\omega $; $OA = {{3R} \over 5}$; OB = r = ?

Moment of inertia about O of point mass at A,

${I_A} = {M \over 8} \times {{9{R^2}} \over {25}}$

Moment of inertia about O of point mass at B, ${I_B} = {M \over 8}{r^2}$

Fina angular momentum of the system

${L_f} = M{R^2}\omega ' + {I_A}\omega ' + {I_B}\omega '$

$ = M{R^2} \times {{8\omega } \over 9} + {M \over 8} \times {{9{R^2}} \over {25}} \times {{8\omega } \over 9} + {M \over 8}{r^2} \times {{8\omega } \over 9}$

As there is no external torque acting on the system so its angular momentum will be conserved, L_i = L_f

$M{R^2}\omega = M{R^2} \times {{8\omega } \over 9} + {M \over 8} \times {{9{R^2}} \over {25}} \times {{8\omega } \over 9} + {M \over 8}{r^2} \times {{8\omega } \over 9}$

${R^2} = {{8{R^2}} \over 9} + {{{R^2}} \over {25}} + {{{r^2}} \over 9} \Rightarrow {{{r^2}} \over 9} = {{16} \over {225}}{R^2}$ $\therefore$ $r = {4 \over 5}R$

The relative velocity of the point $P$ w.r.t. the point $Q$ is given by

$ \vec{v}_r=\vec{v}_P-\vec{v}_Q $ .........(1)

It is easy to see that $\left|\vec{v}_P\right|=\left|\vec{v}_Q\right|=\omega R$ and angle traversed in time $t$ is $\omega t$. Thus, velocities of $P$ and $Q$ are

$ \begin{aligned} & \vec{v}_P=\omega R(-\sin \omega t \hat{\imath}-\cos \omega t \hat{\jmath}) \\ & \vec{v}_Q=\omega R(\sin \omega t \hat{\imath}-\cos \omega t \hat{\jmath}) \end{aligned} $

Substitute $\vec{v}_P$ and $\vec{v}_Q$ in equation (1) to get $\vec{v}_r=$ $-2 \omega R \sin \omega t \hat{\imath}$ and thus $v_r=2 \omega R|\sin \omega t|$.

$\left(\frac{1}{8}\right)^{\mathrm{th}}$ rotation of the disc with a constant $\Omega$ implies that the disc has turned through

$ \frac{1}{8} \times 360^{\circ}=45^{\circ} $

The orientation of the disc now is

For the particle thrown from $Q$ towards $R$, will have only the velocity given to it with respect to disc i.e., it possess a velocity only in the Y-Z plane and covers a range along OR. Hence, it can be seen to land on the disc in the unshaded region. Actually point $O$ of the disc has zero velocity and $Q$ being very close to $O$, will have only the velocity given to it w.r.t disc.

For the particle thrown from P , will have an additional velocity $=R \Omega$ along +X direction i.e., it will cover a range $=\frac{R}{2}$ along PO as well as a distance $=(R \omega) \frac{T}{8}=\frac{R}{8} \times 2 \pi=\frac{\pi R}{4}$ along $+X$ direction

The resultant displacement of particle P is thus obtained as

$\begin{aligned} \Rightarrow \quad d & =\frac{R}{2} \sqrt{1+\frac{\pi^2}{4}} \\ \Rightarrow \quad \pi^2 & =10 \\ d & =\frac{\mathrm{R}}{4} \sqrt{14} \\ \tan \theta & =\frac{\pi \mathrm{R} / 4}{\mathrm{R} / 2}=\frac{\pi}{2}\end{aligned}$

In the diagram shown,

$ \begin{aligned} P S & =R \cos \theta \\ & =R \times \frac{2}{\sqrt{14}} \\ & =\frac{2 R}{\sqrt{14}} \end{aligned} $

Since, $d>$ PS $\quad$ PS $=\frac{R}{7} \sqrt{14}$

Hence, the particle from P also lands in the unshaded region.

As depicted in the figure shown here, when the system, as a whole, turns by 180$^\circ$, the disc also turns by 180$^\circ$ about its vertical axis. Hence, the instantaneous axis that passes through the centre of mass is vertical in both cases (a) and (b).

As depicted in the figure shown here, when the system, as a whole, turns by 180$^\circ$, the disc also turns by 180$^\circ$ about vertical axis. Hence, for both cases (a) and (b), the angular speed about the instantaneous axis that passes through the centre of mass is $\omega$.

Angular momentum of a particle about a point is given by:

L = r $\times$ p = m (r $\times$ v)

For L_O

$\left| L \right| = (mvr\sin \theta ) = m(R\omega )(R)\sin 90^\circ $

$ = m{R^2}\omega $ = constant

Direction of L_O is always upwards. Therefore, complete L_O is constant, both in magnitude as well as direction.

For L_P

$\left| {{L_P}} \right| = (mvr\sin \theta ) = m(R\omega )(l)\sin 90^\circ $

$ = (mRl\omega )$

Magnitude of L_P will remain constant but direction of L_P keeps on changing.

Let M and l be the mass and length of the rod respectively and m be the mass of the insect. Let the insect be at a distance x from O at any instant of time t.

$\therefore$ x = vt ....... (i)

Angular momentum of the system about O,

$L = \left[ {{{M{l^2}} \over {12}} + m{x^2}} \right]\omega $

$ = \left[ {{{M{l^2}} \over {12}} + m{{(vt)}^2}} \right]\omega $ (Using (i))

$ = \left[ {{{M{l^2}} \over {12}} + m{v^2}{t^2}} \right]\omega $

As, $\left| {\overrightarrow \tau } \right| = {{dL} \over {dt}} = {d \over {dt}}\left[ {{{M{l^2}} \over {12}} + m{v^2}{t^2}} \right]\omega $

As $\omega$ and v remain constant

$\therefore$ $\left| {\overrightarrow \tau } \right| = 2m\omega {v^2}t$

$\left| {\overrightarrow \tau } \right| \propto t$

Hence, the graph and t is a straight line passing through the (0, 0). Option (b) represents correct plot.

The outer ring rolls without slipping, meaning the point on the ring in contact with the ground (denoted as point C) is stationary, i.e., $ \vec{v}_{\mathrm{C}} = \overrightarrow{0} $. Therefore, the velocity of point O is :

$ \vec{v}_{\mathrm{O}} = \vec{v}_{\mathrm{C}} + \vec{\omega}_{\mathrm{o}} \times \vec{r}_{\mathrm{CO}} = \overrightarrow{0} + \omega \hat{\jmath} \times 3R \hat{k} = 3R\omega \hat{\imath} $

Point P is located on the inner disc, which has an angular velocity $ \vec{\omega}_{\mathrm{i}} = -\omega / 2 \hat{\jmath} $. The position vector from O to P is given by :

$ \vec{r}_{\mathrm{OP}} = R \cos 30^{\circ} \hat{\imath} + R \sin 30^{\circ} \hat{k} = \frac{\sqrt{3} R}{2} \hat{\imath} + \frac{R}{2} \hat{k} $

Thus, the velocity of point P is calculated as follows :

$ \begin{aligned} \vec{v}_{\mathrm{P}} & = \vec{v}_{\mathrm{O}} + \vec{\omega}_{\mathrm{i}} \times \vec{r}_{\mathrm{OP}} \\ & = 3R\omega \hat{\imath} + \left(-\frac{\omega}{2} \hat{\jmath}\right) \times \left(\frac{\sqrt{3} R}{2} \hat{\imath} + \frac{R}{2} \hat{k}\right) \\ & = \frac{11\omega R}{4} \hat{\imath} + \frac{\sqrt{3} \omega R}{4} \hat{k} . \end{aligned} $