Rotational Motion

The angular momentum is

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

For the particle moving from $D \to A$, we have

$\overrightarrow L = m\left[ {\left( {{R \over {\sqrt 2 }}\widehat i + {R \over {\sqrt 2 }}\widehat j} \right) \times v( - \widehat j)} \right]$

$ = {{mvR} \over {\sqrt 2 }}( - \widehat k)$

For the particle moving from $A \to B$, we have

$\overrightarrow L = m\left[ {\left( {{R \over {\sqrt 2 }}\widehat i + {R \over {\sqrt 2 }}\widehat j} \right) \times v\widehat i} \right]$

$ = {{mvR} \over {\sqrt 2 }}( - \widehat k)$

For the particle moving from $C \to D$, we have

$\overrightarrow L = mv \times $ ($\bot$ distance) $\times$ $(\widehat k)$

$ = mv\left( {{R \over {\sqrt 2 }} + a} \right)(\widehat k)$

For the particle moving from $B \to C$, we get

$\overrightarrow L = mv \times $ ($\bot$ distance) $\times$ $(\widehat k)$

$ = mv\left( {{R \over {\sqrt 2 }} + a} \right)(\widehat k)$

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

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

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

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.

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.

When the disc is at a distance x from the mean position (equilibrium position), the force acting on the disc are give below.

$\therefore -2kx+f=-M_{ac}$ ..... (i)

Where $a_c$ = acceleration of centre of mass.

The torque acting on the disc about its centre of mass C is

$\tau = f \times R = I \times {a_c}$

$\therefore$ $f = {{I\alpha } \over R} = {{{1 \over 2}M{R^2}} \over R} \times {{{a_c}} \over R}$

[$\therefore$ $I = {1 \over 2}M{R^2},{a_c} = R{a_c}$ for rolling without slipping]

$\therefore$ $f = {1 \over 2}M{a_c}$ ...... (ii)

From eq. (i) and (ii),

$ - 2kx + {1 \over 2}M{a_c} = - M{a_c}$

$ \Rightarrow {3 \over 2}M{a_c} = 2kx$

$ \Rightarrow M{a_c} = {{4kx} \over 3}$

$\Rightarrow$ Net external force acting on the disc when its centre of mass is at displacement x with respect to the equilibrium position $ = {{ - 4kx} \over 3}$ directed towards the equilibrium.

When the disc is at a distance x from the mean position (equilibrium position), the force acting on the disc are give below.

$\therefore -2kx+f=-M_{ac}$ ..... (i)

Where $a_c$ = acceleration of centre of mass.

The torque acting on the disc about its centre of mass C is

$\tau = f \times R = I \times {a_c}$

$\therefore$ $f = {{I\alpha } \over R} = {{{1 \over 2}M{R^2}} \over R} \times {{{a_c}} \over R}$

[$\therefore$ $I = {1 \over 2}M{R^2},{a_c} = R{a_c}$ for rolling without slipping]

$\therefore$ $f = {1 \over 2}M{a_c}$ ...... (ii)

From eq. (i) and (ii),

$ - 2kx + {1 \over 2}M{a_c} = - M{a_c}$

$ \Rightarrow {3 \over 2}M{a_c} = 2kx$

$ \Rightarrow M{a_c} = {{4kx} \over 3}$

$\Rightarrow$ Net external force acting on the disc when its centre of mass is at displacement x with respect to the equilibrium position $ = {{ - 4kx} \over 3}$ directed towards the equilibrium.

$ \therefore $ $|{F_{net}}| = {{4k} \over 3}x$

For S.H.M $|{F_{net}}| = M{\omega ^2}x$

$\therefore$ $M{\omega ^2} = {{4k} \over 3}$

$ \Rightarrow \omega = \sqrt {{{4k} \over {3M}}} $ .... (iii)

When the disc is at a distance x from the mean position (equilibrium position), the force acting on the disc are give below.

$\therefore -2kx+f=-M_{ac}$ ..... (i)

Where $a_c$ = acceleration of centre of mass.

The torque acting on the disc about its centre of mass C is

$\tau = f \times R = I \times {a_c}$

$\therefore$ $f = {{I\alpha } \over R} = {{{1 \over 2}M{R^2}} \over R} \times {{{a_c}} \over R}$

[$\therefore$ $I = {1 \over 2}M{R^2},{a_c} = R{a_c}$ for rolling without slipping]

$\therefore$ $f = {1 \over 2}M{a_c}$ ...... (ii)

From eq. (i) and (ii),

$ \Rightarrow - 2kx + f = - 2f$

$ \Rightarrow f = {{2k} \over 3} \times x$

Frictional force depends on x. As x increase friction increase. Frictional force (f) is maximum at $x = A$.

Where A = amplitude of S.H.M

Maximum frictional force,

${f_{\max }} = {{2k} \over 3} \times A$

The force should be almost equal to the limiting friction ($\mu mg$) for rolling without slipping.

$\therefore$ $\mu mg = {{2k} \over 3} \times A$ ..... (iv)

For S.H.M

Velocity amplitude $ = A\omega $

$\therefore$ ${V_0} = A\omega $

$\therefore$ ${V_0} = {{3m\,mg} \over {2k}}w$ (from (iv))

$\therefore$ ${V_0} = {{3\mu mg} \over {2k}} \times \sqrt {{{4k} \over {3M}}} $ (As $ \omega = \sqrt {{{4k} \over {3M}}} $)

$\therefore$ ${V_0} = mg\sqrt {{{3M} \over k}} $

The acceleration of a body rolling down an inclined plane in given by

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

For hollow cylinder,

${I \over {M{R^2}}} = {{M{R^2}} \over {M{R^2}}} = 1$

For solid cylinder,

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

$\therefore$ Acceleration of the solid cylinder is more than that of hollow cylinder and therefore solid cylinder will reach the bottom of the inclined plane first.

$\therefore$ Statement - 1 is False.

Statement - 2 : In case of rolling, there will be no heat losses. Therefore, total mechanical energy remains conserved. The potential energy, therefore gets converted into kinetic energy. In both the cases, since the initial potential energy is same, the final kinetic energy will also be same. Therefore, statement - 2 is correct.

By applying conservation of Energy, we have,

$\frac{1}{2} m v^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=m g h_{\max }$ ..... (i)

For pure rolling,

$v=R\omega$

$\Rightarrow \quad \omega=v / \mathrm{R}$ ..... (ii)

Putting eq. (ii) in eq. (i), we get

$\Rightarrow \frac{1}{2} m v^{2}+\frac{1}{2} \mathrm{I}\left(\frac{v^{2}}{R^{2}}\right)=m g \times \frac{3 v^{2}}{4 g}$ $\left(\because h_{\max } \times \frac{3 v^{2}}{4 g}\right)$

$\Rightarrow \quad \mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}$

Hence, this is a required formula of moment of Inertia of Disc. Therefore, the object a disc.

$2^{\text {nd }}$ Method : For pure rolling, the velocity of point of contact is zero i.e., $v=0$, So, $\Delta x$ (displacement) is zero. Then work done by the frictional force should be

$W=f_{0} \Delta x=0$

If work is done by dissipative force, work done should be zero. It means that work done of system is zero. So energy should be conserved. Initially, the object has both translation as well as rotational motion. If it attains '$h_{\max }$' kinetic energy changes into potential energy.

From energy conservation principle,

$\mathrm{K} . \mathrm{E}_{\text {translational }}+\mathrm{K} . \mathrm{E}_{\text {rotational }}=$ Potential Energy

$\Rightarrow \quad \frac{1}{2} m v^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=m g \times \frac{3 v^{2}}{4 g}$

$\Rightarrow \quad m v^{2}+\mathrm{I} \omega_{2}=\frac{3 v^{2}}{2} m$

$\Rightarrow \quad \mathrm{I} \omega^{2}=\frac{1}{2} m v^{2}$

$\Rightarrow \quad \mathrm{I}=\frac{1}{2} \frac{m v^{2}}{\omega^{2}}$ ..... (i)

For No slipping,

$\Rightarrow$ $\begin{aligned} & v=\omega R \\ & \omega=\frac{v}{R}\end{aligned}$ ..... (ii)

Put eq (ii) in eq (i)

Hence, $\quad \mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}$

$\mathrm{I}=\frac{1}{2} \mathrm{M} \frac{v^{2}}{v^{2}} \times \mathrm{R}^{2}$

this is moment of inertia of disc.

So, object is a Disc.

Statement 1 : For velocity of centre of mass remains constant, net force acting on body must be zero. Therefore the statement 1 is false.

Statement 2 : The linear momentum of an isolated system remains constant. The statement is true.

For disc A,

${1 \over 2}kx_1^2 = {1 \over 2}I{(2\omega )^2}$

$ \Rightarrow kx_1^2 = 2I{\omega ^2}$ ..... (i)

For disc B,

${1 \over 2}kx_2^2 = {1 \over 2}2I{\omega ^2}$

$ \Rightarrow kx_2^2 = I{\omega ^2}$ ..... (ii)

On dividing eq. (i) by (ii)

${{kx_1^2} \over {kx_2^2}} = {{2I{\omega ^2}} \over {I{\omega ^2}}}$

${{{x_1}} \over {{x_2}}} = \sqrt 2 $

When disc B is brought in contact with disc A. Let w' be the final angular velocity of both the disc rotating together.

Applying conservation of angular momentum for the two disc system.

$ \Rightarrow I(2\omega ) + 2I(\omega ) = (I + 2I)\omega '$

$ \Rightarrow \omega ' = {4 \over 3}\omega $

Torque on disc A,

${T_A} = {{\Delta {L_A}} \over t} = {{{L_f} - {L_i}} \over t} = {{I \times {4 \over 3}\omega - I \times 2\omega } \over t}$

$ = {{ - 2I\omega } \over {3t}}$

Note : The negative sign represents that the torque creates angular retardation.

Loss in kinetic energy

= (K.E.)$_{initial}$ $-$ (K.E.)$_{final}$

$ = \left[ {{1 \over 2}I{{(2\omega )}^2} + {1 \over 2}(2I){\omega ^2}} \right] - \left[ {{1 \over 2}(I + 2I){{\left( {{4 \over 3}\omega } \right)}^2}} \right]$

$ = [2I{\omega ^2} + I{\omega ^2}] - \left[ {{3 \over 2}I \times {4 \over 3} \times {4 \over 3}{\omega ^2}} \right]$

$ = 3I{\omega ^2} - {8 \over 3}I{\omega ^2} = {{I{\omega ^2}} \over 3}$

To solve this problem, we need to compare the moments of inertia of the sphere and the disc. For a solid sphere of radius $R$, the moment of inertia about its geometrical axis is given by:

$I_{\text{sphere}} = \frac{2}{5} M R^2$

where $M$ is the mass of the sphere.

When the sphere is melted into a disc of radius $r$ and thickness $t$, the volume (and hence mass) remains the same. So, the volume of the sphere is:

$\frac{4}{3} \pi R^3$

And the volume of the disc is:

$\pi r^2 t$

Equating the volumes, we get:

$\frac{4}{3} \pi R^3 = \pi r^2 t$

This simplifies to:

$t = \frac{4 R^3}{3 r^2}$

Next, we need to find the moment of inertia of the disc about the tangential axis (which is perpendicular to the plane of the disc). The moment of inertia of a disc of mass $M$ and radius $r$ about an axis through its center and perpendicular to the plane of the disc is:

$I_{\text{disc, center}} = \frac{1}{2} M r^2$

Using the parallel axis theorem, the moment of inertia about a tangential axis is:

$I_{\text{disc, tangential}} = I_{\text{disc, center}} + M r^2 = \frac{1}{2} M r^2 + M r^2 = \frac{3}{2} M r^2$

Since we are given that the moment of inertia of the sphere and the disc about their respective axes are equal, we set:

$\frac{2}{5} M R^2 = \frac{3}{2} M r^2$

Canceling the mass $M$ from both sides and solving for $r$, we get:

$\frac{2}{5} R^2 = \frac{3}{2} r^2$

Solving for $r$, we have:

$r^2 = \frac{4}{15} R^2$

Taking the square root of both sides:

$r = \frac{2}{\sqrt{15}} R$

So, the correct answer is:

Option A: $\frac{2}{\sqrt{15}} R$

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 . $

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} $