$$\eqalign{ & K = \frac{{{L^2}}}{{2I}} \Rightarrow {L^2} = 2KI \Rightarrow L = \sqrt {2KI} \cr & \frac{{{L_1}}}{{{L_2}}} = \sqrt {\frac{{{K_1}}}{{{K_2}}} \cdot \frac{{{I_1}}}{{{I_2}}}} = \sqrt {\frac{K}{K} \cdot \frac{I}{{2I}}} = \frac{1}{{\sqrt 2 }} \cr & \Rightarrow {L_1}:{L_2} = 1:\sqrt 2 \cr} $$

$$M.I.$$  of a uniform circular disc of radius $$'R'$$ and mass $$'M'$$ about an axis passing through $$C.M.$$  and normal to the disc is
$${I_{C.M.}} = \frac{1}{2}M{R^2}$$
From parallel axis theorem
$${I_T} = {I_{C.M.}} + M{R^2} = \frac{1}{2}M{R^2} + M{R^2} = \frac{3}{2}M{R^2}$$

Work done by torque is responsible for change in kinetic energy.
$$\eqalign{ & \therefore \tau = \frac{{dE}}{{d\theta }} \cr & \therefore I\alpha = 2K\theta \cr & \therefore \alpha = \frac{{2k\theta }}{I} \cr} $$

The total kinetic energy of the ball rolling on a table without slipping is equal to its rotational kinetic energy and translational kinetic energy.
Total kinetic energy of spherical ball is given by
$$\eqalign{ & K = {\text{Kinetic energy rotational}}\left( {{K_{{\text{rot}}}}} \right) + {\text{Kinetic energy translational}}\left( {{K_{{\text{trans}}}}} \right) \cr & = \frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2} \cr} $$
For sphere, moment of inertia about its diameter $$I = \frac{2}{5}m{r^2}$$
$$\eqalign{ & \therefore K = \frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right){\omega ^2} + \frac{1}{2}m{v^2} \cr & = \frac{1}{5}m{r^2}{\omega ^2} + \frac{1}{2}m{v^2} \cr & = \frac{1}{5}m{v^2} + \frac{1}{2}m{v^2}\,\,\left( {{\text{as}}\,v = r\omega } \right) \cr & = \frac{7}{{10}}m{v^2} \cr & \therefore \frac{{{K_r}}}{K} = \frac{{\frac{1}{5}m{v^2}}}{{\frac{7}{{10}}m{v^2}}} = \frac{2}{7} \cr} $$

$$\eqalign{ & Tr = \frac{{m{r^2}}}{2}{\alpha _1}\,......\left( {\text{1}} \right) \cr & Tr = \frac{{m{r^2}}}{2}\alpha \,......\left( {\text{2}} \right) \cr & {\alpha _1} = \alpha \,......\left( {\text{3}} \right) \cr} $$
Acceleration of point $$b$$ = acceleration of point $$a$$
$$r{\alpha _1} = {a_{cm}} - r\alpha \,......\left( 4 \right)$$
Hence, $$2r\alpha = {a_{cm}}$$

Moment of inertia of a disc and circular ring about a tangential axis in their planes are respectively.
Momentum inertia of disc about tangential axis $${I_d} = \frac{5}{4}{M_d}{R^2}$$
Moment of inertia of ring about a tangential axis $${I_r} = \frac{3}{2}{M_r}{R^2}$$
$$\eqalign{ & {\text{but}}\,\,I = M{k^2} \Rightarrow k = \sqrt {\frac{I}{M}} \cr & \therefore \frac{{{k_d}}}{{{k_r}}} = \sqrt {\frac{{{I_d}}}{{{I_r}}} \times \frac{{{M_r}}}{{{M_d}}}} \cr & {\text{or}}\,\,\frac{{{k_d}}}{{{k_r}}} = \sqrt {\frac{{\left( {\frac{5}{4}} \right){M_d}{R^2}}}{{\left( {\frac{3}{2}} \right){M_r}{R^2}}} \times \frac{{{M_r}}}{{{M_d}}}} = \sqrt {\frac{5}{6}} \cr & \therefore {k_d}:{k_r} = \sqrt 5 :\sqrt 6 \cr} $$

Since $$v$$  is changing (decreasing), $$L$$  is not conserved in magnitude. Since it is given that a particle is confined to rotate in a circular path, it cannot have spiral path. Since the particle has two accelerations $${a_c}$$  and $${a_t}$$   therefore the net acceleration is not towards the centre.

The direction of $${\vec L}$$   remains same even when the speed decreases.

$$\eqalign{ & {a_1} = \frac{{v_1^2}}{{{R_1}}} = \frac{{{\omega ^2}R_1^2}}{{{R_1}}} = {\omega ^2}{R_1} \cr & {a_2} = \frac{{v_2^2}}{{{R_2}}} = {\omega ^2}{R_2} \cr} $$
Taking particle masses equal
$$\frac{{{F_1}}}{{{F_2}}} = \frac{{m{a_1}}}{{m{a_2}}} = \frac{{{a_1}}}{{{a_2}}} = \frac{{{R_1}}}{{{R_2}}}$$

Loss of energy is given by
$$\eqalign{ & \Delta E = \frac{1}{2}{I_t}\omega _i^2 - \frac{1}{2}\frac{{I_t^2\omega _i^2}}{{\left( {{I_t} + {I_b}} \right)}} \cr & = \frac{1}{2}\frac{{{I_b}{I_t}\,\omega _i^2}}{{\left( {{I_t} + {I_b}} \right)}} \cr} $$

As the spheres are smooth there will be no friction (no torque) and therefore there will be no transfer of angular momentum. Thus $$A,$$  after collision will remain with its initial angular momentum. $$i.e.,\,\,{\omega _A} = \omega $$