$$\eqalign{ & \frac{{{\text{Rotational }}KE}}{{{\text{Total}}\,KE}} = \frac{{\frac{1}{2}m{v^2}\left( {\frac{{{K^2}}}{{{R^2}}}} \right)}}{{\frac{1}{2}m{v^2}\left( {1 + \frac{{{K^2}}}{{{R^2}}}} \right)}} \cr & = \frac{{{K^2}}}{{{K^2} + {R^2}}} \cr} $$

Let us consider that solid sphere, disc and solid cylinder are rolling on an inclined plane. $$M, I$$  and $$R$$ be mass, moment of inertia and radius of the rolling section in each case.
(i) Solid sphere The moment of inertia of a solid sphere about its diameter is given by $$I = \frac{2}{5}M{R^2}$$
$${\text{or}}\,\,K = \frac{I}{{M{R^2}}} = \frac{2}{5}$$
As from the concept, acceleration
$$\eqalign{ & a = \frac{{g\sin \theta }}{{1 + K}} \cr & {\text{So,}}\,\,a = \frac{{g\sin \theta }}{{1 + \frac{2}{5}}} = \frac{5}{7}g\sin \theta \cr} $$
(ii) Disc The moment of inertia of disc about an axis perpendicular to the plane of disc and passing through its centre is given by $$I = \frac{1}{2}M{R^2}$$
$${\text{or}}\,\,\frac{I}{{M{R^2}}} = \frac{1}{2}$$
$$\therefore a = \frac{{g\sin \theta }}{{1 + \frac{1}{2}}} = \frac{2}{3}g\sin \theta $$
(iii) Solid cylinder The moment of inertia of a cylinder about the axis passing through its centre and perpendicular to its plane is given by $$I = \frac{1}{2}M{R^2}$$
$$\eqalign{ & {\text{or}}\,\,\frac{I}{{M{R^2}}} = \frac{1}{2} \cr & \therefore a = \frac{{g\sin \theta }}{{1 + \frac{1}{2}}} = \frac{2}{3}g\sin \theta \cr} $$
So, acceleration of solid sphere is more. It implies that solid sphere reaches the bottom first.

Does not shift as no external force acts. The centre of mass of the system continues its original path. It is only the internal forces which comes into play while breaking.

Let the angular velocity of disc after child jumps off. be $${\omega '}$$
$$\therefore $$ From conservation of angular momentum $$\left( {I + m{R^2}} \right)\omega = mvR + I\omega $$
$$\therefore \omega ' = \frac{{\left( {I + m{R^2}} \right)\omega - mvR}}{I}$$

From the definition of angular momentum,
\[L = r \times p = rmv\sin \phi \left( { - k} \right)\,\,\left[ {\begin{array}{*{20}{c}} {r = {\rm{position vector}}}\\ {p = {\rm{momentum}}} \end{array}} \right]\]

Therefore, the magnitude of $$L$$ is $$L = mvr\sin \phi = mvd$$
where, $$d = r\sin \phi $$   is the distance of closest approach of the particle to the origin. As $$d$$ is same for both the particles, hence $${L_A} = {L_B}.$$

Since the objects are placed gently, therefore no external torque is acting on the system. Hence angular momentum is constant.

$$\eqalign{ & i.e.,\,{I_1}{\omega _1} = {I_2}{\omega _2} \cr & M{r^2} \times {\omega _1} = \left( {M{r^2} + 2m{r^2}} \right){\omega _2}\,\,\,\,\,\left( {\because {\omega _1} = \omega } \right) \cr & \therefore {\omega _2} = \frac{{M\omega }}{{M + 2m}} \cr} $$

Linear distance moved by wheel in half revolution $$ = \pi r.$$  Point $${P_1}$$ after half revolution reaches at $${P_2}$$ vertically $$2m$$  above the ground.
$$\therefore $$ Displacement $${P_1}{P_2}$$
$$ = \sqrt {{\pi ^2}{r^2} + {2^\pi }} = \sqrt {{\pi ^2} + 4} \,\,\left[ {\because r = 1m} \right]$$

Centre of mass

$$\eqalign{ & {x_{cm}} = \frac{x}{2}\frac{{\left( {\rho x} \right)\left( {\frac{x}{2}} \right)\frac{1}{2} + \rho y\left( {\frac{y}{2}} \right)}}{{\rho \left( {x + y} \right)}} \Rightarrow \frac{1}{2} + \frac{y}{x} = \frac{{{y^2}}}{{{x^2}}} \cr & \therefore \frac{{BC}}{{AB}} = \frac{y}{x} \cr & = \frac{{1 + \sqrt 3 }}{2} \cr & = 1.37 \cr} $$

Given that $${{I_0}}$$ is the moment of inertia of table and gun and $$m$$ the mass of bullet.
Initial angular momentum of system about centre:
$${L_i} + \left( {{I_0} + m{r^2}} \right){\omega _0}\,......\left( {\text{i}} \right)$$
Let $$w$$ be the nagular velocity of table after the bullet is fired.
Final angular momentum
$$Lf = {I_0}\omega - m\left( {v - r\omega } \right)r\,......\left( {{\text{ii}}} \right)$$
Where $$\left( {v - r\omega } \right)$$   is absolute velocity of bullet to the right.
Equations (i) and (ii); we get
$$\left( {\omega - {\omega _0}} \right) = \frac{{mvr}}{{{I_0} + m{r^2}}}$$

$$\eqalign{ & {X_{cm}} = \frac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + {m_3}}} \cr & {X_{cm}} = \frac{{300 \times \left( 0 \right) + 500\left( {40} \right) + 400 \times 70}}{{300 + 500 + 400}} \cr & {X_{cm}} = \frac{{500 \times 40 + 400 \times 70}}{{1200}} \cr & = 40\,cm \cr} $$