$$\eqalign{ & \tau \times \Delta t = {L_0}\,\,\left\{ {\because {\text{Since}}\,{L_f} = 0} \right\} \cr & {\text{or}}\,\,\tau \times 30 = 2\pi \cr & \tau = \frac{\pi }{{15}}N - m \cr} $$

Moment of inertia of rod about an axis through its centre of mass and perpendicular to rod = (mass of rod) $$ \times $$ (perpendicular distance between two axes)
$$ = \frac{{M{l^2}}}{{12}} + M{\left( {\frac{l}{2}} \right)^2} = \frac{{M{l^2}}}{3}$$
Moment of inertia of the system $$ = \frac{{M{l^2}}}{3} \times 4$$
$$ = \frac{4}{3}M{l^2}$$

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.

Imagine the cylinder to be moving on a frictionless surface. In both the cases the acceleration of the centre of mass of the cylinder is $${\text{g}}\,{\text{sin}}\,\theta .$$   This is also the acceleration of the point of contact of the cylinder with the inclined surface. Also no torque (about the centre of cylinder) is acting on the cylinder since we assumed the surface to be frictionless and the forces acting on the cylinder is $$mg$$  and $$N$$  which pass through the centre of cylinder. Therefore the net movement of the point of contact in both the cases is in the downward direction as shown. Therefore the frictional force will act in the upward direction in both the cases.

Note : In general we find the acceleration of the point of contact due to translational and rotational motion and then find the net acceleration of the point of contact. The frictional force acts in the opposite direction to that of net acceleration of point of contact.

Initially,
$$0 = \frac{{{m_1}\left( { - {x_1}} \right) + {m_2}{x_2}}}{{{m_1} + {m_2}}} \Rightarrow {m_1}{x_1} = {m_2}{x_2}\,\,.....(i)$$
Finally,
The centre of mass is at the origin

$$\eqalign{ & \therefore 0 = \frac{{{m_1}\left( {d - {x_1}} \right) + {m_2}\left( {{x_2} - d'} \right)}}{{{m_1} + {m_2}}} \cr & \Rightarrow 0 = {m_1}d - {m_1}{x_1} + {m_2}{x_2} - {m_2}d' \cr & \Rightarrow d' = \frac{{{m_1}}}{{{m_2}}}d\,\,\,\,\left[ {{\text{from (i)}}} \right] \cr} $$

Problem Solving Strategy
For this type of problem, always apply parallel and perpendicular axes theorem of moment of inertia. Moment of inertia of uniform circular disc about its diameter $$= I$$
According to theorem of perpendicular axes, Moment of inertia of disc about its axis $$= 2I$$
Applying theorem of parallel axes,
Moment of inertia of disc about the given axis $$ = 2I + m{r^2}$$
$$\eqalign{ & = 2I + 4I\,\,\left( {{\text{as}}\,\,2I = \frac{1}{2}m{r^2}\,\therefore m{r^2} = 4I} \right) \cr & = 6I \cr} $$

If the man pulls the cart by the rope, the man and cart will meet at the centre of mass.
$$\therefore {x_{CM}} = \frac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}$$
Taking axis at the point where man is present
\[ = \frac{{M \times 0 + M \times 10}}{{M + M}}\,\,\left[ {\begin{array}{*{20}{c}} {{x_1} = 0,{x_2} = 10}\\ {{m_1} = {m_2} = M} \end{array}} \right]\]
$$ = \frac{{10M}}{{2M}} = 5\,m$$

The moment of inertia of the system about axis of rotation $$O$$ is

$$\eqalign{ & I = {I_1} + {I_2} = 0.3{x^2} + 0.7{\left( {1.4 - x} \right)^2} \cr & = 0.3{x^2} + 0.7\left( {1.96 + {x^2} - 2.8x} \right) \cr & = {x^2} + 1.372 - 1.96x \cr} $$
The work done in rotating the rod is converted into its rotational kinetic energy.
$$\eqalign{ & \therefore W = \frac{1}{2}I{\omega ^2} \cr & = \frac{1}{2}\left[ {{x^2} + 1.372 - 1.96x} \right]{\omega ^2} \cr} $$
For work done to be minimum
$$\eqalign{ & \frac{{dW}}{{dx}} = 0\,\,\,\,\, \Rightarrow 2x - 1.96 = 0 \cr & \Rightarrow x = \frac{{1.96}}{2} = 0.98\,m \cr} $$

In a hollow sphere, the mass is distributed away from the axis of rotation. So, its moment of inertia is greater than that of a solid sphere.

When no external torque acts on system then, angular momentum of system remains constant.
Angular momentum before contact $$ = {I_1}{\omega _1} + {I_2}{\omega _2}$$
Angular momentum after the discs brought into contact $$ = {I_{{\text{net}}}}\omega = \left( {{I_1} + {I_2}} \right)\omega $$
So, final angular speed of system $$ = \omega $$
$$ = \frac{{{I_1}{\omega _1} + {I_2}{\omega _2}}}{{{I_1} + {I_2}}}$$
Now, to calculate loss of energy, we subtract initial and final energies of system.
⇒ Loss of energy
$$\eqalign{ & = \frac{1}{2}I\omega _1^2 + \frac{1}{2}I\omega _2^2 - \frac{1}{2}\left( {2I} \right){\omega ^2} \cr & = \frac{1}{4}I{\left( {{\omega _1} - {\omega _2}} \right)^2} \cr} $$