There is no external force so centre of mass of the system will not shift.

A rod lying along any of coordinate axes serves for us as continuous body.
Suppose a rod of mass $$M$$ and length $$L$$ is lying along the $$x$$-axis with its one end at $$x = 0$$  and the other at $$x = L.$$
Mass per unit length of the rod $$ = \frac{M}{L}$$
Hence, the mass of the element $$PQ$$  of length $$dx$$  situated at $$x = x$$  is $$dm = \frac{M}{L}dx$$

The coordinates of the element $$PQ$$  are $$\left( {x,0,0} \right).$$  Therefore, $$x$$-coordinate of centre of gravity of the rod will be
$${x_{CG}} = \frac{{\int_0^L x dm}}{{\int {dm} }} = \frac{{\int_0^L {\left( x \right)\left( {\frac{M}{L}} \right)} dx}}{M} = \frac{1}{L}\int_0^L x dx = \frac{L}{2}$$
but as given, $$L = 3\,m$$
$$\therefore {x_{CG}} = \frac{3}{2} = 1.5\,m$$
The $$y$$-coordinate of centre of gravity
$${y_{CG}} = \frac{{\int {y\,dm} }}{{\int {dm} }} = 0\,\,\left( {{\text{as}}\,y = 0} \right)$$
Similarly, $${z_{CG}} = 0$$
i.e., the coordinates of centre of gravity of the rod are $$\left( {1.5,0,0} \right)$$   or it lies at the distance $$1.5\,m$$  from one end.

Moment of inertia of a thin rod of length $$l$$ about an axis passing through centre and perpendicular to the rod $$ = \frac{1}{{12}}M{l^2}.$$
Thus moment of inertia of the frame.
$$\frac{{m{l^2}}}{{12}} + \frac{{m{l^2}}}{4} = \frac{{4m{l^2}}}{{12}} = \frac{{m{l^2}}}{3}$$
Total $$M.I. = 4 \times \frac{{m{l^2}}}{3}$$

Given, $$m = 50\,kg,\,r = 0.5\,m,\,\alpha = 2\,rev/{s^2}$$
⇒ Torque produced by the tension in the string
$$ = T \times r = T \times 0.5 = \frac{T}{2}N - m\,......\left( {\text{i}} \right)$$
We know $$\tau = I\alpha \,......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), $$\frac{T}{2} = I\alpha $$
$$\eqalign{ & = \left( {\frac{{M{R^2}}}{2}} \right) \times \left( {2 \times 2\pi } \right)rad/{s^2}\,\,\left[ {{\text{because }}{I_{{\text{Solid}}\,{\text{cylinder}}}} = \frac{{M{R^2}}}{2}} \right] \cr & \frac{T}{2} = \frac{{50 \times {{\left( {0.5} \right)}^2}}}{2} \times 4\pi \cr & T = 50 \times \frac{1}{4} \times 4\pi = 50\,\pi = 157\,N \cr} $$

When the thread is pulled, the bobbin rolls to the right.

Resultant velocity of point $$B$$ along the thread is $$v = {v_0}\sin \alpha - \omega r,$$   Where $${v_0}\sin \alpha $$  is the component of translational velocity along the thread and $$\omega r$$ linear velocity due to rotation. As the bobbin rolls without slipping, $${v_0} = \omega R.$$   Solving the obtained equations, we get $${v_0} = \frac{{vR}}{{R\sin \alpha - r}}$$

From conservation of angular momentum about any fix point on the surface,

$$\eqalign{ & m{r^2}{\omega _0} = 2m{r^2}\omega \cr & \Rightarrow \omega = \frac{{{\omega _0}}}{2} \cr & \Rightarrow v = \frac{{{\omega _0}r}}{2}\,\,\,\,\,\,\left[ {\because v = r\omega } \right] \cr} $$

Let $$r$$ be the perpendicular distance of $${F_1},{F_2}$$  and $${F_3}$$ from $$O$$ as shown in figure.

The torque of force $${F_3}$$ about $$O$$ is clockwise, while torque due to $${F_1}$$ and $${F_2}$$ are anticlockwise.
For total torque to be zero about $$O,$$ we must have $${F_1}r + {F_2}r - {F_3}r = 0$$
$$ \Rightarrow {F_3} = {F_1} + {F_2}$$

Using parallel axes theorem, moment of inertia about $$‘O'$$
$$\eqalign{ & {I_o} = {I_{cm}} + m{d^2} \cr & = \frac{{7M{R^2}}}{2} + 6\left( {M \times {{\left( {2R} \right)}^2}} \right) = \frac{{55M{R^2}}}{2} \cr} $$

Again, moment of inertia about point $$P,\,{I_p} = {I_o} + m{d^2}$$
$$ = \frac{{55M{R^2}}}{2} + 7M{\left( {3R} \right)^2} = \frac{{181}}{2}M{R^2}$$

Apply parallel axes theorem of moment of inertia.
According to question by applying conservation of angular momentum $${I_1}{\omega _1} = {I_2}{\omega _2}$$
In the given case $${I_1} = M{R^2}$$
$$\eqalign{ & {I_2} = M{R^2} + 2m{R^2} \cr & {\omega _1} = \omega \cr} $$
Then, $${\omega _2} = \frac{{{I_1}}}{{{I_2}}}\omega = \frac{M}{{M + 2m}}\omega $$

Kinetic energy of rotation of a body is the energy possessed by the body on account of its rotation about a given axis. If $$I$$ is the moment of inertia of the body about the given axis of rotation, $$\omega $$ is angular velocity of the body, then kinetic energy of rotation $${K_{{\text{rot}}}} = \frac{1}{2}I{\omega ^2}$$
Moment of inertia of the ring about an axis perpendicular to the plane of the ring and passing through its centre is $$I = m{r^2}$$
So, $${K_{{\text{rot}}}} = \frac{1}{2}m{r^2}{\omega ^2}$$