\(Given,\ { m }_{ 1 }=1kg,{ m }_{ 2 }=2kg\)
\( { x }_{ 1 }=1m,{ x }_{ 2 }=-1m\)
\( { y }_{ 1 }=2m,{ y }_{ 2 }=3m\)
\(\\ \therefore \quad { x }_{ CM }=\frac { { m }_{ 1 }{ x }_{ 1 }+{ m }_{ 2 }{ x }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { 1\times 1+2\times -1 }{ 1+2 } \)
\(=\frac { 1-2 }{ 3 } =\frac { -1 }{ 3 } =-0.33\)
\(\\ and\ { y }_{ CM }=\frac { { m }_{ 1 }{ y }_{ 1 }+{ m }_{ 2 }{ y }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { 1\times 2+2\times 3 }{ 1+2 } \)
\(\\=\frac { 2+6 }{ 3 } =\frac { 8 }{ 3 } =2.66\)
Thus, the coordinates of the centre of mass are (-0.33, 2.66).

Given, \(radius,r=\frac { diameter }{ 2 } =\frac { 4 }{ 2 } =2m\) 
\({ n }_{ 1 }=120rpm=\frac { 120 }{ 60 } rps=2rps\)
\(\\ { n }_{ 1 }=660rpm=\frac { 660 }{ 60 } rps=11rps\) 
 \(\therefore \ Angular\ acceleration,\alpha =\frac { { \omega }_{ 2 }-{ \omega }_{ 1 } }{ t } =\frac { 2\pi \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ t } \) 
\(=\frac { 2\pi \left( 11-2 \right) }{ 9 } =2\pi rad/{ s }^{ 2 }\)  
Linear acceleration, \(\alpha =r\alpha =2\times 2\pi =4\pi m/s^{ 2 }\).

Centre of mass and centre of gravity are two different concepts. But if g goes not vary from one part of body to other than CG and CM coincides.
So, when vertical height of the object is very small compared to a radius of earth, we call object small, otherwise, we call it extended. In above context, building and pond are small objects and a deep lake and a mountain are large extend objects.

Let the axis of rotation be at a distance x from mass m₁. Therefore, the distance of the axis of rotation from mass m₂ is (L-x)

When the rod is set into rotation, the increase in the rotational KE. is given by
K.E = \(\frac { 1 }{ 2 } I{ _{ 1 }\omega }_{ 0 }^{ 2 }=\frac { 1 }{ 2 } I{ _{ 1 }\omega }_{ 0 }^{ 2 }=\frac { 1 }{ 2 } I{ _{ 2 }\omega }_{ 0 }^{ 2 }\)
= \(\frac { 1 }{ 2 } { m }_{ 1 }{ x }^{ 2 }{ \omega }_{ 0 }^{ 2 }+\frac { 1 }{ 2 } { m }_{ 2 }=(L-x)^{ 2 }{ \omega }_{ 0 }^{ 2 }\) 
According to work-energy theorem
Work done W =increase in rotational K.E
i.e W = \(\frac { 1 }{ 2 } { m }_{ 1 }{ x }^{ 2 }{ \omega }_{ 0 }^{ 2 }+\frac { 1 }{ 2 } { m }_{ 2 }=(L-x)^{ 2 }{ \omega }_{ 0 }^{ 2 }\)
Work will be minimum if \(\frac { dW }{ dx } \)= 0
or \(\frac { d }{ dx } \left[ \frac { 1 }{ 2 } { m }_{ 1 }{ x }^{ 2 }{ \omega }_{ 0 }^{ 2 }+\frac { 1 }{ 2 } { m }_{ 2 }(L-{ x })^{ 2 }{ \omega }_{ 0 }^{ 2 } \right] \)=0
\({ m }_{ 1 }{ x }{ \omega }_{ 0 }^{ 2 }-{ m }_{ 2 }(L-x){ \omega }_{ 0 }^{ 2 }=0\)
\({ \omega }_{ 0 }^{ 2 }[{ m }_{ 1 }x-{ m }_{ 2 }(L-x)]=0\)
Since  \({ \omega }_{ 0 }^{ 2 }\neq 0\quad \) ∴  m₁x -m₂(L-x) = 0
or (m₁ + m₂)x = m₂L
 ∴      x = m₂L/(m₁+m₂)

Consider an equilateral triangle of side 1 m as shown in fig. Take X and Y axes as shown in fig.
By the definition of centre of mass, we have
\(\vec { x } \) =\(\frac { { m }_{ 1 }{ x }_{ 1 }{ +m }_{ 2 }{ x }_{ 2 }+{ m }_{ 3 }{ x }_{ 3 } }{ { m }_{ 1 }{ +m }_{ 2 }+{ m }_{ 3 } } \)
and  \(\vec { y } =\frac { { m }_{ 1 }y_{ 1 }{ +m }_{ 2 }y_{ 2 }+{ m }_{ 3 }y_{ 3 } }{ { m }_{ 1 }{ +m }_{ 2 }+{ m }_{ 3 } } \)
Here, m₁ = 1 kg, m₂= 2kg and m₃ = 3 kg
[x₁= 0,y₁= 0] [x₂  =1,y₂= 0] and [x₃= 0.5 y₃ = \(\frac { \sqrt { 3 } }{ 2 } \)]
Here y₃ = CD = AC sin 60^o = \(\frac { \sqrt { 3 } }{ 2 } \) 
\(\bar { x } \) = \(\frac { 1\times 0+2\times 1+3\times 0.5 }{ 1+2+3 } =\frac { 3.5 }{ 6 } m\)
and  \(\vec { y } =\frac { 1\times 0+2\times 0+3\times \left( \frac { \sqrt { 3 } }{ 2 } \right) }{ 1+2+3 } =\frac { \sqrt { 3 } }{ 4 } m\)
The co-ordinates of centre of mass are \(\left( \frac { 3.5 }{ 6 } ,\frac { \sqrt { 3 } }{ 4 } \right) \)

Consider a rigid body rotating about a given axis with a uniform angular acceleration a, under the action of a torque.
Let the body consist of particles of masses m₁, m₂, m₃ ... m_n at perpendicular distances r₁, r₂, r₃ ... r_n respectively from the axis of rotation,
As the body is rigid, angular acceleration a of all the particles of the body is the same. However, their linear accelerations are different because of different distances of the particles from the axis.
If a₁,a₂,a₃...a_n are the respective linear accelerations of the particles, then a₁=r₁∝,a₂=r₂∝ ,a₃∝=r₃∝
Force on particle of mass m₁ is
 f₁ =m₁a₁ = m₁r₁∝
Moment of this force about the axis of rotation
=  f₁r₁=(m₁r₁∝)r₁ =m₁r₁²∝
Similarly, moments of forces on other particles about the axis of rotation are m₁r₁²∝,
\({ m }_{ 3 }{ r }_{ 3 }^{ 2 }\alpha ,{ m }_{ n }{ r }_{ n }^{ 2 }\alpha \)
∴ Torque acting on the body, ፒ
=\({ m }_{ 1 }{ r }_{ 1 }^{ 2 }\alpha +{ m }_{ 2 }{ r }_{ 2 }^{ 2 }\alpha +{ m }_{ 3 }{ r }_{ 3 }^{ 2 }\alpha +...{ m }_{ n }{ r }_{ n }^{ 2 }\alpha \) 
= (\({ m }_{ 1 }{ r }_{ 1 }^{ 2 }+{ m }_{ 2 }{ r }_{ 2 }^{ 2 }+{ m }_{ 3 }{ r }_{ 3 }^{ 2 }+...{ m }_{ n }{ r }_{ n }^{ 2 }\))\(\alpha \)
ፒ = \(\left( \sum _{ i=1 }^{ i=n }{ { m }_{ i }{ r }_{ i }^{ 2 } } \right) \alpha \)
or  ፒ​= I or ∝ \(\vec { \tau } =I\vec { \alpha } \)

The radius of gyration of a body about the axis of rotation of a body is the point at which the weighed mass of the body acts. It is also equal to the square root of moment of inertia of all particles of the body about the axis of rotation divided by the total mass of the body i.e.,
\(k=\sqrt {\frac{m_1r_1^2+m_2r_2^2+m_3r_3^2....+m_nr_n^2}{m_1+m_2+m_3...+m_n}}\)
\(\sqrt {\frac{\sum mR}{M}}=\sqrt {\frac{I}{M}}\)
Its dimensions are those of length and its is measured in metre in SI units. The moment of a body is a quantity which comes in rotational motion and plays same role in rotational motion as does mas in translational motion. Thus a body continues of rotate or be at rest in the absence of any external torque. This is similar to the law of inertia in translational motion. This aspect is used in over creasing the dead points in the engines and crankshafts. Similarly, the kinetic energy of rotation is dependent on the moment of inertia of the body in the same manner the kinetic energy of translation of motion depends, on the mass of the body. For a given angular velocity (\(\omega\)) kinetic energy of rotation \(\propto\) I. If equal torques are applied I₁ and I₂ the their angular acceleration are inversely proportional to the moments of inertia of the bodies.
\(\frac{I_1\alpha_1}{I_2\alpha_2}=1\)
\(\frac{\alpha_1}{\alpha_2}=\frac{I_2}{I_1}\)
\(\alpha\propto\frac{1}{I}\)
Similarly if two bodies have same angular acceleration, then the moments of inertia are directly proportional to the torque applied on then.
\(\frac{\tau_1}{\tau_2}=\frac{I_1}{I_2}\ or \alpha_1=\alpha_2\)
The linear momentum of a body depends on its mass and velocity. If two bodies have same velocity, then their moments are proportional to their masses i.e.,
\(\frac{\rho_1}{\rho_2}=\frac{m_1}{m_2}\)
Similarly for angular momentum
\(\frac{L_1}{L_2}=\frac{I_1}{I_2}\)
Thus, moment of inertia of a body plays same role in the rotational motion as does mass in translational motion.
The moment of inertia determines the amount of torque to be applied to produce desired angular acceleration.

KE is not conserved in process.
K_final > K_initial
Muscular work done by the man in folding his arms is converted into KE.

Work done by torque in turing rotor by angle d \(\theta \) is
= \(\tau \)  d \(\theta \)
So, power delivered by engine
 P = \(\frac { Work\ done }{ Time\ taken } =\tau \frac { d\theta }{ dt } \)
[dt= time for turing by angle dθ]     
or   P = \(\tau \omega \)
So, power required = 180 x 200 = 36000 W
= 36 k W     [ 1 k W = 1000 W]

(i) We shall use ω = ω₀ + αt
ω₀ = initial angular speed in rad/s
= 2π × angular speed in rev/s
\(=\frac{2 \pi \times \text { angular speed in rev } / \mathrm{min}}{60 \mathrm{~s} / \mathrm{min}} \)
\(=\frac{2 \pi \times 1200}{60} \mathrm{rad} / \mathrm{s} \)
\(=40 \pi \mathrm{rad} / \mathrm{s}\)
Similarly \(\omega=\) final angular speed in rad / s
\( =\frac{2 \pi \times 3120}{60} \mathrm{rad} / \mathrm{s} \)
\(=2 \pi \times 52 \mathrm{rad} / \mathrm{s} \)
\(=104 \pi \mathrm{rad} / \mathrm{s}\)
Therefore Angular acceleration
\(\alpha=\frac{\omega-\omega_0}{t} =4 \pi \mathrm{rad} / \mathrm{s}^2\)
The angular acceleration of the engine \(=4 \pi \mathrm{rad} / \mathrm{s}^2\)
(ii) The angular displacement in time t is given by
\( \theta=\omega_0 t+\frac{1}{2} \alpha t^2\)
\(=  \left(40 \pi \times 16+\frac{1}{2} \times 4 \pi \times 16^2\right) \mathrm{rad}\)
\(=  (640 \pi+512 \pi) \mathrm{rad} \)
\(=  1152 \pi \mathrm{rad}\)
Number of revolutions \(=\frac{1152 \pi}{2 \pi}=576\)