g = \(\frac{GM_e}{R^2_e}\)
M_e = 2 M_e      R_e= 2 R_e then,
g' = \(\frac{G \times 2M_e}{(2R_e)^2}\) = \(2 \frac{Gm_e}{4R_e^2}\)
\(2 \frac{Gm_e}{4R_e^2}\) = \(\cfrac g2\)

Let x be the distance of each scale of the \(\Delta\)ABC from the centre 6. Then total torque about O will be
\(F_{1} \times x+F_{2} \times x-F_{3} \times x=\mathrm{O} \text { or } F_{1}+F_{2}=F_{3} \)

Potential energy = Translational kinetic energy + Rotational kinetic energy
\(m g h=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2} \)
\(=\frac{1}{2} m v^{2}+\frac{1}{2} \times \frac{2}{5} M R^{2} \times \frac{v^{2}}{R^{2}}\left[\omega=\frac{v}{R}\right] \)
\(=\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2} \)
\(=\frac{5 m v^{2}+2 m v^{2}}{10}=\frac{7 m v^{2}}{10} \)
\(m g h=\frac{7 m v^{2}}{10} \)
\(g h=\frac{7 v^{2}}{10} \)
\(\therefore v^{2}=\frac{10 g h}{7} \)
\(\therefore v=\frac{\sqrt{10 g h}}{7} \)

Acceleration of the solid sphere while rolling down without slipping
\(a_{1}=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}}\)
Acceleration developed while slipping down \(a_{2}=g \sin \theta\)
\(\text { Required ratio } \frac{a_{1}}{a_{2}}=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}} / g \sin \theta\)
\(\frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{k^{2}}{r^{2}}}\)
\(\text { For a solid sphere } \frac{k^{2}}{r^{2}}=\frac{2}{5}\)
\(\therefore \text { Ratio of accelerations } \frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{2}{5}}\)
\(=\frac{1}{5+\frac{2}{5}}=\frac{1}{\frac{7}{5}}=\frac{5}{7}\)
\(\therefore a_{1}: a_{2}=5: 7 \)

The moments of inertia of two discs are I_a and I_b respectively The angular velocity of the disc A is \(\omega\).
The sum of kinetic energies of two discs before coming in contact is \(k_{1}=\frac{1}{2} I_{a} \omega_{1}^{2}+\frac{1}{2} I_{b} \omega_{2}^{2}\)
\(\text { But angular velocity of the disc be is } \omega_{2}=0 \ \text {(rest)}\)
\(\therefore k_{1}=\frac{1}{2} I_{a} \omega_{1}^{2}\)
The final kinetic energy of the two discs system \(k_{2}=\frac{1}{2} \frac{I_{a}^{2} \omega_{1}^{2}}{I_{a}+I_{b}}\)
The loss of kinetic energy is
\(k_{1}-k_{2} =\frac{1}{2} I_{a} \omega^{2}-\frac{1}{2}\left[\frac{I a^{2} \omega_{1}^{2}}{I_{a}+I_{2 b}}\right] \)
\(=\frac{1}{2} \frac{\left[I_{1}\left(I_{a}+I_{b}\right) \omega^{2}-I_{a}^{2} \omega^{2}\right]}{I_{a}+I_{b}} \)
\(k_{1}-k_{2} =\frac{1}{2} \frac{I_{a} b}{\left(I_{a}+I_{b}\right)} \omega^{2} \)

\(m=3 \mathrm{~kg} \quad r=40 \times 10^{-2} \mathrm{~m}=0.4 \mathrm{~m}\)
\(\text { Force }=30 N\)
Moment of inertia of a hollow-cylinder I= MR²
\(=3 \times\left(40 \times 10^{-2}\right)^{2} \)
\(=3 \times 0.4 \times 0.4=0.48 \mathrm{kgm}^{2} \)
\( F R =I \alpha \)
\(30 \times 40 \times 10^{-2}=0.48 d \)
\(d=\frac{12}{0.48}=\frac{1200}{48}=25 \mathrm{rad} \mathrm{s}^{-2} \)
\(\alpha=25 \mathrm{rad} \mathrm{s}^{-2} \)

The gravitational potential energy
\(\mathrm{V} =-\frac{G M^{2}}{R}\left[1+\frac{4}{\sqrt{2}}\right]  \)
\(\mathrm{V} =-\frac{G M^{2}}{R}[1+2 \sqrt{2}]  \)
\(\text {Gravitational potential } \mathrm{V}_{\mathrm{o}}(\mathrm{r}) =-\frac{4 G M}{R}\)

Centripetal acceleration \(a=\frac{V^{2}}{R}\)
Centripetal force \(=\frac{M V^{2}}{R}\)
\(\therefore\) Speed \(V=\frac{1}{2} \sqrt{\frac{G M}{R}(1+2 \sqrt{2})}\)

If the Earth has us tilt then there would not be seasons of the Earth.

The gravitational potential at a distance r due to a mass is defined as the amount of work required to bring unit mass from infinity to the distance r and it is denoted as V(r).
\(\mathrm{V}(\mathrm{r})=-\frac{\mathrm{Gm}}{r}\)

Radial Component.

No. It can produce only rotatory motion.

\(\sum { \bar { \tau } } =0\)

A rigid body is said to be in mechanical equilibrium when both its linear momentum and angular momentum remain constant.

\({x}_{CM}={m(2.5)+m(3.5)+2m(3)\over m+m+2m}=3\)
\({y}_{CM}={m(1.5)+m(1.5)+2m(3)\over m+m+2m}=2.25\)

The centripetal force cannot do work on the object since it is always perpendicular to the velocity.

(i) The torque of a force about an axis is independent of the choice of the origin as long as it is chosen on that axis itself.
(ii) Let O be the origin on the axis AB, which is the rotational axis of a rigid body. F is the force acting at the point P. Now, choose another point O' anywhere on the axis.
(iii) The torque of F about O' is,
\(\overset { \rightarrow }{ O'P } \times \overset { \rightarrow }{ F } =(\overset { \rightarrow }{ O'O } +\overset { \rightarrow }{ OP } )\times \overset { \rightarrow }{ F } \)
=\((\overset { \rightarrow }{ O'O } \times \overset { \rightarrow }{ F } )+(\overset { \rightarrow }{ OP } \times \overset { \rightarrow }{ F } )\)
(iv) As \(\overset { \rightarrow }{ O'O } \times \overset { \rightarrow }{ F } \)  is perpendicular to \(\overset { \rightarrow }{ O'O } \)  this term will not have a component along AB. Thus, the component of \(\overset { \rightarrow }{ O'P } \times \overset { \rightarrow }{ F } \)  is equal to that of \(\overset { \rightarrow }{ OP } \times \overset { \rightarrow }{ F } \).

At Earth's surface \(\mathrm{KE}=\frac{1}{2} \mathrm{M} v_{\mathrm{e}}^{2}, \mathrm{P} . \mathrm{E}=-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\)
At infinity K.E = \(\frac{1}{2} M V_{\infty}^{2}, \mathrm{P} . \mathrm{E}=\mathrm{O}\)
By law of conservation of energy,
\(\frac{1}{2} \mathrm{Mv}_{\mathrm{e}}^{2}-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}} =\frac{1}{2} \mathrm{M} v_{\infty}^{2}  \)
\(v_{\mathrm{e}}^{2} =v_{\infty}^{2}+\frac{2 \mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}  \)
\(\text {But } \mathrm{GM}_{\mathrm{E}} =\mathrm{qR_{ \textrm {E } } ^ { 2 }}  \)
\(\therefore v_{\mathrm{e}}^{2} =v_{\infty}^{2}+2 \mathrm{gR}_{\mathrm{E}}  \)
\(v_{\mathrm{c}} =\sqrt{\mathrm{v}_{\infty}^{2}+2 \mathrm{gR}}\)

Consider an object of mass m at a height h from the surface of the Earth. Acceleration experienced by the object due to Earth is
g' = \(\frac { GM }{ ({ R }_{ e }+h)^{ 2 } } \)       ..........(1)
g'= \(\frac { GM }{ { R }_{ e }^{ 2 }\left( 1+\frac { h }{ { R }_{ e } } \right) ^{ 2 } } \)
g'=\(\frac { GM }{ { R }_{ e }^{ 2 } } \left( 1+\frac { h }{ { R }_{ e } } \right) ^{ 2 }\)
If h << R_e
We can use Binomial expansion. Taking the terms up to first order
g'= \(\frac { GM }{ { R }_{ e }^{ 2 } } \left( 1-2\frac { h }{ { R }_{ e } } \right) \)
g' = \(g\left( 1-2\frac { h }{ { R }_{ e } } \right) \)  .....(2)
We find that g'< g. This means that as altitude h increases, the acceleration due to gravity g decreases.

Due to Earth's spinning motion, the stars in sky appear to move in circular motion about the pole star.

Consider 'n' particles of masses \(m_{1}, m_{2}, \ldots \ldots m_{n}\), distributed in space at positions \(\overrightarrow{r_{1}}, \overrightarrow{r_{2}}, \overrightarrow{r_{3}}, \ldots .\) etc, with respect to point P. The total gravitational field at a point P due to all the masses is given by the vector sum of the gravitational field due to the individual masses. This principle is known as superposition of gravitational fields.
\(\vec{E}_{\text {total }} =\vec{E}_{1}+\vec{E}_{2}+\ldots \vec{E}_{n}  \)
\(=-\frac{G m_{1}}{r_{1}^{2}} \hat{r}_{1}-\frac{G m_{2}}{r_{2}^{2}} \hat{r}_{2}-\ldots-\frac{G m_{n}}{r_{n}^{2}} \hat{r}_{n}=-\sum_{i=1}^{n} \frac{G m_{i}}{r_{i}^{2}} \hat{r}_{i}\)

Let the mass of the disc be ml and its radius R. The mass of the suspended body is m².
m₁ = 500 g = 500\(\times\)10^-3 kg =0.5 kg
m2 = 100 g = 100\(\times\)10^-3 kg = 0.1 kg
R= 10cm= 10\(\times\)10^-2m=0.1 m

As the light inextensible string is wound around the disc several times it makes the disc rotate without slipping over it. The translational acceleration of m₂ and tangential acceleration of m₁ will be the same. Let us draw the free body diagram (FBD) of m₁ and m₂ separately.

FBD of the disc: Its gravitational force (m₁g) acts downward and normal force N exerted by the fixed support at the center acts upward. The tension T acts downward at the edge. The gravitational force (m₁g) and the normal force (N) cancel each other. (m₁g) = N. The tension T produces a torque (RT), which produces a rotational motion in the disc with angular acceleration,
\((\alpha=\frac{a}{R})\). Here, a is the linear acceleration of a point at the edge of the disc. If the moment of inertia of the disc is I and its radius of gyration is K, then
\(RT=I\alpha; RT=(m_1 K^{2})\frac{a}{R}\)
\(T=(m_1K^{2})\frac{a}{R^{2}}\)
FBD of the body: Its gravitational force (m₂g) acts downward and the tension T acts upward. As (T < m₂g), there is a resultant force (m₂a) acting on it downward.
m₂g - T= m₂a

\(m_2g-(m_1 K^{2})\frac{a}{R^{2}}=m_1 a\)
\(m_2 g=(m_1 k^{2})\frac{a}{R^{2}}+m_2 a\)
\(m_2 g=(m_1 k^{2})\frac{a}{R^{2}}+m_2 a\)
\(m_2 g=[(m_1 \frac{K^{2}}{R^{2}})+m_2]a\)
\(a=\frac{m_2}{(m_1 \frac{k_2}{R_2})+m_2}g\)
The expression \((\frac{K^{2}}{R^{2}})\) for a disc rotating about an axis passing through the center and perpendicular to the plane is, \(\frac{K^{2}}{R^{2}}=\frac{1}{2}\). Now the expression for acceleration further simplifies as
\(a=\frac{m}{[(\frac{m_1}{2})+m_2]}g; a=\frac{2m_2}{[m_1+2m_2]}g\)
substituting the values  \(a=\frac{2\times 0.1}{[0.5+0.2]}\times 10=\frac{0.2}{0.7}\times 10\)
a = 2.857 ms^-2.

(i) Torque about point D is zero. (as F passes through D).
Torque about point C is zero. (as F passes through C).
Torque about point O is \((\overrightarrow {OC})\times \overrightarrow{F}\) and direction is along GH.
Torque about point B is \((\overrightarrow {BD})\times \overrightarrow{F}\) and direction is along GH.
(The perpendicular distance of \(\overrightarrow {BD}\) with respect to \(\overrightarrow {F}\) is \(\overrightarrow {OC}\).
(ii) Torque about axis CD is zero (as F is parallel to CD).
Torque about axis OC is zero (as F intersects OC).
Torque about axis AB is zero (as F is parallel to AB).
Torque about axis GH is  \((\overrightarrow {OC})\times \overrightarrow{F}\) and direction is along GH.

For translational equilibrium of car
N_F + N_B = W = \(1800 \times 9.8=17640 N\)
For rotational equilibrium of car
1.05 N_F = 0.75 N_B
1.05 N_F = 0.75 (17640-N_F)
1.8 N_F = 13230
N_F = 13230/1.8 = 7350 N
N_B = 17640 - 7350 = 10290 N
Force on each front wheel = \(\frac{7350}{2}\) = 3675 N
Force on each back wheel = \(\frac {10290}{2}=5145 N\)

During collapse, the total angular momentum of an isolated star is conserved, hence
l₁ω₁ = I₂ω₂
or\(\frac{2}{5}\) MR₁²ω_{1 = \(\frac{2}{5}\)}MR₂²ω₂  [∵ I=\(\frac{2}{5}\)MR²]
or R₁²ω₁ = R₂²ω₂  ∴ ω₂ = \(\frac { R_{ 1 }^{ 2 } }{ R^{ 2 }_{ 2 } } \) ω₁  
But R₁ = 10^-6km, R₂ = 10⁴ , ω_{1 =} 10^-6s^-1
∴ ω₂ = \(\frac { \left( { 10 }^{ 6 } \right) ^{ 2 } }{ \left( { 10 }^{ 4 } \right) ^{ 2 } } \times \)10^-6 =0.01 rad s^-1

Here R = 0.05m, l = 1.5 m,
\(\rho=8\times10^3\) kgm^-3
Mass of cylinder M = Volume \(\times\) density
\(=\pi R^2l.\rho\)
= 3.14 \(\times\) (0.05)² \(\times\) 1.5 \(\times\) 8 \(\times\) 10³
= 94.2 kg
M.I. of the cylinder about its own axis,  \(I={1\over2}MR^2\)
\(={1\over2}\times94.2\times(0.05)^2=0.1175\) kgm²

Given: diameter = 0.5m, Mass = 16 kg
Formula: Moment of inertia of solid disc
I = \(\frac { 1 }{ 2 } \)MR² = \(\frac { 1 }{ 2 } \times 16\times \left[ \frac { { 0.5 } }{ 2 } \right] ^{ 2 }\)= \(\frac { 1 }{ 2 } \)
Angular velocity \(\omega \) = \(\frac { 120\times 2\pi }{ 60 } \) = \(4\pi \)
And change in angular momentum \(\Delta L=I\omega -0=I\omega \)
Angular impulse \(\tau \times t=\Delta L\Rightarrow \tau =\frac { \Delta L }{ t } =\frac { 1 }{ 8 } \times \frac { 1 }{ 2 } \times 4\pi =\frac { \pi }{ 4 } Nm\quad \)

It is the combination of pure translational and pure rotational motion.
1. The rolling motion is the most commonly observed motion in daily life. The motion of wheel is an example of rolling motion.
2. Round objects like ring, disc, sphere etc. are most suitable for rolling. Let us study the rolling of a disc on a horizontal surface. Consider a point P on the edge of the disc.
3. While rolling, the point undergoes translational motion along with its center of mass and rotational motion with respect to its center of mass.

(i) For direction, the vector rule or right hand rule is used
ፒ = r Fsin\(\theta\)
(ii) The expression for the magnitude of torque can be written in two different ways by associating sinፀ either with r or F in the following manner
\(\tau =r(F\ sin\theta )=r\times (F\bot )\)
\(\tau =(r\ sin\theta )F=(r\bot )\times F\)
(iii) Here, (F sinፀ) is the component of \(\overset { \rightarrow }{ F } \) perpendicular \(\overset { \rightarrow }{ r } \). Similarly, (r sinፀ) is the component of \(\overset { \rightarrow }{ r } \) perpendicular to \(\overset { \rightarrow }{ F } \).

During noon time of summer solstice the Suns rays cast no shadow in the city Syne which was located 500 miles away from Alexandria. At the same day and same time in Alexandria the Suns rays made 7.2 degree with local vertical as shown in the figure. This difference of 7.2 degree was due to the curvature of the Earth.

The angle 7.2 degree is equivalent to radian
So  rad
If S is the length of the arc between the cities. of Syne and Alexandria; and if R is radius of Earth, then
S  = R\(\theta \) = 500 miles,
So radius of the Earth
R =\(\frac { 500 }{ \theta } \)miles
R =500 \(\frac { miles }{ \frac { 1 }{ 8 } } \)
R = 4000 miles
1 mile is equal to 1.609 km. So, he measured the radius of the Earth to be equal to R = 6436 km, which is amazingly close to the correct value of 6378 km.

Newton's inverse square Law:
Newton considered the orbits of the planets as circular. For circular orbit of radius r,' the centripetal acceleration towards the center is
\(a=\frac { { V }^{ 2 } }{ r } \) ...(1)

Here v is the velocity and r, the distance of the planet from the center of the orbit.
The velocity in terms of known quantities r and T, is
v = \(\frac { 2\pi r }{ T } \)    ...(2)
Here T is the time period of revolution of the planet. Substituting this value of v in equation we get,
a = \(\frac { \left( \frac { 2\pi }{ T } \right) ^{ 2 } }{ r } =-\frac { 4\pi ^{ 2 }t }{ { T }^{ 2 } } \)   ...(3)
Substituting the value of 'a' from (3) in Newton's second law, F = ma, where 'm' is the mass of the planet
F = \(\frac { 4\pi mr }{ { T }^{ 2 } } \)   ...(4)
From Kepler's third law
 \(\frac { r^{ 3 } }{ { T }^{ 2 } } \) = k(constant)   ...(5)
\(\frac { r }{ { T }^{ 2 } } =\frac { k }{ { r }^{ 2 } } \)   ....(6)
By substituting equation (6) in the force expression, we can arrive at the law of gravitation
\(F=\frac { 4\pi ^{ 2 }mk }{ { r }^{ 2 } } \) ...(7)
Here negative sign implies that the force is attractive arid it acts towards the center. In equation (7), mass of the planet 'm'. comes explicitly. But Newton strongly felt that according to his third law, if Earth is attracted by the Sun, then the Sun must also be attracted by the Earth with the same magnitude of force. So he felt that the Sun's mass (M) should also occur explicitly in the expression for force. From this insight, he equated the constant 4π²k to GM which turned out to be the law of gravitation
F = \(-\frac { GMm }{ { r }^{ 2 } } \)
Again the negative sign in the above equation implies that the gravitational force is attractive.

Let R be the radius of the ring and M be the total mass of the complete ring.
Let m be the mass of the section removed from the ring then, mass of the incomplete ring is M-m
Let us introduce a positive integer (n), such that, \(n\theta=360^{0} \), or \(n=\frac{360^{0}}{\theta}\)

mass of incomplete ring=M - m
\(m=\frac{M}{360}\times \theta\)
∴ Mass of complete ring = \(M-\frac{M}{360}\times \theta\)
Mass of incomplete ring = \(M-\frac{M}{n}=M(\frac{n-1}{n})\)
For example, (a) when \(\theta=60^{0}; n=\frac{360^{0}}{60^{0}}=6\)
∴ n-1 =5
Mass of incomplete ring = \(\frac{5}{6}M\)
(b)when \(\theta=30^{0}; n=\frac{360^{0}}{30^{0}}=12\)
n-1=11
Mass of incomplete ring = \(\frac{11}{12}M\)
The moment of inertia of the incomplete ring is, I=\(M \frac{(n-1)}{n}R^{2}.\)

A point mass is a hypothetical point particle which has non-zero mass and no size or shape. To find the center of mass for a collection ofn point masses, say, m₁, m₂, m₃ ... m_n we have to first choose an origin and an appropriate co-ordinate system as shown in Figure. Let, x₁, x₂, x₃ ... x_n be the X-coordinates of the positions of these point masses in the X direction from the origin.

The equation for the X coordinate of the center of mass is,
\({ x }_{ cm }=\frac { \sum { { m }_{ i }{ x }_{ i } } }{ \sum { { m }_{ i } } } \)
where, \(\sum { { m }_{ i } } \) is the total mass M of all the particles, (\(\sum { { m }_{ i } } =M\)). Hence, 
\({ x }_{ CM }=\frac { \sum { { m }_{ i }{ x }_{ i } } }{ M } \)
Similarly, we can also find y and z coordinates of the center of mass for these distributed point masses as indicated in Figure.
\({ y }_{ CM }=\frac { \sum { { m }_{ i }{ y }_{ i } } }{ M } \)
\({ z }_{ CM }=\frac { \sum { { m }_{ i }{ z }_{ i } } }{ M } \)
Hence, the position of center of mass of these point masses in a Cartesian coordinate system is (x_CM, y_CM z_CM) In general, the position of center of mass can be written in a vector form as,
\(\overset { \rightarrow }{ r_{ CM } } =\frac { \sum { { m }_{ i } } \overset { \rightarrow }{ r } i }{ M } \)
where,  \(\overset { \rightarrow }{ r_{ CM } } ={ x }_{ CM }\hat { i } +{ y }_{ CM }\hat { j } +{ z }_{ CM }\hat { k } \) is the position vector of the center of mass and \(\overset { \rightarrow }{ { r }_{ i } } ={ x }_{ i }\hat { i } +{ y }_{ i }\hat { j } +{ z }_{ i }\hat { k } \) is the position vector of the distributed point mass; where \(\hat { i } ,\hat { j } \) and \(\hat { k } \) are the unit vectors along X, Y and Z-axes respectively.

We can also determine the centre of gravity of a uniform lamina of even an irregular shape by pivoting it at various points by trial and error. The lamina remains horizontal when pivoted at the point where the net gravitational force acts, which is the centre of gravity as shown in the figure. When a body is supported at the centre of gravity, the sum of the torques acting on all the point masses of the rigid body becomes zero. Moreover, the weight is compensated by the normal reaction force exerted by the pivot. The body is in static equilibrium and hence it remains horizontal.

Mass of the disc m = 100 g = 6.1 kg.
Diameter of the disc d = 10 cm.
Radius of the disc r = 5 cm = 0.05m
Rolling with a velocity v = 20 cms^-1 = 0.20 ms^-1
Total energy of the disc E_Tot = ?
E_Tot = Translational K energy + rotational K.E
M.I of the disc about its own axis,
\(I=\frac { 1 }{ 2 } { mr }^{ 2 }\)
\(v=r\omega \ \therefore { \omega }^{ 2 }=\frac { { v }^{ 2 } }{ { r }^{ 2 } } \)
Rotational K.E = \(I=\frac { 1 }{ 2 } I{ \omega }^{ 2 }=\frac { 1 }{ 2 } \times \left( \frac { 1 }{ 2 } { mr }^{ 2 } \right) \times \left( \frac { { v }^{ 2 } }{ { r }^{ 2 } } \right) \)
\(=\frac { 1 }{ 4 } { mv }^{ 2 }\)
T.E. \(=\frac { 1 }{ 2 } { mv }^{ 2 }+\frac { 1 }{ 4 } { mv }^{ 2 }=\frac { 3 }{ 4 } { mv }^{ 2 }\)
T.E. of the disc E_Tot = \(\frac{3}{4}\) \(\times\) 0.1 \(\times\) 0.20 \(\times\) 0.20
= 0.003 J

(i) The total kinetic energy (KE) can be written as the sum of kinetic energy due to translational motion (KE_TRANS) and kinetic energy due to rotational motion (KE_ROT) 
KE = KE_TRANS + KE_ROT
(ii) If the mass of the rolling object is M, the velocity of center of mass is v_CM its moment of inertia about center of mass is ICM and angular velocity is ω, then
KE = \(\frac { 1 }{ 2 } Mv^{ 2 }_{ CM }+\frac { 1 }{ 2 } { I }_{ CM }\omega ^{ 2 }\)
(iii) With center of mass as reference:
The moment of inertia (lCM) of a rolling object  about the center of mass is,
I_CM= MK² and vCM= Rω. Here, K is radius of gyration.
\(KE=\frac { 1 }{ 2 } Mv^{ 2 }_{ CM }+\frac { 1 }{ 2 } \left( MK^{ 2 } \right) \frac { v^{ 2 }_{ CM } }{ R^{ 2 } } \\ \\ \)
\(KE=\frac { 1 }{ 2 } Mv^{ 2 }_{ CM }+\frac { 1 }{ 2 } Mv^{ 2 }_{ CM }\left( \frac { { K }^{ 2 } }{ R^{ 2 } } \right) \)
\(KE=\frac { 1 }{ 2 } Mv^{ 2 }_{ CM }\left( 1+\frac { K^{ 2 } }{ R^{ 2 } } \right) \)
(iv) With point of contact as reference: We can also arrive at the same expression by taking the momentary rotation happening with respect to the point of contact (another approach to rolling). if we take the point of contact as 0, then,
\(KE=\frac { 1 }{ 2 } { I }_{ 0 }\omega ^{ 2 }\)
Here, I_o is the moment of inertia of the object about the point of contact. By parallel axis theorem, I_o = I_CM+ MR². Further we can write, I_o = MK² + MR². With v_CM= Rω or 
\(\omega =\frac { { v }_{ CM } }{ R } \)
\(KE=\frac { 1 }{ 2 } \left( MK^{ 2 }+MR^{ 2 } \right) \frac { v^{ 2 }_{ CM } }{ { R }^{ 2 } } \)
\(KE=\frac { 1 }{ 2 } Mv^{ 2 }_{ CM }\left( 1+\frac { { K }^{ 2 } }{ R^{ 2 } } \right) \)
(vi) K.E. in pure rolling can be determined by anyone of the following two cases.
(a) The combination of translational motion and rotational motion about the center of mass. (or)
(b) The momentary rotational motion about the point of contact.

(i) Consider a rigid body rotating about a fixed axis. A point mass m in the body will execute a circular motion about a fixed axis.
(ii) A tangential force \(\vec F\) acting on the point mass produces the necessary torque for this rotation. This force \(\vec F\) is perpendicular to the position vector \(\vec r\) of the point mass.

iii) The torque produced by the force on the point mass m about the axis can be written as,
\(\tau =rFsin90=rF\quad \left[ \because sin90=1 \right] \)
\(\tau =rma\ \left[ \because \left( F=ma \right) \right] \)
\(\tau =rmr\alpha =mr^{ 2 }\alpha \ \left[ \because \left( a=r\alpha \right) \right] \)
\(\tau =\left( mr^{ 2 } \right) \alpha \)
(iv) Hence, the torque of the force acting on the point mass produces an angular acceleration (a) in the point mass about the axis of rotation.
In vector notation,
\(​​\vec \tau =\left( mr^{ 2 } \right) \vec \alpha\)
(v) The directions of \(​​\tau \)  and \(\alpha\) are along the axis of rotation. If the direction of \(\tau\) is in the direction of \(\alpha\), it produces angular acceleration. On the other hand if \(​​\tau \) is opposite to \(\alpha\) angular deceleration or retardation is produced on the point mass.
(vi) The term mr² in equations is called moment of inertia (I) of the point mass. A rigid body is made up of many such point masses. Hence, the moment of inertia of a rigid body is the sum of moments of inertia of all such individual point masses that constitute the body \(\left( I=\sum { { m }_{ i }{ r }_{ i }^{ 2 } } \right) \) Hence, torque for the rigid body can be written as,
\(\vec \tau =\left( \sum { m_{ i }{ r }_{ i }^{ 2 } } \right) \vec a\)
\(\vec \tau = I \vec a\)