\(v_{0}=r \omega ; \quad \therefore v_{0} \alpha r\)
For a wheel (uniform ring) the distance of a point on the rim in level with the center 
\(\text { [i.e., radius] is } \sqrt{2} r\)
\(\therefore \text { The speed of the center is } \sqrt{2} v_{0}\)

Moment of inertia of a disc
\(\mathrm{I}_{1}=\frac{M R^{2}}{2}\)
\(\text { Mass of small disc }=\frac{M}{\pi R^{2}} \times \pi \times\left(\frac{R}{2}\right)^{2}\)
\(=\frac{M}{\pi R^{2}} \times \frac{\pi R^{2}}{4}=\frac{M}{4}\)
By the theorem of parallel axis, the moment of inertia of the small disc. About an axis passing through 0 is
\(I_{2} =\frac{1}{2} \times \frac{M}{4}\left(\frac{R}{2}\right)^{2}+\frac{M}{4}\left(\frac{R}{2}\right)^{2} \)
\(=\frac{M}{8} \times \frac{R^{2}}{4}+\frac{M}{4} \times \frac{R^{2}}{4} \)
\(=\frac{M R^{2}}{32}+\frac{M R^{2}}{16}=\frac{M R^{2}+2 M R^{2}}{32} \)
\(I_{2} =\frac{3 M R^{2}}{32} \)
Moment of inertia of the remaining part is I= I₁ - I₂
\(=\frac{M R^{2}}{2}-\frac{3 M R^{2}}{32} \)
\(=\frac{16 M R^{2}-3 M R^{2}}{32}=\frac{13 M R^{2}}{32}\)
\(I =\frac{13 M R^{2}}{32} \)

\(K \cdot E=\frac{1}{2} I \omega^{2} ; \quad L=I \omega ; \quad K \cdot E=\frac{2^{2}}{2^{2}} \)
\(\therefore K \cdot E \alpha L^{2} \quad E_{1} \alpha L_{1}^{2} \quad E_{2} \alpha L_{2}^{2}\)
\(\frac{E_{1}}{E_{2}}=\left(\frac{L_{1}}{L_{2}}\right)^{2} \)
\(\text { Here } E_{1}=E \quad E_{2}=\frac{E}{2} \)
\(L_{1}=L \quad \quad L_{2}=? \)
\(\frac{E}{\frac{E}{2}}=\left(\frac{L}{L_{2}}\right)^{2} \)
\(\frac{2 E}{E}=\left(\frac{L}{L_{2}}\right)^{2}\left(\frac{L_{1}}{L_{2}}\right)^{2}=2 \)
\(\therefore \frac{L}{L_{2}}=\sqrt{2} \)
\(L_{2}=\frac{L}{\sqrt{2}} \)

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

Mass of the sphere, m = 20 kg
Radius r = 0.25 m
Angular velocity 0 = 5 rad s^-1
Angular momentum \(L=I\omega =\frac{2}{5} mr^{2}\omega\)
\(=\frac{2}{5}\times 20 \times (0.25)^{2}\times 5 = 40 \times (0.0625)=2.5\)
L = 2.5 kg m² s^-1.

Lifting a table from the floor by two persons. Pushing the car by two persons.

The weight of tree exerts a torque about the point where the cut is made. This causes rotation of the tree about the cut has to be made at say point A to weaken the tree trunk and to shift the centre of mass to the right and eventually move towards the ground on the right.

The centre of mass of a body is defined as a point where the entire mass of the body appears to be concentrated.

(i) A rigid body which has non-zero. external torque (ፒ) about the axis of rotation would have an angular acceleration (α) about that axis.
(ii) The scalar relation between the torque and angular acceleration is,
ፒ=Iα
where, I is the moment of inertia of the rigid body. The torque in rotational motion is equivalent to the force in linear motion.

The direction of torque helps us to find the type of rotation caused by the torque. For example, if the direction of torque is out of the paper, then the rotation produced by the torque is anticlockwise. On the other hand, if the direction of the torque is into the paper, then the rotation is clockwise.

(i) Consider a rigid body rotating about a fixed axis. A point the body rotating about an axis perpendicular to the plane of the page. A tangential force F is applied on the body.
(ii) It produces a small displacement ds on the body. The work done (dw) by the force is,
dw=Fds
(iii) As the distance ds, the angle of rotation d\(\theta\) and radius r are related by the expression
ds=r d\(\theta\)
The expression for work done now becomes,
dw=F ds; dw=F r d\(\theta\)
(iv) The term (Fr) is the torque ፒ produced by the force on the body.
dw=\(\tau\)d\(\theta\)
This expression gives the work done by the external torque ፒ, which acts on the body rotating about a fixed axis through an angle d\(\theta\).

(i) A rotating rigid body with respect to any axis, is considered to be made up of point masses m₁, m₂, m₃, ... m_n at perpendicular distances (or positions) r₁, r₂, r₃....r_n respectively.
(ii) The moment of inertia of that object can be written as,
I=\(\sum { { m }_{ i }{ r }_{ i }^{ 2 } } ={ m }_{ 1 }{ r }_{ 1 }^{ 2 }+{ m }_{ 2 }{ r }_{ 2 }^{ 2 }+{ m }_{ 3 }{ r }_{ 3 }^{ 2 }+....+{ m }_{ n }{ r }_{ n }^{ 2 }\)
If all the n number of individual masses to be equal,
m=m₁=m₂=m₃=....=m_n
then,
I=\({ mr }_{ 1 }^{ 2 }+{ mr }_{ 2 }^{ 2 }+{ mr }_{ 3 }^{ 2 }+...+{ mr }_{ n }^{ 2 }\)
=\(m({ r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 3 }^{ 2 }+...+{ r }_{ n }^{ 2 })\)
=\(nm\left( \frac { { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 3 }^{ 2 }+...+r_{ n }^{ 2 } }{ n } \right) \)
I=MK²
where, nm is the total mass M of the body and K is the radius of gyration.
K=\(\sqrt { \frac { { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 3 }^{ 2 }+...+{ r }_{ n }^{ 2 } }{ n } } \)
(iii) The expression for radius of gyration indicates that it is the root mean square (rms) distance of the particles of the body from the axis of rotation.

1. Greater the mass concentrated away from the axis, greater the moment of inertia.
2. Moment of inertia of a body about an axis of rotation resists a change in it. In rotational motion, moment of inertia increases with increase in torque to change it's rotation.

(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.

Consider a disc of mass M and radius R. This disc is made up of many infinitesimally small rings as shown in Figure. Consider one such ring of mass (dm) and thickness (dr) and radius (r). The moment of inertia (dI) of this small ring is,
dI = (dm)r²

Moment of inertia of a uniform disc
As the mass is uniformly distributed, the mass per unit area (\(\alpha\)) is, \(\alpha=\frac{mass}{area}=\frac{M}{\pi R^2}\)
(iii) The mass of the infinitesimally small ring is,
dm = \(\alpha\)2\(\pi\)rdr = \(\frac { M }{ \pi R^{ 2 } } \) 2\(\pi\)rdr
where, the term (2\(\pi\)rdr) is the area of this elemental ring (2\(\pi\)r is the length and dr is the thickness) dm = \(\frac{2M}{R^2}rdr\)
dI = \(\frac{2M}{R^2}r^3dr\)
The moment of inertia (I) of the entire disc is,
\(I=\int { dI } \)
\(I=\int _{ 0 }^{ R }{ \frac { 2M }{ { R }^{ 2 } } } { r }^{ 3 }dr=\frac { 2M }{ R^{ 2 } } \int _{ 0 }^{ R }{ r^{ 3 }dr } \)
\(I=\frac { 2M }{ R^{ 2 } } \left[ \frac { { r }^{ 4 } }{ 4 } \right] ^{ R }_{ 0 }=\frac { 2M }{ R^{ 2 } } \left[ \frac { { R }^{ 4 } }{ 4 } -0 \right] \)
\(I=\frac { 1 }{ 2 } MR^{ 2 }\)

Let us consider a uniform rod of mass (M) and length (1) as shown in Figure. Let us find an expression for moment of inertia of this rod about an axis that passes through the center of mass and perpendicular to the rod. First an origin is to be fixed for the coordinate system so that it coincides with the center of mass, which is also the geometric center of the rod. The rod is now along the x axis. We take an infinitesimally small mass (dm) at a distance (x) from the origin. The moment of inertia (dI) of this mass (dm) about the axis is,

dI = (dm) x²
As the mass is uniformly distributed, the mass per unit length (λ) of the rod is, \(\lambda =\frac { M }{ l } \)
The (dm) mass of the infinitesimally small length as, dm = λ dx = \(\frac { M }{ l } dx\)
The moment of inertia (I) of the entire rod can be found by integrating dI,
\(I=\int { dI } =\int { \left( dm \right) { x }^{ 2 } } =\int { \left( \frac { M }{ l } dx \right) { x }^{ 2 } } \)
\(I=\frac { M }{ l } \int { { x }^{ 2 }dx } \)
As the mass is distributed on either side of the origin, the limits for integration are taken from -1/2 to 1/2.
\(I=\frac { M }{ l } \int _{ -t/2 }^{ t/2 }{ { x }^{ 2 }dx=\frac { M }{ l } } \left[ \frac { { x }^{ 3 } }{ 3 } \right] ^{ t/2 }_{ -t/2 }\)
\(I=\frac { M }{ l } \left[ \frac { { l }^{ 3 } }{ 24 } -\left( -\frac { { l }^{ 3 } }{ 24 } \right) \right] =\frac { M }{ l } \left[ \frac { { l }^{ 3 } }{ 24 } +\frac { { l }^{ 3 } }{ 24 } \right] \)
\(I=\frac { M }{ l } \left[ 2\left( \frac { { l }^{ 3 } }{ 24 } \right) \right] \)
I = \(\frac { 1 }{ 12 } \) ml²