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

\(\overrightarrow{r}=7\hat{i}+4\hat{j}-2\hat{k}\)
\(\overrightarrow{F}=4\hat{i}-3\hat{j}+5\hat{k}\)
Torque,\(\overrightarrow{\tau}=\overrightarrow{r}\times \overrightarrow{F}\)
\(\overrightarrow{\tau}=\left| \begin{matrix} \hat { i } & \hat {j } & \hat {k } \\ 7 &4 & -2 \\ 4 & -3 & 5 \end{matrix} \right| \)
\(\overrightarrow{\tau}=\hat{i}(20-6)-\hat{j}(35+8)+\hat{k}(-21-16)\)
\(\overrightarrow {\tau}=(14\hat{i}-43\hat{j}-37\hat{k})Nm\)

The K.E. of A without friction E_A = \(\frac{1}{2}\) m(2gh).
The K.E. of B undergoes pure rolling E_B
\(= \frac{1}{2}m \left( \frac { 2gh }{ 1+\frac { { K }^{ 2 } }{ { R }^{ 2 } } } \right) \)
The K.E. of C rolling with slipping E_c
=\(\frac{1}{2}\) m(2gh).

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

When the round object moves, it always tends to roll on any surface which has a coefficient of friction any value greater than zero (μ > 0). The friction that enabling the rolling motion is called rolling friction.

(i) The -torque about the axis will rotate the object about it and the torque perpendicular to the axis will turn the axis of rotation.
(ii) When both exist simultaneously on a rigid body, the body will have a precession.

A rigid body is the one which maintains its definite and fixed shape even when an external force acts on it.

Internal forces are the forces acting among the particles within a system that constitute the body. External forces are the forces acting on the particles of a system from outside.

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.

According to parallel axis theorem,
I = I₀ + mx²
=\(\frac { M{ l }^{ 2 } }{ 12 } +M\left( \frac { l }{ 2 } \right) ^{ 2 }\)
=\(\frac { { Ml }^{ 2 } }{ 12 } +\frac { Ml^{ 2 } }{ 4 } ={ Ml }^{ 2 }\left[ \frac { 1 }{ 12 } +\frac { 1 }{ 4 } \right] \)
=\({ Ml }^{ 2 }\left[ \frac { 1 }{ 3 } \right] \)
I=\(\frac { { Ml }^{ 2 } }{ 3 } \) .

(i) If the radius of the rolling object is R, in one full rotation, the center of mass is displaced by 2\(\pi \)R (its circumference).
(ii) Not only the center of mass, but all the points on the disc are displaced by the same 2\(\pi \)R after one full rotation. The only difference is that the center of mass takes a straight path but, all the other points.
(iii) Undergo a path which has a combination of the translational and. rotational motion. Especially the point on the edge undergoes a path of a cycloid.
(iv) As the center of mass takes only a straight line path, its velocity v_CM is only translational velocity v_TRANS  (v_CM = v_TRANS). All the other points have two velocities. One is the translational velocity v_TRANS (which is also the velocity of center of mass) and the other is the rotational velocity v_ROT (V_ROT = rω).
(v) Here, r is the distance of the point from the center of mass and 0 is the angular velocity. The rotational velocity vROT is perpendicular to the instantaneous position vector from the center of mass.
(vi) The resultant of these two velocities is v. This resultant velocity v is perpendicular to the position vector from the point of contact of the rolling object with the surface on which it is rolling as shown in Figure.

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,
\(I=\int { dI } \)
\(I=\int _{ 0 }^{ R }{ \frac { 2M }{ { R }^{ 2 } } } { r }^{ 3 }dr=\frac { 2M }{ R^{ 2 } } \int _{ 0 }^{ R }{ r^{ 3 }dr } \)
\(I=\frac { 2M }{ R^{ 2 } } \left[ \frac { { r }^{ 4 } }{ 4 } \right] ^{ R }_{ 0 }=\frac { 2M }{ R^{ 2 } } \left[ \frac { { R }^{ 4 } }{ 4 } -0 \right] \)
\(I=\frac { 1 }{ 2 } MR^{ 2 }\)

If we suspend the lamina from different points like P, Q, R as shown in Figure, the vertical lines PP', QQ', RR' all pass through the centre of gravity Here, reaction force acting at the point of suspension and the gravitational force acting at the centre of gravity cancel each other and the torques caused by them also cancel each other.