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

Let the particle of mass m cross a horizontal distance x in time t.

Angular momentum \(\overrightarrow{L}=\int{\overrightarrow{\tau}}{dt}\)
But \(\overrightarrow{\tau}=\overrightarrow{r}\times \overrightarrow{F}\)
\(\overrightarrow{r}=x\hat { i } +y\hat { j} \) and \(\hat { F }=-mg \hat { j } \)
\(\therefore \overrightarrow { \tau }=(x\hat { i }+y\hat { j } ) \times (-mg \hat { j } )\)
\(\overrightarrow { \tau } =-mgx(\hat { i } \times \hat { j } )=-mgx\hat { k } \)
\(\overrightarrow { L } = -mg \int{(xdt)}\hat {k} =-gv\ cos\theta (\int t dt)\hat { k } \)
Let initial time t = 0 and final time t = t_f
\(\overrightarrow {L}=-mg \cos \theta \left( ^{ t }\int _{ 0 }^{ f }{ tdt } \right) \hat k = -\frac{1}{2}mgv\ cos \theta\ t^{2}_{f} \hat{k}\)
Negative sign indicates, \(\overrightarrow {L}\) point inwards. 

Mass of the electron, m = 9 \(\times\)10^-31 kg
Radius of the electron, r = 0.53 \(\overset { 0}{ A } \) = 0.53 \(\times\)10^-10 m
Velocity of the electron, v = 2.2 \(\times\)10⁶ ms^-1
Angular momentum of electron is, L = I\(\omega\)
Electron is considered as a point mass. Hence, its moment of inertia is, I= mr²
The relation, \(\omega=\frac{v}{r}\)could be used.
Angular momentum, \(L=mr^{2}\times \frac{v}{r}=mvr\)
=\(9.1 \times 10^{-31}\times 2.2 \times 10^{6}\times 0.53\times 10^{-10}\)
\(L=1.06\times 10^{-34}\)kg m² s^-1. 

Relation between angular momentum and rotational K.E.
Angular momentum, L = IW.
Rotational K.E., EK = \(\frac{1}{2}\)
\(2I={ (I\omega ) }^{ 2 }\)
\({ x }_{ k }=\frac { { L }^{ 2 } }{ 2I } \)
\(\sqrt { { E }_{ k } } =\frac { L }{ \sqrt { 2I } } \)
\(\frac { \sqrt { { E }_{ k } } }{ L } =\frac { I }{ \sqrt { 2I } } =constant\)
If I = 1, graph will be hyperbola.

The conditions for pure rolling we
1. The total kinetic energy is the sum of kinetic energies of translational and rotational motions.
2. There is no relative motion of the point of contact with the surface.
3. When the rolling object speeds up or slower down, it must accelerate or decelerate respectively. If this suddenly happens the rolling object to ship or slide.

The conditions are
(i) When position vector \(\vec{r}\) and force \(\vec{F}\) are parallel or antiparallel.
(ii) If the force acts at the reference point.

Sum of the clockwise moments is equal to sum of the anticlockwise moments when a body is in rotational equilibrium or algebraic sum of moments at any point is zero.

Since the blocks are identical, the centre of mass of each block is located at its midpoint. Let us take the origin to be at the midpoint [or centre of mass ] of block at the bottom and find out the shift in the position of centre of mass of the stack on the addition of the block of the block each time.
The stack will not fall over till the shift in the position of its centre of mass is less than 1/2 cm. When the stack contains two blocks: if Δx₁ and Δx₂ are the shifts in the positions of the centre, of the mass of the blocks with respect to origin, then shift in the position of the centre of mass of the stack
\(\Delta X=\frac { M\Delta x_{ 1 }+M\Delta x_{ 2 } }{ M+M } \)
\(=\frac { M\left[ \Delta x_{ 1 }+\Delta x_{ 2 } \right] }{ 2M } \)
\(\Delta X=\frac { \Delta x_{ 1 }+\Delta x_{ 2 } }{ M+M } \)
Here Δx₁= 0 and Δx₂= d cm
Therefore, ΔX = 0+\(\frac{d}{2}\) = \(\frac{d}{2}\) cm
Here, If \(\frac{d}{2}\) cm < \(\frac{1}{2}\) cm
Then the stack will not fall over.
In the above way, Suppose that let the maximum n block can be stacked before the stack falls over. If Δx₁, Δx₂, Δx₃ .......,Δx_n are the shifts in the position of the n blocks then shift in the position of the centre of mass of the stack
ΔX = Δx₁ + Δx₂ + Δx₃ + .... + Δx_n
So that the stack does not fall
 Δx₁ + Δx₂ + Δx₃ + .... + Δ\(\frac {x^n}{n}\) < \(\frac{1}{2}\)
S_n = Δx₁ + Δx₂ + Δx₃ + ..... + Δx_n
Then the above condition becomes,
\(\frac { S_{ n } }{ n }<\frac{1}{2}\)

(i) Let us assume a round object of mass m and radius R is rolling down an inclined, plane without slipping as shown in Figure. There are two forces acting on the object along the inclined plane.
(ii) One is the component of gravitational force (mg sinθ) and the other is the static frictional force (f). The other component of gravitation force (mg cosθ) is cancelled by the normal force (N) exerted by the plane. As the motion is happening along the incline, we shall write the equation for motion from the free body diagram (FBP) of the object.

(iii) For translational motion, mg sinθ is the supporting force and f is the opposing force,
mg sinθ - f = ma
For rotational motion, let us take the torque with respect to the center of the object. Then mg sinθ cannot cause torque as it passes through it but the frictional force f can set torque of Rf.
Rf = Iα
(iv) By using the relation, a = rα, and moment of inertia I = mK², we get,
Rf = mK² \(\frac{a}{R}\); f = ma\(\left( \frac { { K }^{ 2 } }{ R^{ 2 } } \right) \)
K.E \(=\frac{1}{2}MV^2_{CM}\left( 1+\frac { { K }^{ 2 } }{ R^{ 2 } } \right) \) becomes,
mg sinθ - ma \(\left( \frac { K^{ 2 } }{ R^{ 2 } } \right) \) = ma
mg sinθ = ma + ma \(\left( \frac { K^{ 2 } }{ R^{ 2 } } \right) \)
\(a\left( 1+\frac { { K }^{ 2 } }{ { R }^{ 2 } } \right) =gsin\theta \)
After rewriting it for acceleration, we get,
\(a=\frac { gsin\theta }{ \left( 1+\frac { { K }^{ 2 } }{ R^{ 2 } } \right) } \)