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,
\(\frac { S_{ n } }{ n }<\frac{1}{2}\)

Perpendicular axis theorem:
(i) The theorem states that the moment of inertia of a plane laminar body about an axis perpendicular to its plane is equal to the sum of moments of inertia about two perpendicular axes lying in the plane of the body such that all the three axes are mutually perpendicular and have a common point.
(ii) Let the X and Y-axes lie in the plane and Z-axis perpendicular to the plane of the laminar object. If the moments of inertia of the body about X and Y-axes are Ix and Iy respectively and It is the moment of inertia about Z-axis, then the perpendicular axis theorem could be expressed as,
 I_z = I_x + I_y
 (iii) To prove this theorem, let us consider a plane laminar object of negligible thickness on which lies the origin (0). The X and Y-axes lie on the plane and Z-axis is perpendicular to it as shown in Figure. The lamina is considered to be made up of a large number of particles of mass m. Let us choose one such particle at a point P which has coordinates (x, y) at a distance r from O.

(iv) The moment of inertia of the particle about Z axis is mr². The summation of the above expression gives the moment of inertia of the entire lamina about Z-axis as, I_z = \(\Sigma \)mr²
Here r² = x² + y²
Then, I_z = \(\Sigma \)(x² +y²)
I_z = \(\Sigma \)mx² + \(\Sigma \)my²
(v) In the above expression, the term \(\Sigma \)mx² is the moment of inertia of the body about the Y-axis and similarly the term \(\Sigma \)my² is the moment of inertia about X-axis. Thus,
I_x= \(\Sigma \)my²and I_y= \(\Sigma \)mx²
Substituting in the equation for I_z gives,
I_z = I_x + I_y
Y-axis and similarly the term \(\Sigma \)my² is the moment of inertia about X-axis. Thus,
I_X = \(\Sigma \)my²and I_y= \(\Sigma \)mx²
Substituting in the equation for I_z gives,
I_z = I_x + I_y
Thus, the perpendicular axis theorem is proved.

(i) Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.
(ii) If I_C is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is given by the relation,
I = I_C + Md²
(iii) Let us consider a rigid body as shown in Figure. Its moment of inertia about an axis AB passing through the center of mass is I_C DE is another axis parallel to AB at a perpendicular distance d from AB. The moment of inertia of the body about DE is I. We attempt to get an expression for I in terms of I_C For this, let us consider a point mass m on the body at position x from its center of mass.

(iv) The moment of inertia of the point mass about the axis DE is, m(x + d)². The moment of inertia I of the whole body about DE is the summation of the above expression.
\(I=\sum { m\left( x+d \right) ^{ 2 } } \)
This equation could further be written as,
\(I=\sum { m\left( { x }^{ 2 }+{ d }^{ 2 }+2xd \right) } \)
\(I=\sum { \left( { mx }^{ 2 }+m{ d }^{ 2 }+2dmx \right) } \)
\(I=\sum { { mx }^{ 2 }+\sum { m{ d }^{ 2 } } +2d\sum { mx } } \)
(v) Here, \(\sum { mx^{ 2 } } \) is the moment of inertia of the body about the center of mass. Hence, 
I_C = \(\sum { mx= } 0\) because, x can take positive and negative values with respect to the axis AB. The summation \(\left( \sum { mx } \right) \) will be zero.
Thus, I = I_c + \(\sum { md^{ 2 } } \) = I_C + \(\left( \sum { m } \right) d^{ 2 }\)
(vi) Here, \(\sum { m } \) is the entire mass M of the object \(\left( \sum { m=M } \right) \)
I = I_C + Md²
Hence the parallel axis theorem is proved.

Let us consider a uniform ring of mass and radius R. To find the moment of inertia of the ring about an axis passing through its center and perpendicular to the plane, let us take an infinitesimally small mass (dm) of length (dx) of the ring. This (dm) is located at a distance R, which is the radius of the ring from the axis as shown in Figure.

The moment of inertia (dI) of this small mass (dm) is,
dI = (dm)R²
The length of the ring is its circumference (2\(\pi\)R). As the mass is uniformly distributed, the mass per unit length (λ) is,
\(\lambda =\frac { mass }{ length } =\frac { M }{ 2\pi R } dx\)
The mass (dm) of the infinitesimally small length is, dm = λ.dx = \(\frac { M }{ 2\pi R } dx\)
Now, the moment of inertia (I) of the entire ring is
\(I=\int { dI } =\int { (dm) } R^{ 2 }=\int { \left( \frac { M }{ 2\pi R } dx \right) } R^{ 2 }\)
\(I=\frac { MR }{ 2\pi } \int { dx } \)
To cover the entire length of the ring, the limits of integration are taken from 0 to 2\(\pi\)R.
\(I=\frac { MR }{ 2\pi } \int _{ 0 }^{ 2\pi R }{ dx } \)
\(I=\frac { MR }{ 2\pi } \left[ x \right] ^{ 2\pi R }_{ 0 }=\frac { MR }{ 2\pi } \left[ 2\pi R-0 \right] \)
\(I=MR^{ 2 }\)

For a body to move in a circular path, there has, to be a centripetal force \((F=\frac {mv^2}{r})\) that keeps it in the circular path. When a cyclist bends while negotiating a round path, the direction of the normal force tilts which results in a vertical as well as the horizontal component of the force.
The forces acting on the cyclist are not balanced in a turn, since the cyclist is not having a uniform motion. The sum of the forces on the cyclist is oriented inside the turn, this force is created by the tire friction. When a cyclist bends inward on a curve to create the centripetal force needed to pull him in the curved path by the horizontal component of the normal reaction.
Let us consider a cyclist negotiating a circular level road (not banked) of radius r with a speed v. The cycle and the cyclist are considered as one system with mass m. The centre gravity of the system is C and it goes in a circle of radius r with center at O. Let us choose the line OC as X-axis and the vertical line through O as Z-axis as shown in Figure.
The system as a frame is rotating about Z-axis. The system is at rest in this rotating frame.
To solve problems in rotating frame of reference, we have to apply a centrifugal force (pseudo force) on the system which will be \(\frac{mv^2}{r}\). This force will act through the center of gravity. The forces acting on the system are, (i) gravitational force (mg), (ii) normal force (N), (iii) frictional force (f) and (iv) centrifugal force \(\left( \frac { mv^{ 2 } }{ r } \right) \).  As the system is in equilibrium in the rotational frame of reference, the net external force and net external torque must be zero. Let us consider all torques about the point A in Figure.
The torque due to the gravitational force about point A is (mg AB) which causes a clockwise turn that is taken as negative. The torque due to the centripetal force is
\(\left( \frac { mv^{ 2 } }{ r } BC \right) \)which causes an anti-clockwise turn that is taken as positive.
-mg AB + \(\frac { mv^{ 2 } }{ r } BC=0\)
mg AB \( = \ \frac { mv^{ 2 } }{ r } BC\)
From ΔABC,
AB = AC sin θ and BC = AC cos θ
mg AC sin θ = \(\frac{mv^2}{r}\) AC cos θ
tan θ = \(\frac{v^2}{rg}\)
θ = tan^-1\(\left( \frac { { v }^{ 2 } }{ rg } \right) \)


While negotiating a circular level road of radius r at velocity v, a cyclist has to bend by an angle θ from vertical given by the above expression to stay in equilibrium (i.e. to avoid a fall).