Parallel Axes Theorem
The theorem of parallel axes states that the moment of inertia of a rigid body about any axis is equal to its moment of inertia about a parallel axis through its centre of mass plus the product of the mass of the body and the square of the perpendicular distance between the two parallel axes.
If M is the total mass of the body and h the distance between the two parallel axes, the relation is
I = I_{CM }+ Mh^{2 }(7.41)
This relation is known as the parallel axes theorem.
The parallel axis theorem 
The theorem holds irrespective of whether the centre of mass C is chosen to coincide with the origin of the reference frame.
Perpendicular Axes Theorem
The theorem of perpendicular axes for a plane laminar body states that the moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia of the lamina about any two mutually perpendicular axes in its own plane and intersecting each other at the point where the perpendicular axis passes through it.
If I_{x} and I_{y} are the moments of inertia of a plane lamina (Fig) about the perpendicular axes x and y respectively which lie in the plane of the lamina and intersect each other at O, then the moment of inertia I of the lamina about an axis passing through O and perpendicular to its plane is given by
I = I_{x} + I_{y }

The two theorems described above help us in determining the moment of inertia of a body about any other axis.
NEWTON'S SECOND LAW FOR A SYSTEM OF PARTICLES
If you roll a cue ball at a second billiard ball that is at rest, you expect that the twoball system will continue to have some forward motion after impact. You would be surprised, for example, if both balls came back toward you and if both moved to the right or to the left.
What continues to move forward, its steady motion completely unaffected by the collision, is the center of mass of the two balls. If you focus on this point  which is always halfway between these bodies of identical mass  you can easily convince yourself by trial at a pool table that this is so. No matter whether the collision is glancing, head on, or somewhere in between, the center of mass moves majestically forward, as if the collision had never occurred. Let us look into this in more detail.
To do so, we replace the pair of billiard balls with an assemblage of n particles of (possibly) different masses. We are interested not in the individual motions of these particles but only in the motion of their center of mass. Although the center of mass is just a point, it moves like a particle whose mass is equal to the total mass of the system; we can assign a position, a velocity, and an acceleration to it. We state (and shall prove below) that the (vector) equation that governs the motion of the center of mass of such a system of particles is
âˆ‘ F_{ext} = Ma_{cm} (system of particles) .............(7.7)
Equation 7.7 is Newton's second law governing the motion of the center of mass of a system of particles. Remarkably, it retains the same form ( âˆ‘ F = ma) that holds for the motion of a single particle. In using eq. 7.7, the three quantities that appear in it must be evaluated with some care:
 âˆ‘ F_{ext} is the vector sum of all the external forces that act on the system. Forces exerted by one part of the system on another are called internal forces, and you must be careful not to include them when using eq.7.7.
 M is the total mass of the system. We assume that no mass enters or leaves the system as it moves, so that M remains constant. The system is said to be closed.
 a_{cm} is the acceleration of the center of mass of the system. Equation 7.7, like all other vector equations, is equivalent to three equations involving the components of âˆ‘ F_{ext} and a_{cm} along the three coordinate axes. These equations are
Equation 916, like all other vector equations, is equivalent to three equations involving the components of âˆ‘ F_{ext} and a_{cm} along the three coordinate axes. These equations are
âˆ‘ F_{ext,x} = Ma_{cm,xâ€²}
âˆ‘ F_{ext y }= Ma_{cm,yâ€²}
âˆ‘ F_{ex,z }= Ma_{cm,z}_{ â€²} ..........(7.8)
Now we can go back and examine the behaviour of the billiard balls. Once the cue ball has begun to roll, no net external force acts on the (twoball) system. And thus, because âˆ‘ F_{ext} = 1, Eq. 7.7 tells us that a_{cm} = 0 also. Because acceleration is the rate of change of velocity, we conclude that the velocity of the center of mass of the system of two balls does not change. When the two balls collide, the forces that come into play are internal forces, exerted by one ball on the other. Such forces doe not contribute to âˆ‘ F_{ext}, which remains at zero. Thus, the center of mass of the system, which was moving forward before the collision, must continue to move forward after the collision, with the same speed and in the same direction.
Equation 7.7 applies not only to a system of particles but also to a solid body, such as a bat. In that case, M in Eq. 7.7 is the mass of the bat and âˆ‘ F_{ext} is the weight Mg of the bat. Equation 7.7 then tells us that a_{cm} = g. In other words, the center of mass of the bat moves as if the bat were a single particle of mass M, with force Mg acting on it.
Kepler's Second Law Planetary Motion. The most famous example of the above result is Kepler's second law of the motion of planets round the sun. Kepler proposed his second law in 1602 and months later he proposed his first law which states that each planet moves on an ellipse with the sun at one of its foci. His second law states that the straight line from the sun to the planet sweeps out equal areas in equal intervals of time. Two such areas are shown shaded in Fig. Thus the planet has greater speed when it is closer to the sun as indicated by the figure.
This law implies that the angular momentum of the planet is conserved during its motion and the sun's force of attraction which keeps the planet in orbit must be radial, always pointing towards the sun. A radial force

Kepler's second law of planetary motion (v_{1} < v_{2}) 
A particle of mass m is released form rest from point P at x = x_{0 }on the xaxis from origin O and falls vertically along the yaxis as shown in Fig.
 Find the torque Ï„ acting on the particle at a time t when is a point Q with respect to O.
 Find the angular momentum L of the particle about O at this time t.
 Show that Ï„ = in this example.
The torque is produced by the force of gravity F = mg. Let r be the position vector at Q.
( a ) The magnitude of the torque is given by
= rF sin Ï…
= r x mg x ()
= mgx_{0 } (i)
The direction of the torque is shown in the figure; it is directed into the page and perpendicular to it.
(b) The magnitude of the angular momentum is
L = rp sinÏ…
Where p = m Ï…, Ï… being the velocity at Q which is given by
Ï… = u + at = 0 + gt = gtâˆ´ L = r x mgt x mgx_{0}t (ii)
The direction of angular momentum is the same as that of torque.
(c) Differentiating (ii) with respect to t, we have
= Ï„ [see eq. (i)]
Hence Ï„ = is obeyed in this example.
Angular Momentum and Energy for a System of Particles
Now we turn our attention to the motion of a system of particles, with respect to an origin. Note that "a system of particles" includes a rigid body as a special case. The total angular momentum L of a system of particles is the (vector) sum of the angular momenta l of the particles:L = l_{1} + l_{2} + l_{3} + ...... + l_{n} = (7.9)
in which i (= 1, 2, 3, .....) labels the particles.
With time, the angular momenta of individual particles may change, either because of interactions within the system (between the individual particles) or because of influences that may act on the system from the outside. We can find the change in L as these changes take place by taking the time derivative of eq.7.9. Thus
Some torques are internal, associated with forces that the particles within the system exert on one another; other torques are external, associated with forces that act from outside the system. The internal forces, because of Newton's law of action and reaction, cancel in pairs. So, to add the torques, we need to consider only those associated with external forces. Equation 7.10 then becomes