Centre of mass for different bodies are given below

• For uniform rod, `x_(CM) = 1/2 , y_(CM) = 0` and `z_(CM) = 0`
• Centre of mass of a uniform rectangular square or circular plate lies at its centre.
• For uniform semicircular ring `y_(CM) = (2R)/pi , x_(CM) = 0` and `z_(CM) = 0`
• For uniform semicircular disc , `y_(CM) = ( 4R)/(3pi) , x_(CM) = 0` and `z_(CM) = 0`
• For hemispherical shell , `y_(CM) = R/2, x_(CM) = 0` and `z_(CM) = 0`
• For solid hemisphere, `y_(CM) = (3R)/8` and `x_(CM) = 0` and `z_(CM) = 0`
• If a body falling under gravity explodes into pieces, then the centre of mass of the system remains along the same vertical line.
• Centre of mass and centre of gravity of a regular body coincide.

Motion of rigid body have two types

(i) `text(Transnational Motion)`
Transnational motion is the motion by which a body shifts from one point in space to another point. One example of transnational motion is the motion of bullet fired from a gun.

(ii) `text(Rotational Motion)`
A body is said to possess rotational motion, if all its particles move dong circles in parallel lines, The centres of these circle lie on a fixed line perpendicular to the parallel lines and this fixed line is called the: axis of rotation.

The property by virtue of which a body opposes its change in its state of motion is called its inertia.
In similar manner, a body have the tendency to oppose its rotational motion or change in rotational motion about its axis is known as moment of inertia.
Mathematically, it is given as the product of mass and square of the perpendicular distance of a particle from the axis of rotation.

Moment of inertia, `I = mr^2`.
where, `m=` mass of the particle
`r=` perpendicular distance from axis of rotation

The moment of inertia of the rigid body, `I = sum mr^2`. Its SI unit is `kgm^2`. Moment of inertia is the property of a body due to which it opposes any change its state of rest or uniform motion.

The moment of inertia of a body depends on
(i) Mass of the body.
(ii) Size and shape of the body.
(iii) Distribution of mass about the axis of rotation.
(iv) Position and orientation of the axis of rotation w.r.t the body.

Radius of gyration of a given body about a given axis of rotation is the normal distance of a point from the axis, where if whole mass of the body is placed, then its moment of inertia will be exactly same as it has with its actual distribution of mass.
Thus, radius of gyration , `K = sqrt(I/M)`

or `K = [ (r_1^2 + r_2^2 + r_3^2 + .... r_n^2 )/n ]^(1/2)`

The SI unit of radius of gyration is metre.

Radius of gyration depends upon the shape and size of the body, position and configuration of the axis of rotation and also on distribution of mass of body w.r.t axis of rotation.

There are two theorems in moment of inertia, i.e. theorem of parallel axes and theorem of perpendicular axes.

1. Theorem of Parallel Axes :

Moment of inertia of a body about a given axis I is equal to the sum of moment of inertia of the body about an axis parallel to given axis and passing through centre of mass of the body `I_g` and `Ma^2`, where M is the mass of the body and a is the perpendicular distance between the two axes. as shown in fig 1

`I = I_g + Ma^2`

e.g. Moment of inertia of a disc about an axis through its centre and perpendicular to the plane is `1/2 MR^2` , so moment of inertia of the disc about an axis through its tangent and perpendicular to the plane will be

`I = I_g + Ma^2`

`I = 1/2 MR^2 + MR^2`

`:. I = 3/2 MR^2` which is shown in fig 1.1

2. Theorem of Perpendicular Axes :

According to this theorem, the sum of moment of inertia of a plane lamina about two mutually perpendicular axes lying in its plane is equal to its moment of inertia about an axis perpendicular to the plane of lamina and passing through the point of intersection of first two axes, as shown in Fig 2

`I_z = I_x + I_y`

e.g. Moment of inertia of a disc about an axis through its centre of mass and perpendicular to its plane is `1/2M R^2`, if the disc is in xy plane, then by theorem of perpendicular axes, fig 2.1

`I_z = I_x + I_y`

where , `I_z = (MR^2)/2 => 1/2 MR^2 = 2 I_D`

as disc is symmetrical body , so `I_x = I_y = I_D`

Torque is a quantity which measures the capability of a force to rotate a body. It is a vector quantity which is directly perpendicular to both position vector (r) and force (F), Its SI unit is Nm.
Torque, `tau = vecr xx vecF=rF sin theta hatn`
where, `theta=` angle between `vecr` and `vecF`
and `hatn=` unit vector which shows the direction `tau`

Torque due to a force is also known as the moment of a force.
► Angular Momentum :
The moment of linear momentum of a given body about an axis of rotation is called as its angular momentum. If `p` is the linear momentum of a particle and `r` is its position vector from the point of rotation, then angular momentum `(vecL) = vecr xx vecp = rp sin theta hat n`.

• `text(Relation between Torque and Angular Momentum)`
We know that,
torque, `tau = vecr xx vecF`
angular momentum, `vecL = vecr xx vecp`

On differentiating both sides w.r .t time t, we get
`(dvecL)/(dt) = d/(dt) ( vecr xx vecp) = (dvecr)/ (dt) xx vecp + vecr xx (dvecp)/(dt)`
`= vecv xx vecp + vecr xx vecF \ \ \ \ \ \ [ because (dvecp)/(dt) = vecF]`
`= 0 + tau \ \ \ \ \ \ \ \ [ because vecv xx vecp = vecv xx mvecv = 0 ]`
`:. tau = ( dL) /(dt)`

Thus, the torque acting on a particle is equal to its rate of change of angular momentum.

► `text(Conservation of Angular Momentum)`
According to the law of conservation of angular momentum, if no external torque is acting on a system, then total vector sum of angular momentum of different particles of the system remains constant.
We know that, `(dvecL)/(dt)=tau_(ext)`
Hence, if `tau_(ext)=0`, then `(dvecL)/(dt)=0=>L=` constant

Therefore, in the absence of any external torque, total angular momentum of a system must remain conserved.

The energy, which is possessed by the rotation of body, known as rotational kinetic energy. It is represented by `K_R`.

`K_R = 1/2 I omega^2` , where `omega =` angular velocity

In this type of rotational motion, axis of rotation is in motion. Plane motion can be considered as combination of transnational motion of the centre of mass and rotational motion of the body about an axis passing through the centre of mass.

Body possess both translational and rotational kinetic energy.

Net kinetic energy = (Translatory + Rotatory) kinetic energy

`K_N = K_T + K_R = 1/2 mv^2 + 1/2 I omega^2 = 1/2 mv^2 + 1/2mv^2 K^2/R^2`

`:. K_N =1/2 mv^2 (1 + K^2/R^2)`

If the radius and mass of an object are R and m and the vertical height of inclined plane is h. It performs rolling motion on inclined plane, then following conditions occur