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Rigid Body
A rigid body is defined as a solid body in which the particles are compactly arranged so that the inter-particle distance is small and fixed, and their positions are not disturbed by any external forces applied on it. A rigid body can undergo both translational and rotational motion.
A rigid body is said to have translator motion if it moves bodily from one plate to another. The motion of a car is the translator in nature. A rigid body is said to be in rotational motion about a fixed axis when its particles generate concentric circles with the same angular velocity but different linear velocities. The motion of a wheel of a train about its axle is rotational motion.
Moment of Inertia
The inertness or inability of a body to change its state of rest or uniform motion by itself is called inertia. In the case of linear motion, the mass of the object determines the inertia of the body. As the mass of the body is high the inertia is also high and hence requires more force to move it (to change the state of that body).
Similarly in case of rotational motion if the moment of inertia is the high large amount of rotational force i.e. torque is applied on the rigid body which means ‘moment of inertia’ describes the state of change body in rotational motion.
Similarly in rotational motion, a body rotating about an axis opposes any change desired to be produced in its state. This inertia is the rotational motion of the body is called rotational inertia.
Consider a rigid body consisting a large number of small particles of mass m_{1}, m_{2}, m_{3}, ….. Suppose the body be rotating about axis YY’ and the distance of these particles from this axis is r_{1}, r_{2}, r_{3} ... As shown in the figure. The moment of inertia of these particles about the axis of rotation YY’ will be m_{1}r_{1}^{2}, m_{2}r_{2}^{2}, m_{3}r_{3}^{2} , … respectively. The moment of inertia, I of the body about the axis YY’ of which the body is made.
$$I=m_1r^2_1 +m_2r^2_2+m_3r^2_3 +...................................+m_nr^2_n$$
$$=\sum_{i=1}^nm_ir^2_1$$
Where, m_{i} is the mass of the i^{th} particle and r_{i} is the perpendicular distance from the axis of rotation.
Hence, the moment of inertia of the rigid body about a given axis of rotation is the sum of the product of the masses of the various particles and square of their perpendicular distances from the axis of rotation. In SI-units, the unit of moment of inertia is kg m^{2}. In CGS-system, its unit is g cm^{2}. The dimensional formula of the moment of inertia is [ML^{2}T^{0}].
Parallel theorem of moment of inertia states that, “Moment of inertia of any rotating body about same axis parallel to the axis passing through centre of mass is equal to the sum of the moment of inertia about centre of mass (C.M.) and product of the total mass of the rotating body and square of the distance between two parallel axis.”
$$ I =\ I_{cm} + Mr^2 $$
Where I_{cm} is the moment of inertia about the parallel axis through the centre of mass, M is the total mass of the body and r is the distance between two axes.
Proof
Consider CD be the axis passing through the centre of the mass and perpendicular to the plane of the rotating body AB be the same axis at a distance ‘r’ from CD and is parallel to CD. We have to calculate the moment of inertia of the rotating body of mass (M) about AB. Suppose m be the mass of point abject which is at a distance x from the centre of mass.
Total moment of inertia about PQ is
$$I=\sum m(x+r)^2$$
$$\sum mx^2+\sum mr^2+2\sum mxr$$
$$since,I_{cm}=\sum mx^2 ,the\; moment\;of\;inertia\;of\;the\;body\;about\;the\;axis\;CD'\;so\;$$
$$I= I-{cm}+\sum mr^2 +2 \sum mxr$$
Similarly, \(\sum mr^2 = r^2, \sum m = Mr^2\) , where M is the mass of the whole body and \(\sum mx = \) the sum of the moments of all the particles of the body about the axis CD passing through its centre of mass and therefore equal to zero i.e. \(\sum mix = 0 \). Then, equation (i) can be written as
$$ I = I_{cm} + Mr^2 $$
That is the moment of inertia of the body about the axis AB is the sum of its moment of inertia about the parallel axis CD through its centre of mass and the product of the mass of the body and square of the distance between the two axes. This is the theorem of the parallel axis.
Theorem of Perpendicular Axis
This theorem states, “The sum of the moment of inertia of a laminar body about any two mutually perpendicular axes in the plane is equal to its moment of inertia about an axis perpendicular to its plane and passing through the point of intersection of the two axes.”
Let I_{x}, I_{y} and I_{z} be the moments of inertia of plane lamina three mutually perpendicular axes passing through the point O. OX and OY axes are in the plane and axis OZ is perpendicular to the plane, then
$$ I_z = I_x + I_y $$
Proof
Let OX and OY be two mutually perpendicularly axes in the plane of the lamina and OZ be an axis passing through O and perpendicular to the plane of lamina. If a particle P of mass m is at a distance r from O, the moment of inertia of this particle about the axis OZ = mr^{2}. The moment of inertia of the inertia of the entire body about the OZ-axis is given by
\begin{align*} I_z &= \sum mr^2 \\\end{align*} The moment of inertia of the body about OX-axis is\begin{align*} I_X = \sum my^2 \text {and about OY-axis,} I_y = \sum mr^2. \\ \text {then,} \\ I_x + I_y &= \sum m (x^2 + y^2) \\ I_x + I_y &= \sum mr^2 \dots (ii) \\ \text {where } r^2 = x^2 + y^2 \\ \text {therefore, from equation} (i) \text {and equation} (ii), \text {we have} \\\ I_x + I_y &= I_z \dots (iii) \end{align*}
Which proves the theorem of perpendicular axis for a laminar body.
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