# 18.4 Rotation of rigid body  (Page 4/4)

 Page 4 / 4

$\begin{array}{l}I=\int {r}^{2}đm\end{array}$

Evaluation of this integral for a given body is a separate task in itself. It is mathematically possible to evaluate this integral for bodies of regular shape. It would, however, be very difficult to evaluate the same for irregularly shaped rigid body. In such cases, it is pragmatic to resort to experimental methods to calculate moment of inertia. Mathematical evaluation of moment of inertia even for regularly shaped bodies would require specialized analysis and evaluation.

Two theorems, pertaining to moment of inertia, are of great help in the mathematical evaluation of moment of inertia of regularly shaped bodies. They are (i) parallel axes theorem and (ii) perpendicular axes theorem. These theorems extend the result of moment of inertia of basic geometric forms of rigid bodies to other axes. We shall cover these aspects and shall evaluate moment of inertia of certain important geometric rigid bodies in separate modules.

## Kinetic energy of rigid body in rotation

Kinetic energy of a particle or body represents the form of energy that arises from motion. We are aware that kinetic energy of a particle in translation is given by the expression :

$\begin{array}{l}K=\frac{1}{2}m{v}^{2}\end{array}$

In pure rotation, however, the rigid body has no "over all" translation of the body. However, the body in rotation must have kinetic energy as it involves certain motion. A closer look on the rotation of rigid body reveals that though we may not be able to assign translation to the rigid body as a whole, but we can recognize translation of individual particles as each of them rotate about the axis in circular motion with different linear speeds. The speed of a particle is given by :

$\begin{array}{l}{v}_{i}=\omega {r}_{i}\end{array}$

Thus, kinetic energy of an individual particle is :

$\begin{array}{l}{K}_{i}=\frac{1}{2}{m}_{i}{{v}_{i}}^{2}\end{array}$

where " ${K}_{i}$ " is the kinetic energy of "i" th particle having a speed " ${v}_{i}$ ". In terms of angular speed, the kinetic energy of an individual particle is :

$\begin{array}{l}{K}_{i}=\frac{1}{2}{m}_{i}{\left(\omega {r}_{i}\right)}^{2}\end{array}$

Now, the kinetic energy of the rigid body is sum of the kinetic energies of the particles constituting the rigid body :

$\begin{array}{l}K=\sum {K}_{i}=\sum \frac{1}{2}{m}_{i}{\omega }^{2}{{r}_{i}}^{2}\end{array}$

We note here that angular speeds of all particles constituting the body are same. Hence, the constant "1/2" and " ${\omega }^{2}$ " can be taken out of the summation sign :

$\begin{array}{l}K=\frac{1}{2}{\omega }^{2}\sum {m}_{i}{{r}_{i}}^{2}\end{array}$

However, we know that :

$\begin{array}{l}I=\sum {m}_{i}{{r}_{i}}^{2}\end{array}$

Combining two equations, we have :

$\begin{array}{l}K=\frac{1}{2}I{\omega }^{2}\end{array}$

This is the desired expression of kinetic energy of a rigid body rotating about a fixed axis i.e. in pure rotational motion. The form of expression of the kinetic energy here emphasizes the correspondence between linear and angular quantities. Comparing with the expression of kinetic energy for translational motion, we find that "moment of inertia (I)" corresponds to "mass (m)" and "linear speed (v)" corresponds to "angular speed (ω)".

We can also interpret the result obtained above from a different perspective. We could have directly inferred that expression of kinetic energy in rotation should have an equivalent form as :

$\begin{array}{l}\mathrm{K \left(Kinetic energy\right)}=\frac{1}{2}x\mathrm{\left(inertia\right)}x{\mathrm{\left(speed\right)}}^{2}\end{array}$

In rotation, inertia to the rotation is "moment of inertia (I)" and speed of the rigid body is "angular speed (ω)". Substituting for the quantities, we have the expression for kinetic energy of rigid body in rotation as :

$\begin{array}{l}K=\frac{1}{2}I{\omega }^{2}\end{array}$

Comparing this equation with the expression of the sum of kinetic energy of individual particles as derived earlier :

$\begin{array}{l}K=\frac{1}{2}{\omega }^{2}\sum {m}_{i}{{r}_{i}}^{2}\end{array}$

Clearly,

$\begin{array}{l}I=\sum {m}_{i}{{r}_{i}}^{2}\end{array}$

This conclusion, thus, clearly establishes that the expression as given by $\sum {m}_{i}{{r}_{i}}^{2}$ indeed represents the inertia of the rigid body in rotation.

Though it is clear, but we should emphasize that expressions of kinetic energy of rotating body either in terms of angular speed or linear speed are equivalent expressions i.e. two expressions measure the same quantity. Two expressions do not mean that the rotating body has two types of kinetic energy.

## Summary

1. Moment of inertia is the inertia of an object against any change in its state of rotation. This quantity corresponds to “mass”, which determines inertia of an object in translational motion.

2. Moment of inertia is defined for rotation about a fixed axis for a particle, a system of particles and a rigid body.

3. Expressions of Moment of inertia

(i) For a particle

$\begin{array}{l}I=m{r}^{2}\end{array}$

(ii) For a system of particles

$\begin{array}{l}I=\sum {m}_{i}{{r}_{i}}^{2}\end{array}$

(iii) For a rigid body

$\begin{array}{l}I=\int {r}^{2}đm\end{array}$

4. Moment of inertia, also referred in short as MI, is a scalar quantity. The directional positions (angular positions) of the particles/objects with respect to axis of rotation does not matter. The unit of MI is $\mathrm{kg}-{m}^{2}$ .

5. Objects in rotation have rotational kinetic energy of rotation due to rotational motion of individual particles, constituting the object.

6. Object in pure rotation has only rotational kinetic energy i.e. no translational kinetic energy is involved.

7. The expression of rotational kinetic energy is given by :

$\begin{array}{l}K=\frac{1}{2}I{\omega }^{2}\end{array}$

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