Curriculum Objectives
- understand and use the definition of the moment of inertia of a system of particles about a fixed axis as $\sum {m{r^2}} $ and the additive property of moment of inertia for a rigid body composed of several parts (the use of integration to find moments of inertia will not be required);
- use the parallel and perpendicular axes theorems (proofs of these theorems will not be required);
- recall and use the equation of angular motion $C = I\ddot \theta $ for the motion of a rigid body about a fixed axis (simple cases only, where the moment C arises from constant forces such as weights or the tension in a string wrapped around the circumference of a flywheel; knowledge of couples is not included and problems will not involve consideration or calculation of forces acting at the axis of rotation);
- recall and use the formula ${\textstyle{1 \over 2}}I{\omega ^2}$ for the kinetic energy of a rigid body rotating about a fixed axis;
- use conservation of energy in solving problems concerning mechanical systems where rotation of a rigid body about a fixed axis is involved.
Note:
The following chapter is going to involve angular and linear acceleration and velocity. The notation used up till now has been such that linear velocity and acceleration are given by the Latin alphabets, v and a, while the angular counterparts have been denoted by Greek alphabets, $\omega $ and $\alpha $. There is an alternate notation that is given precedence by many, including it seems your examiners. This notation uses concept of differentiation. We begin by fixing the following convention: the linear displacement is given by s; the angular displacement is given by $\theta $; and the time-derivative (the differentiation of a function with respect to time) of is denoted by a dot on the top. With this convention in place linear velocity becomes $\dot s$, linear acceleration becomes $\ddot s$, angular velocity $\dot \theta $, and the angular acceleration is given $\ddot \theta $
Moment of Inertia
Imagine a body being rotated around an axis, O, and further imagine a point, P, on it, such that the distance from O is given by ${r_i}$, its mass by ${m_i}$, and the its velocity is given by ${v_i}$. You should recall from your previous work on classical mechanics that the kinetic energy of the particle is given by ${\textstyle{1 \over 2}}{m_i}{v_i}^2$. From the work that we have done up till now, we know that ${v_i} = {r_i}\omega $ where $\alpha $ is the angular velocity of the particle. Thence we deduce that the Kinetic Energy of that single particle is ${\textstyle{1 \over 2}}{m_i}{\left( {{r_i}\omega } \right)^2} = {\textstyle{1 \over 2}}{m_i}{r_i}^2{\omega ^2}$.
If we now model that body as a collection, or matrix, of such points, then the total rotational kinetic energy for the body is given by:
\[\sum {{\textstyle{1 \over 2}}{m_i}{r_i}^2{\omega ^2}} \]
As ${\textstyle{1 \over 2}}$ is a constant and so is angular acceleration, we can say that:
\[Total\,Rotational\,Kinetic\,Energy = {\textstyle{1 \over 2}}\left( {\sum {{m_i}{r_i}^2} } \right){\omega ^2} = {\textstyle{1 \over 2}}I{\omega ^2} \]
If we compare this with its linear counterpart, under the pretence that these are similar formulae, we can see that the role of linear velocity is played by its angular counterpart, and that role of mass is played by ${\sum {{m_i}{r_i}^2} }$. This quantity is particularly important in our study. To see how it is important we shall draw parallels from mass in the Linear Kinetic Energy formula. The mass (of an object) can be understood to a measure of inertia, or the ability of the body to oppose changes in its state of motion. Which is to say that you will require a greater amount of force to cause the same acceleration (or impart the same amount of velocity) to a more massive body. Think how it easier for you to push a toddler than a full grown man. In rotational motion that we are dealing with here, not only the mass but the distance between the particle is also important to that particles inertia, because as you will come to see a particle at a greater distance to the origin will require more torque to be applied to it, than a particle of same mass that is closer to cause the same measure of angular acceleration. Hence the quantity, ${{m_i}{r_i}^2}$, is the rotational inertia of the particle P, and ${\sum {{m_i}{r_i}^2} }$, is the rotational inertia of the entire body. We call this quantity the moment of inertia.
\[I = \sum {{m_i}{r_i}^2} \]
The SI unit of this quantity is \[kg\,{m^2}\]
The Equation of Rotational Motion
For an object in rotational motion, we have the following equation. The derivation of this is not particularly difficult, though it does require a result we haven't yet used. For this reason and that it is not required by your board the proof is put in the relevant post and you need only note the following:
\[C = I\ddot \theta \]
or alternatively you may note:
\[\tau = I\alpha \]
The two equation differ only in the notation used. C and $\tau $ is the torque, I is the moment of inertia and $\ddot \theta \,and\,\alpha $ refer to angular acceleration. One can again draw comparisons with the linear case (Newton's Second Law) $F = m\ddot s$.
Angular Momentum (Optional)
It's optional to note, but instructive nonetheless that the angular momentum is given by:
\[L = I\dot \theta \]
The reader is encouraged to deduce this by again modelling an object as a matrix of points.
Here I will suggest to the reader a recourse to the second chapter of our mechanics to revise the equations regarding constant angular acceleration.
Ex 1. A flywheel, such that it has a moment of inertia of 300 $kg\,{m^2}$, is rotating at $2\,rad{\kern 1pt} {s^{ - 1}}$. A constant torque of 150 $N{\kern 1pt} m$ is applied to the flywheel for 10 seconds, in the sense that it increases its rate of rotation. Find
a) Angular Velocity at the end of this time
b) Angular Displacement during this time
c) Increase in the Kinetic Energy of the flywheel
Sol.
a) We note to begin with $C = 150;\,I = 300$
From this we have: $\ddot \theta = \frac{C}{I} = \frac{{150}}{{300}} = 0.5\,rad{\kern 1pt} {s^{ - 2}}$
The relevant equation then is:
$\omega = {\omega _0} + \alpha t$
$\omega = {\omega _0} + \alpha t = 2 + 0.5\left( {10} \right) = 7$
b) The equation to be used here is:
$\theta = {\theta _0} + {\omega _0}t + {\textstyle{1 \over 2}}\alpha {t^2} = 0 + 2\left( {10} \right) + {\textstyle{1 \over 2}}\left( {0.5} \right)\left( {100} \right) = 45$
c) $RE = {\textstyle{1 \over 2}}I{\omega ^2}$
Increase in RE is then given by:
$\Delta RE = {\textstyle{1 \over 2}}I{\omega ^2} - {\textstyle{1 \over 2}}I{\omega _0}^2 = {\textstyle{1 \over 2}}I\left( {{\omega ^2} - {\omega _0}^2} \right) = {\textstyle{1 \over 2}}300\left( {{7^2} - {2^2}} \right) = 6750$
