Theorems in Rotational Dynamics

Theorems in Rotational Dynamics

We can find out the moment of inertia of different bodies with the help of following theorems:

1) Theorem of parallel axis:

It states that " The moment of inertia of a rigid body about an axis is equal to the algebraic sum of moment of inertia about an axis parallel to the given axis and passing through the C.G of the body and the product of mass of the body and square of distance between the two parallel axes. i.e,\begin{align*}I_{AB}=I_{C}+MD^{2}\end{align*}Where,

\(I_{AB}\)= M.I about the axis AB

\(I_{C}\)=M.I about the axis through C.G and parallel to AB.

\(M\)= Mass of the body.

\(D\)= Perpendicular distance the two parallel axes.

Proof:Consider a rigid body of mass M. Suppose we are going to find the M.I of the body about an axis AB. Consider an elementary particle P on the rigid body, with mass 'm'. the M.I of the particle P about AB is given by \(\begin{align*}m\times OP^{2}\end{align*}\). Then, total M.I of the rigid body about AB is given by\begin{align*}I=\sum m\times (OP^{2})\end{align*}\begin{align*}I=\sum m\times (OQ^{2}+QP^{2})\end{align*}\begin{align*}I=\sum m\times [(OC+CQ)^{2}+QP^{2}]\end{align*}\begin{align*}I=\sum m\times [OC^{2}+2\times OC\times CQ+ CQ^{2}+QP^{2}]\end{align*}\begin{align*}I=\sum m\times [OC^{2}+2\times OC\times CQ+CP^{2}]\end{align*}\begin{align*}I=\sum m\times [D^{2}+2\times D\times CQ+CP^{2}]\end{align*}\((\because OC=D^{2})\)\begin{align*}I=\sum mD^{2}+2D\sum mCQ+\sum mCP^{2}\end{align*}\begin{align*}I=MD^{2}+2D\sum mCQ+I_{C} (\because \sum m=M)\end{align*}Where, \(I_{C}\) is the M.I of the rigid body about an axis through C.G. Since the algebraic sum of moment about centre of m is zero.\begin{align*}\sum (m.CQ)=0\end{align*}\begin{align*}\therefore I=I_{C}+MD^{2}\end{align*} \(proved.\)

2)Theorem of perpendicular axis:

It states that," The moment of inertia of a plane lamina (a plane body) about an axis perpendicular to the plane is equal to the sum of the moments of inertia about the mutually perpendicular axes in the plane of the lamina which intersect the axis perpendicular to the plane. i.e, the M.I of a plane lamina about z-axis(Iz) is given by,\begin{align*}I_{Z}=I_{X}+I_{Y}\end{align*}Where, \(\begin{align*}I_{_{X}}\end{align*}\) and \(\begin{align*}I_{_{Y}}\end{align*}\) are the M.I of the body about X and Y- axes respectively.

Proof:Consider a plane lamina of mass 'm' lying in Xy-plane. Let OZ is the axis which is perpendicular to the plane of the lamina and passes through the point of intersection 'O' of X-axis and Y-axis.
Now, let us consider an elementary particle of mass 'm' in the lamina at point P(x,y) such that OP=r Then, M.I of the particle about Z-axis is given by,\begin{align*}I=mr^{2}\end{align*}\begin{align*}I=m(x^{2}+y^{2})\left [ \because r^{2}=x^{2}+y^{2} \right ]\end{align*}Thus, the MI of the plane lamina about z-axis is given by,\begin{align*}I_{Z}=\sum m(x^{2}+y^{2})\end{align*}\begin{align*}I_{Z}=\sum mx^{2}+\sum my^{2}-------(1)\end{align*}Here, \(\begin{align*}\sum mx^{2}=I_{Y}
\end{align*}\)= MI of the lamina about Y-axis.\begin{align*}\sum my^{2}=I_{X}\end{align*}=MI of te lamina about X-axis.
thus from equation (1)\begin{align*}I_{Z}=I_{X}+I_{Y}\end{align*}

References:

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014