Moment of Inertia of a Rectangular Lamina

Moment of Inertia of a Rectangular Lamina

In the case of continuous,homogenous body of a definite geometrical shape,its M.I. is calculated by

(1) the first obtaining an expression for moment of inertia of an infinitesimal element of mass (dm) about the given axis i.e. dm.r² where r is the distsnce of that element from the axis

(2) integrating this expression over appropriate limits so as to cover the entire body ^.

1) About an axis perpendicular to the plane and passing through Centre of gravity. [-l/2 to +l/2]:

Consider a rectangular lamina of mass ‘M’ length ‘l’ and breadth ‘b’. let YOY’ be the axis, perpendicular to the plane of the body and passing through the C.G of the body, about which MI of he lamina is to be calculated.

Let us divide the lamina into large no. of such elementary piece (along the length). Consider any one of such elementary piece of thickness ‘dx’ at a distance ‘x’ from the axis of rotation YY’.

Mass per unit length of the lamina= \(\begin{align*}\frac{M}{l}\end{align*}\)

Mass of the elementary piece= \(\begin{align*}\frac{M}{l}dx\end{align*}\)

So, moment of inertia of that elementary piece about Y axis is given by

\begin{align*}z\end{align*}\begin{align*}=\frac{M}{l}dx\times x^{2}\end{align*}

Since there are such pieces in the lamina varying from x= -l/2 to x= l/2 on both sides of YY’ axis, so total M.I of the whole body is given by, (about the length):

\begin{align*}I_{l}=\int dI\end{align*}\begin{align*}=\int_{-l/2}^{+l/2}\frac{M}{l}x^{2}dx\end{align*}\begin{align*}=\frac{M}{l}(\frac{2l^{3}}{8})\end{align*}\begin{align*}I_{l}=\frac{Ml^{2}}{12}\end{align*}

Similarly, M.I about the breadth is given by,

\begin{align*}I_{b}=\frac{Mb^{2}}{12}\end{align*}

Thus, total moment of inertia of the rectangular lamina about perpendicular to the plane of the lamina axis passing through ‘O’ is

\begin{align*}I=I_{l}+I_{b}[\because using perpendicular theorem]\end{align*}\begin{align*}I=\frac{m(l^{2}+b^{2})}{12}\end{align*}

2) About an axis through any one side of the lamina [0 to l]:

Consider a rectangular lamina of mass M, length ‘l’ and breadth ‘b’. Let YOY’ be the axis passing through one end of the lamina, about which we are going to find the M.I of the lamina. Let us divide the lamina into small elementary pieces (along the length). Let us take anyone of such pieces having thickness ‘dx’ at a distance x from the axis of rotation.

Mass per unit length of the lamina= \(\begin{align*}\frac{M}{l}\end{align*}\)

Mass of the elementary piece=\(\begin{align*}(\frac{M}{l})dx\end{align*}\)

Moment of inertia of the elementary piece is given by

\begin{align*}dI=mass\times (distance)^{2}\end{align*}\begin{align*}=\frac{M}{l}dx\times x^{2}\end{align*}

In the lamina such elementary pieces are varying from x=0 to l. so total M.I of the lamina about the length is given by,

\begin{align*}I_{l}=\int dI\end{align*}\begin{align*}=\int_{0}^{l}\frac{M}{l}x^{2}dx\end{align*}\begin{align*}I_{l}=\frac{Ml^{2}}{3}\end{align*}

Similarly M.I of the body about the breadth is given by,

\begin{align*}I_{b}=\frac{Mb^{2}}{3}\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