Moment of Inertia of an Annular disc
MI of annular disc about an axis perpendicular to the plane and passing through C.G:\begin{align*}I=\frac{M(R^{2}+r^{2})}{2}\end{align*} MI about an axis along the plane and passing through C.G:\begin{align*}I_{d}=\frac{M(R^{2}+r^{2})}{4}\end{align*}
Summary
MI of annular disc about an axis perpendicular to the plane and passing through C.G:\begin{align*}I=\frac{M(R^{2}+r^{2})}{2}\end{align*} MI about an axis along the plane and passing through C.G:\begin{align*}I_{d}=\frac{M(R^{2}+r^{2})}{4}\end{align*}
Things to Remember
MI of the annular disc.
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Moment of Inertia of an Annular disc
Moment of Inertia of an Annular disc:
A circular disc from which a small disc is removed is called annular disc. For eg: CD
1) About an axis perpendicular to the plane and passing through CG(centre): (r to R):
Consider an annular disc of mass ‘M’, external radius ‘R’ and internal radius ‘r’. Let us take an axis AB, perpendicular to the plane of the disc and passing through its CG about which moment of inertia is to be determined. Let us divide the disc into large no. of concentric elementary rings. Let us take any one of the elementary rings of radius x and thickness dx.
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Mass per unit surface area of the disc= \(\begin{align*}\frac{M}{\pi (R^{2}-r^{2})}\end{align*}\)
Mass of the elementary ring, m=\(\begin{align*}\frac{M}{\pi (R^{2}-r^{2})}\times 2\pi xdx\end{align*}\)\begin{align*}m=\frac{2M}{R^{2}-r^{2}}xdx\end{align*}
Moment of inertia of that elementary ring is given by,
\begin{align*}dI=mass\times (distance)^{2}\end{align*}\begin{align*}=(\frac{2M}{R^{2}-r^{2}})xdx\times x^{2}\end{align*}
Since, such elementary rings vary from x= r to x= R the total MI of the whole disc is given by,
\begin{align*}I=\int dI\end{align*}\begin{align*}=\int_{r}^{R}\frac{2M}{R^{2}-r^{2}}x^{3}dx\end{align*}\begin{align*}=(\frac{2M}{R^{2}-r^{2}})(\frac{R^{4}-r^{4}}{4})\end{align*}\begin{align*}I=\frac{M(R^{2}+r^{2})}{2}\end{align*}
2) About an axis along the plane and passing through CG (i.e. about a diameter):
To determine the MI of an annular disc about one of its diameter we have to apply the theorem of perpendicular axis. For this le us consider two diameters XOX’ and YOY’ perpendicular to each other about which M.I is Id.
consider an axis AOB perpendicular to the plane of the ring and passing through its centre. Then by applying the perpendicular axis theorem, we get
\begin{align*}I_{AB}=I_{XOX'}+I_{YOY'}\end{align*}\begin{align*}I_{d}=\frac{M(R^{2}+r^{2})}{4}\end{align*}
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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
Lesson
Dynamics of Rigid Bodies
Subject
Physics
Grade
Bachelor of Science
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