Moment of Inertia of a Circular Disc
This note provides us information about moment of inertia on circular disc on about an axis perpendicular to the plane and passing through CG and about an axis along the plane passing through CG. This note provides us an informaion about moment of inertia of a circular disc.
Summary
This note provides us information about moment of inertia on circular disc on about an axis perpendicular to the plane and passing through CG and about an axis along the plane passing through CG. This note provides us an informaion about moment of inertia of a circular disc.
Things to Remember
MI of circular disc about an axis perpendicular to the plane and passing through CG:\begin{align*}I=\frac{MR^{2}}{2}\end{align*}
MI about an axis along the plane passing through the C.G:\begin{align*}I_{d}=\frac{MR^{2}}{4}\end{align*}
MCQs
No MCQs found.
Subjective Questions
No subjective questions found.
Videos
No videos found.

Moment of Inertia of a Circular Disc
Moment of Inertia of a Circular Disc:
1) About an axis perpendicular to the plane and passing through CG (or centre)-(o to R):
Consider a circular disc of mass ‘M’ and radius ‘R’. Let us take an axis AB perpendicular to the plane of the disc and passing through CG, about which moment of inertia is to be determined. Let us divide the disc into large no. of elementary concentric circular rings. Consider any one of such ring of radius x and thickness dx.
_.jpg)
Here, Mass of the disc= M
Mass per unit surface area of the disc= \(\begin{align*}\frac{M}{\pi R^{2}}\end{align*}\)
Mass of the elementary ring, m=\(\begin{align*}\frac{M}{\pi R^{2}}2\pi xdx\end{align*}\)\begin{align*}=\frac{2M}{R^{2}}xdx\end{align*}
Now, the moment of inertia of the elementary ring is given by,
\begin{align*}dI=mass\times (radius)^{2}\end{align*}\begin{align*}=\frac{2M}{R^{2}}x^{3}dx\end{align*}
Since, such elementary rings are varying from x= 0 to x= R total M.I of the disc is given by,
\begin{align*}I=\int dI\end{align*}\begin{align*}=\int_{0}^{R}\frac{2M}{R^{2}}x^{3}dx\end{align*}\begin{align*}I=\frac{MR^{2}}{2}\end{align*}
2) About an axis along the plane passing through the CG (i.e. about a diameter):
To determine the M.I. of a circular disc about the diameter, we have to apply the theorem of perpendicular axis.
Let us consider two diameters XOX’ and YOY’ perpendicular to each other, about which the M.I is Id. Also, consider an axis AB perpendicular to the plane of the disc and passing through its centre.
_.jpg)
Then by applying theorem of perpendicular axis, we can write,
\begin{align*}I_{AB}=I_{XOX'}+I_{YOY'}\end{align*}\begin{align*}I_{d}=\frac{MR^{2}}{4}\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
Lesson
Dynamics of Rigid Bodies
Subject
Physics
Grade
Bachelor of Science
Recent Notes
No recent notes.
Related Notes
No related notes.