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*}

 

 

 

 

 

 

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Moment of Inertia of a Circular Disc

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.

Fig: Axis perpendicular to the plane and passing through C.G
Fig: Axis perpendicular to the plane and passing through C.G

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.

Fig: Axis along the plane passing through the C.G
Fig: Axis along the plane passing through the C.G

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

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