JEE Class main Answered
moment of Inertia
![question image](http://images.topperlearning.com/topper/new-ate/top_mob1673280684564809254IMG_20230109_214001.jpg)
Asked by harshveer262006 | 09 Jan, 2023, 21:41: PM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/dd8dac7b0793e2b67f2061924ee2c98463bc5b3d3c1849.79633440f4.png)
Fig.(1) shows the moment of inertia of disk of radius R with respect to the rotation axis perpendicular to the
plane of disk and passing through centre of mass.
Let the axis of rotation in fig.(1) coincide with z-axis of cartesian coordinate system.
Then ![begin mathsize 14px style I subscript z space equals space 1 half M space R squared end style](https://images.topperlearning.com/topper/tinymce/cache/34d24e3034cc98728ac811176dce0913.png)
![begin mathsize 14px style I subscript z space equals space 1 half M space R squared end style](https://images.topperlearning.com/topper/tinymce/cache/34d24e3034cc98728ac811176dce0913.png)
where M is mass of disc .
If our axis of rotation is either x-axis or y-axis as shown in fig.(2) ,
by perpendicular axis theorem
![begin mathsize 14px style I subscript z equals space I subscript x space plus space I subscript y end style](https://images.topperlearning.com/topper/tinymce/cache/0aeb099b5003693d7d7212dd8dbcbefc.png)
By symmetry Ix = Iy
Hence Iy = (1/2) Iz = (1/2) (1/2) M R2 = (1/4) M R2
If axis of rotation is shifted to the edge of disc as shown in figure ,
Then by parallel axis theorem ,
![begin mathsize 14px style I space equals space I subscript y space plus space M space R squared space equals space M R to the power of 2 to the power of space end exponent left parenthesis space 1 space plus space 1 fourth space right parenthesis space space equals space 5 over 4 M space R squared end style](https://images.topperlearning.com/topper/tinymce/cache/71faff108bb5e3f16a794924b6fc757c.png)
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