CBSE Class 12-science Answered
pls explain
![question image](http://images.topperlearning.com/topper/new-ate/top_mob158719995723806994569dc5e7d-b275-45d8-96de-32edd1cb02c5.jpg)
Asked by mufeedatvp2000 | 18 Apr, 2020, 14:22: PM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/b7ef4a15919154b30998bbb760589b6b5e9bab480117a2.22184393ring1.png)
Let us find the centre of mass of semi circular ring of radius R .
From the given figure above, if we consider origin of X-Y coordinate system is at center of semi-circular ring,
then centre of mass is in y-axis by symmetry, i.e. x-coordinate Cx of centre of mass is zero.
y-coordinate Cy of centre of mass is given by
![begin mathsize 14px style C subscript y space equals space 1 over M integral d m space y space equals space 1 over M integral subscript 0 superscript theta rho space R d theta space R sin theta space equals fraction numerator rho space R squared over denominator M end fraction space integral subscript 0 superscript theta sin theta space d theta space equals space fraction numerator 2 space rho space R squared over denominator M end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/a8d1ac579cd17059d5a0962251c78101.png)
where M is mass of semi-circular ring, dm is mass of a small element of length dl .
Let radial line connecting the element dl makes angle θ with x-axis and dθ be angle subtended by element dl at centre.
ρ is mass per unit length of wire. Mass M of wire is obtained as M = ρ ( π R )
Using the substitution ( ρ/M ) = 1/( π R ) , equation (1) for y-coordinate of centre of mass becomes
Cy = 2R/π
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/7c57b8813e729db341e032f27487cedf5e9bb31087a5b6.00269947ring2.png)
Moment of inertia I about centre of mass is given by
![begin mathsize 14px style I space equals space integral d m space r squared space equals space integral subscript 0 superscript pi R space d theta space rho space open curly brackets R squared cos squared theta space plus space open parentheses R space sin theta space minus space fraction numerator 2 R over denominator pi end fraction close parentheses squared close curly brackets end style](https://images.topperlearning.com/topper/tinymce/cache/3156235c1cb8464d08406a8863cd5673.png)
we get from above integration, ![begin mathsize 14px style I space equals space M space open curly brackets 1 space minus space 4 over pi squared close curly brackets R squared end style](https://images.topperlearning.com/topper/tinymce/cache/c3d17bfaf17367e90d9e67f3132f96e7.png)
![begin mathsize 14px style I space equals space M space open curly brackets 1 space minus space 4 over pi squared close curly brackets R squared end style](https://images.topperlearning.com/topper/tinymce/cache/c3d17bfaf17367e90d9e67f3132f96e7.png)
Answered by Thiyagarajan K | 19 Apr, 2020, 07:54: AM
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