A disc of mass m and radius R is attached to a rectangular plate of the same mass breadth and length 2R . Find moment of inertia of this system about diameter of of disc.

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TopNewestplease specify how it is attached as it would be of great help in solving the problem... – Nishant Sharma · 4 years, 3 months ago

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– Zahid Shekh Mohammed · 4 years, 3 months ago

the disc and rectangular are attached about their endsLog in to reply

Hey you have mentioned that it is a \(\underline{rectangular}\) \(\underline{plate}\) of same mass, breadth and length 2R which actually contradicts that it is a rectangle. I have solved the problem taking it to be a square.

Here is how I got it: \(\I_{of disc about diameter}\)=\(\frac {1}{4}\)

M\(R^{2}\) and \(\I_{of square about diameter of plate}\)= \(\frac {4}{3}\)M\(R^{2}\) . So \(\I_{required}\)=\(\frac {19}{12}\)M\(R^{2}\) – Nishant Sharma · 4 years, 3 months agoLog in to reply

– Nishant Sharma · 4 years, 3 months ago

Is it attached about its length or width ?Log in to reply

– Zahid Shekh Mohammed · 4 years, 3 months ago

about lengthLog in to reply

– Nishant Sharma · 4 years, 3 months ago

Hey dear, please post complete question because incomplete ques. causes undesired anxiety. You have not mentioned about which diameter ? Please post the complete ques. once more.Log in to reply

– Zahid Shekh Mohammed · 4 years, 3 months ago

Can I email the photo of question please give your emailLog in to reply

– Zahid Shekh Mohammed · 4 years, 3 months ago

but answer is 31MRsqaure\12Log in to reply

– Zahid Shekh Mohammed · 4 years, 3 months ago

No actually breadth is R and length is 2RLog in to reply

Therefore, ip = 29Mr^2/12 Adding this to Mr^/4 , we get 32mr^2/12 – Devashish Mehta · 3 years, 3 months ago

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