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.

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\(R^{2}\)

@Zahid Shekh Mohammed
–
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.

@Zahid Shekh Mohammed
–
Ip=I (com) + Md^2 , p= moment of inertia of axis through p , parallel to centre of mass and at a distance d from centre of mass. Icom = M/6(R)^2 for rectangular sheet of breadth R
Therefore , Taking p is the centre of disk and axis parallel to centre of mass of rectangular sheet axis, We get Ip=M/6R^2 + M(3R/2)^2 {AS DISTANCE BETWEEN COM OF RECTANGLE AND CIRCLE IS 3R/2}
Therefore, ip = 29Mr^2/12 Adding this to Mr^/4 , we get 32mr^2/12

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## Comments

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TopNewestplease specify how it is attached as it would be of great help in solving the problem...

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the disc and rectangular are attached about their ends

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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\(R^{2}\)Log in to reply

Is it attached about its length or width ?

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about length

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but answer is 31MRsqaure\12

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No actually breadth is R and length is 2R

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Therefore, ip = 29Mr^2/12 Adding this to Mr^/4 , we get 32mr^2/12

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