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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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
–
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

@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.

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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

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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Is it attached about its length or width ?

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about length

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