# Rotational Mechanics question

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.

Note by Zahid Shekh Mohammed
5 years, 2 months ago

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

- 5 years, 2 months ago

the disc and rectangular are attached about their ends

- 5 years, 2 months ago

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$$R^{2}$$

- 5 years, 2 months ago

Is it attached about its length or width ?

- 5 years, 2 months ago

- 5 years, 2 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.

- 5 years, 2 months ago

Can I email the photo of question please give your email

- 5 years, 2 months ago

but answer is 31MRsqaure\12

- 5 years, 2 months ago

No actually breadth is R and length is 2R

- 5 years, 2 months ago

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

- 4 years, 2 months ago