# calculating mi of ring whose axis passes through its diameter.please show thw steps.

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### 1 Answer

Let us take a random plane shape (as in the first figure below) having 2 axis Ox1 and Ox2 in the plane of the figure and the Ox3 axis perpendicular to the plane of the figure (perpendicular to the paper).

For the moments of inertia of the shape in the figure about axis Ox1 and Ox2 we can write

`I_1 =int(y^2*dm)`

`I_2 =int(x^2*dm)`

As said, the third axis Ox3 is perpendicular to the plane of the figure, therefore

`I_3 = int(r^2*dm) =int(x^2+y^2)*dm = I_1+I_2`

Thus for a **circular crown** of mass m and radius R1 and R2 (see the second figure) we have

`I_1=I_2= (1/2)*I_3`

`I_3 =int(x^2+y^2)*dm`

and if we write

`sigma = m/S`

we have

`I_3 =int_(R1)^(R2)(sigma*r^2*2pir*dr) =(pi*sigma/2)*(R2^4-R1^4) =`

`=(1/2)*m*(R1^2+R2^2)`

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**For a ring** of radius R (=R1=R2) we have

`I_3 =I_z =m*R^2`

`I_1=I_2 (=I_x=I_y) =(1/2)*I_3 =(m*R^2)/2`

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