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Derive the expression for moment of iner...

Derive the expression for moment of inertia of a circular disc
about an axis passing through its centre and perpendicular to its plane

Text Solution

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Consider a circular disc of mass M and radius R.
Area of circular disc = `piR^2`
Mass per unit area of disc = `M/(piR^2)`
Now, let us consider a ring of radius x and thickness dx.
Then, Area of the ring `= 2pix dx`
Mass of the ring `=M/(piR^2)xx2pixxdx = (2M)/(R^2)x dx`
Moment of inertia of the ring about ZOZ axis
= Mass `xx" (radius)"^2 = (2M)/(R^2)xd"x" xx x^(2) = (2M)/(R^2)x^3 dx`
`:.` Moment of inertia of the disc about ZoZ axis
`I = int_(0)^(2)(2M)/R^2x^2dx =(2M)/R^2[x^4/4]_0^R`
`I=(2M)/(R^2)xxR^2/4=(MR^2)/2 implies I = (MR^2)/2`
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