Question

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

Answer 1

Answer:

MR^2/2

Explanation:

as above in image

Answer 2

moment of inertia of disc about an axis passing through its centre and perpendicular to its plane is MR²/2

cut an ring of thickness dx and radius x as shown in figure.

let mass of disc is M.

so, the mass of ring, dm = M/πR² × 2πx dx

= 2Mx/R² dx

now moment of inertia of disc about an axis passing through its centre, ∫dI = (dm)x²

I = (2Mx/R² dx )x²

= 2M/R² ∫x³ dx

= 2M/R² × [x⁴/4]

= 2M/R² × R⁴/4

= MR²/2

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