Q.3 Obtain an expression for moment of inertia of a uniform disc of mass ‘M’ and radius ‘R’ rotating about an axis passing through its centre and perpendicular to its plane.
Best answer
Answer: \(I = \dfrac{1}{2}MR^2\)
Mass per unit area of the disc: \(\sigma = \dfrac{M}{\pi R^2}\). Divide the disc into thin concentric rings. Consider a ring of radius \(x\) and width \(dx\).
Area of the ring \(= 2\pi x\,dx\); its mass \(dm = \sigma\,2\pi x\,dx = \dfrac{2M}{R^2}x\,dx\).
All of this mass is at distance \(x\) from the axis, so its moment of inertia is \(dI = dm\,x^2 = \dfrac{2M}{R^2}x^3\,dx\).
\[I = \int_0^R \frac{2M}{R^2}x^3\,dx = \frac{2M}{R^2}\left[\frac{x^4}{4}\right]_0^R = \frac{2M}{R^2}\cdot\frac{R^4}{4} = \frac{1}{2}MR^2\]Log in or create a free account to answer this question.
