A semicircular plate of mass \(m\) has radius \(r\) and centre \(c\). The centre of mass of the plate is at…

A semicircular plate of mass \(m\) has radius \(r\) and centre \(c\). The centre of mass of the plate is at a distance \(x\) from its centre \(c\). Its moment of inertia about an axis passing through its centre of mass and perpendicular to its plane is
  1. \(\frac{m r^2}{2}\)
  2. \(\frac{m r^2}{4}\)
  3. \(\frac{m r^2}{2}+m x^2\)
  4. \(\frac{m r^2}{2}-m x^2\)

Solution

Given, mass of a semicircular plate \(=m\) radius of semicircular plate \(=r\) According to the question we can drawn the following diagram,
Now, from the parallel axis theorem, \(\begin{aligned} I_c & =I_{\mathrm{cm}}+m x^2 \\ I_{\mathrm{cm}} & =I_c-m x^2 \\ \text{Here, } I_c & =\frac{m r^2}{2} \end{aligned}\) Hence, moment of inertia of semicircular plate about an axis passing through its centre of mass and perpendicular to it's plane is \(I_{\mathrm{cm}}=\frac{m r^2}{2}-m x^2\)

Asked in: AP EAMCET 2019 (20 Apr Shift 1)

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