Question

Find out the moment of inertia of a semicircular disc about an axis passing through its centre of mass and perpendicular to the plane? (mass=M and radius=R)

Answer 1

Hint: Find the moment of inertia of the semicircular disc about the axis passing through the centre of its base and perpendicular to the plane. Then use the parallel axis theorem to find the moment of inertia about the axis passing through its centre of mass and perpendicular to its plane.

Complete step by step answer:
Let us first find the moment of inertia of the semi-circular disc about an axis passing through its centre and perpendicular to the plane. Then we will use the parallel axis theorem to find the moment of inertia about an axis passing through its centre of mass and perpendicular to the plane. Let the moment of inertia about its centre be I. Now, imagine that you joined another semi-circular disc of the same mass and size and completed the disc. The moment of inertia of this imaginary disc will also be I. Therefore, the moment of inertia of the complete will be 2I. But we already know the moment of inertia of the complete circular disc about the same axis i.e. $\dfrac{M{{R}^{2}}}{2}$.
Therefore, $2I=\dfrac{M{{R}^{2}}}{2}\Rightarrow I=\dfrac{M{{R}^{2}}}{4}$
We found the moment of inertia about the centre of the disc. Now, use the parallel axis theorem to find the moment of inertia about its centre of mass. According to the parallel axis theorem, $I={{I}_{c}}+M{{d}^{2}}$ ….(1), where I is the moment of inertia about any axis parallel to the axis passing through the centre of mass, Ic is the moment of inertia about its centre of mass and d is the perpendicular distance between the two axes.
Here, $I=\dfrac{M{{R}^{2}}}{4}$ , $d=\dfrac{4R}{3\pi }$ (the distance between the centre and centre of mass of the semicircular disc) and we have to find ${{I}_{c}}$.
Substitute the values of I and d in equation (1).
$\Rightarrow \dfrac{M{{R}^{2}}}{4}={{I}_{c}}+M\left( \dfrac{4R}{3\pi } \right)$
$\Rightarrow {{I}_{c}}=M\left( \dfrac{16{{R}^{2}}}{9{{\pi }^{2}}} \right)-\dfrac{M{{R}^{2}}}{4}=\dfrac{\left( 64-9{{\pi }^{2}} \right)M{{R}^{2}}}{36{{\pi }^{2}}}$
Therefore, the moment of inertia of a semi-circular disc about an axis passing through its centre of mass and perpendicular to the plane has been found.

Note: Students may make mistakes in using the parallel axis theorem. It is used to find the moment of inertia of a body about an axis parallel to the axis passing through the centre of mass of the body by knowing the perpendicular distance between the two axes. Therefore, it is must that one of the axes be the axis passing through the centre of mass otherwise, we cannot use the theorem.