Question

The moment of inertia of a uniform semicircular disc of mass M and radius R about a line perpendicular to the plane of disc and passing through the center is:
A.$\dfrac{{M{R^4}}}{4}$
B.$\dfrac{2}{5}M{R^2}$
C.$M{R^2}$
D.$\dfrac{{M{R^2}}}{2}$

Answer 1

Moment of inertia is defined as the product of the mass of section and the square of the distance between the reference axis and the centroid of the section.
The approach for solving this question should be to find the moment of inertia of a circular disc first, and then the moment of inertia of a semicircular disc is found.

Complete Step by Step Answer:
Mass of disc= M and the Radius= R
Basically, the moment of inertia for a rotating object is calculated by taking the distance of each particle from the axis of rotation and squaring that value and multiplying it with the mass of that particle.

The moment of inertia of a circular disc of mass 2M will be equal to

$${I_{circular disc}} = \dfrac{{M{R^2}}}{2} \\\ = \dfrac{{2M{R^2}}}{2} \\\ = M{R^2} \\\$$

So the moment of inertia of a circular disc will be$= M{R^2}$
The moment of inertia of a semicircular disc will be half of the moment of inertia of a circular disc, therefore,

$${I_{semicircular disc}} = \dfrac{1}{2}{I_{circular disc}} \\\ = \dfrac{1}{2}M{R^2} \\\$$

Hence the moment of inertia of a semicircular disc $= \dfrac{1}{2}M{R^2}$
Option (D) is correct.

Note: Moment of inertia, also known as the mass moment of inertia, angular mass of a rigid body is a quantity that determines the torque needed for angular acceleration about a rotational axis.