Question

Given a uniform disc of mass M and radius R. A small disc of radius R/2 is cut from this disc in such a way that the distance between the centers of the two-disc is R/2. Find the moment of inertia of the remaining disc about a diameter of the original disc perpendicular to the line connecting the centers of the two discs.
A. $\dfrac{{3M{R^2}}}{{12}}$
B. $\dfrac{{5M{R^2}}}{{16}}$
C. $\dfrac{{11M{R^2}}}{{64}}$
D. None of these

Answer 1

Moment of inertia of an object is defined as its tendency to resist angular acceleration. Moment of inertia of a disc about its centre of mass is given as,
${I_{Disc}} = \dfrac{{M{R^2}}}{4}$
Where,
M is the mass of disc
R is the radius of disc.

Complete step by step solution:
It is given in the Question that the mass of a disc of radius R is equal to M.
We know the fact that for two dimensional objects,
$Mass{\text{ }}\alpha {\text{ }}Area$
Now,
Area of disc of radius R is $\pi {R^2}$
Area of disc of radius $\dfrac{R}{2}$ is $\dfrac{{\pi {R^2}}}{4}$
Since the area of the disc is one – fourth, the mass of cut out region will be equal to $\dfrac{M}{4}$
Moment of inertia of the uncut (original) disc about the axis passing through its centre and perpendicular to its plane will be given by,
${I_{Original}} = \dfrac{{M{R^2}}}{4}$
Moment of inertia of a ring of radius $\dfrac{R}{2}$about an axis at a distance of $\dfrac{R}{2}$from its centre and perpendicular to its plane is given by,
${I_{cut}} = (\dfrac{M}{4})\dfrac{{{{(\dfrac{R}{2})}^2}}}{4} + (\dfrac{M}{4}){(\dfrac{R}{2})^2}$
${I_{cut}} = \dfrac{M}{4} \times \dfrac{{{R^2}}}{{16}} + \dfrac{M}{4} \times \dfrac{{{R^2}}}{4}$
${I_{cut}} = M{R^2}(\dfrac{1}{{64}} + \dfrac{1}{{16}})$
${I_{cut}} = \dfrac{5}{{64}}M{R^2}$
Now we need to calculate the moment of inertia of the cut disc. For this, we will subtract Icut from Ioriginal (Since mass is being removed)
${I_{final}} = {I_{original}} - {I_{cut}}$
Inserting the values of Ioriginal and Icut in the above equation,
We get,
${I_{final}} = \dfrac{{M{R^2}}}{4} - \dfrac{{5M{R^2}}}{{64}}$
${I_{final}} = \dfrac{{(16 - 5)M{R^2}}}{{64}}$
${I_{final}} = \dfrac{{11M{R^2}}}{{64}}$

Hence Option (C) is correct.

Note: We have used a parallel axis theorem in order to find out the moment of inertia of Icut. According to parallel axis theorem,
$I = {I_{cm}} + M{D^2}$
Where,
I is the moment of inertia about its any axis parallel to the axis through the center of mass of the object
Icm is the moment of inertia of the object about its center of mass
M is the mass of object
D is the distance of center of mass from the arbitrary axis.