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

Hint: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 I_cut. 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
I_cm 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.