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. ${\displaystyle \frac{3M{R}^2}{12}}$
B. ${\displaystyle \frac{5M{R}^2}{16}}$
C. ${\displaystyle \frac{11M{R}^2}{64}}$
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,
 $$I_{Disc}={\displaystyle \frac{M{R}^2}{4}}$$
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,
$$Mass\alpha Area$$
Now,
Area of disc of radius R is $\pi R^2$
Area of disc of radius ${\displaystyle \frac{R}{2}}$ is ${\displaystyle \frac{\pi R^2}{4}}$
Since the area of the disc is one – fourth, the mass of cut out region will be equal to ${\displaystyle \frac{M}{4}}$
Moment of inertia of the uncut (original) disc about the axis passing through its centre and perpendicular to its plane will be given by,
$$I_{Original}={\displaystyle \frac{M{R}^2}{4}}$$
Moment of inertia of a ring of radius ${\displaystyle \frac{R}{2}}$about an axis at a distance of ${\displaystyle \frac{R}{2}}$from its centre and perpendicular to its plane is given by,
$$I_{cut}=({\displaystyle \frac{M}{4}}){\displaystyle \frac{{({\displaystyle \frac{R}{2}})}^2}{4}}+({\displaystyle \frac{M}{4}})({\displaystyle \frac{R}{2}}{)}^2$$
$$I_{cut}={\displaystyle \frac{M}{4}}\times {\displaystyle \frac{R^2}{16}}+{\displaystyle \frac{M}{4}}\times {\displaystyle \frac{R^2}{4}}$$
$$I_{cut}=M{R}^2({\displaystyle \frac{1}{64}}+{\displaystyle \frac{1}{16}})$$
$$I_{cut}={\displaystyle \frac{5}{64}}M{R}^2$$
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,
$$I_{final}={\displaystyle \frac{M{R}^2}{4}}-{\displaystyle \frac{5M{R}^2}{64}}$$
$$I_{final}={\displaystyle \frac{(16-5)M{R}^2}{64}}$$
$$I_{final}={\displaystyle \frac{11M{R}^2}{64}}$$

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.