Question

A very thin, uniform disc is made by removing a small disc of diameter \(2R\) from a bigger (thin, uniform) disc of uniform mass density, mass \(M\), and radius \(2R\), as shown in the figure. The M.I. of this remaining part about an axis passing through point O and perpendicular to the plane of the figure is

• A. \(\frac{9MR^{2}}{8}\)
• B. \(\frac{11MR^{2}}{9}\)
• C. \(\frac{13MR^{2}}{8}\)
• D. \(\frac{15MR^{2}}{9}\)

Hint:

Subtract the moment of inertia of the removed disc, found with the parallel-axis theorem, from that of the full disc.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Remaining part = full disc minus the removed disc, both taken about the axis through O perpendicular to the plane.

Step 2: Full disc
Mass \(M\), radius \(2R\): \(I_1=\dfrac{1}{2}M(2R)^2=2MR^2\).

Step 3: Removed disc
Radius \(R\) means its mass is \(\dfrac{M}{4}\), and its centre is at distance \(R\) from O.
By the parallel-axis theorem: \(I_2=\dfrac{1}{2}\dfrac{M}{4}R^2+\dfrac{M}{4}R^2=\dfrac{3MR^2}{8}\).

Step 4: Result
\(I=I_1-I_2=2MR^2-\dfrac{3}{8}MR^2=\dfrac{13MR^2}{8}\).

Final Answer:
The moment of inertia of the remaining part is 13MR squared divided by 8. \[ \boxed{\text{(C) }\dfrac{13MR^{2}}{8}} \]