Question

Seven identical discs are arranged in a planar pattern, so as to touch each other as shown in the figure. Each disc has mass ‘m’ radius R. What is the moment of inertia of system of six discs about an axis passing through the centre of central disc and normal to plane of all discs?

• A. 100 mR²
• B. \(55\ \frac{mR^2}{2}\)
• C. \(85\ \frac{mR^2}{2}\)
• D. 100 mR²

Answer 1

The Correct Option is B

Given: 

• Mass of each disc: \( m \)
• Radius of each disc: \( R \)
• Six surrounding discs with a central disc

Step 1: Moment of Inertia of a Single Disc

The moment of inertia of a disc about its central axis is:

\[ I_c = \frac{1}{2} m R^2 \]

Step 2: Moment of Inertia of Six Outer Discs

Each outer disc is at a distance \( r = 2R \) from the central axis. Using the parallel axis theorem:

\[ I = I_c + m r^2 \]

\[ I_{ ext{outer}} = \frac{1}{2} m R^2 + m (2R)^2 \]

\[ I_{ ext{outer}} = \frac{1}{2} m R^2 + 4 m R^2 \]

\[ I_{ ext{outer}} = \frac{1}{2} m R^2 + 4 m R^2 = \frac{9}{2} m R^2 \]

Step 3: Total Moment of Inertia

Since there are 6 outer discs:

\[ I_{ ext{total}} = I_{ ext{central}} + 6 I_{ ext{outer}} \]

\[ I_{ ext{total}} = \frac{1}{2} m R^2 + 6 \times \frac{9}{2} m R^2 \]

\[ I_{ ext{total}} = \frac{1}{2} m R^2 + \frac{54}{2} m R^2 \]

\[ I_{ ext{total}} = \frac{55}{2} m R^2 \]

Answer: The moment of inertia of the system is \( \frac{55}{2} m R^2 \) (Option B).