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).

Answer 2

The total moment of inertia (I_total) is the sum of the MoI of the central disc (I_central) and the MoI of the six outer discs (I_outer) about the specified axis.

I_total = I_central + I_outer

1. Moment of Inertia of the Central Disc (I_central):

The axis of rotation passes through the center of the central disc and is perpendicular to its plane. The MoI of a disc about such an axis is given by:

I_CM = (1/2) mR²

So, I_central = (1/2) mR²

2. Moment of Inertia of the Six Outer Discs (I_outer):

Consider one of the outer discs. Its center is at a distance 'd' from the axis of rotation (the center of the central disc). Since the discs are touching, the distance 'd' is the sum of the radius of the central disc and the radius of the outer disc:

d = R + R = 2R

We need to use the Parallel Axis Theorem to find the MoI of one outer disc about the axis passing through O. The theorem states:

I = I_CM + md²

Where I_CM is the MoI about the center of mass of the outer disc, which is (1/2)mR², and d = 2R.

I_{one_outer_disc} = (1/2) mR² + m(2R)²

I_{one_outer_disc} = (1/2) mR² + m(4R²)

I_{one_outer_disc} = (1/2) mR² + (8/2) mR²

I_{one_outer_disc} = (9/2) mR²

Since there are six identical outer discs, all at the same distance 'd' from the axis, the total MoI for the six outer discs is:

I_outer = 6 * I_{one_outer_disc}

I_outer = 6 * (9/2) mR²

I_outer = (54/2) mR² = 27 mR²

3. Total Moment of Inertia (I_total):

Now, sum the MoI of the central disc and the six outer discs:

I_total = I_central + I_outer

I_total = (1/2) mR² + 27 mR²

I_total = (1/2) mR² + (54/2) mR²

I_total = (55/2) mR²

Answer: Based on the likely intended question (MoI of all seven discs), the correct option is (B) 55 mR² / 2.