Question

Disk \( m_1 = 5 \, \text{kg} \) \& radius \( r_1 = 10 \, \text{cm} \) and disk \( m_2 = 10 \, \text{kg} \) \& radius \( r_2 = 50 \, \text{cm} \) are arranged as shown in the figure. Find the moment of inertia about an axis through the common tangent and parallel to the plane of the disks.

• A. \( \frac{31}{8} \, \text{kg} \, \text{m}^2 \)
• B. \( \frac{57}{64} \, \text{kg} \, \text{m}^2 \)
• C. \( \frac{41}{8} \, \text{kg} \, \text{m}^2 \)
• D. \( \frac{51}{16} \, \text{kg} \, \text{m}^2 \)

Hint:

To calculate the moment of inertia for an axis not passing through the center, use the parallel axis theorem to account for the offset distance.

Answer 1

The Correct Option is D

Step 1: Contribution of each disk
Each disk contributes to the total moment of inertia due to its own rotation about its center as well as due to its position relative to the given axis. The rotational inertia of a disk depends on its mass and radius.

Step 2: Effect of the axis position
The axis of rotation does not pass through the centers of the disks but is parallel to the central axes and tangential to them. Hence, the distance of each disk’s center from the axis must be taken into account while computing the total rotational effect.

The contribution from each disk is obtained by combining its rotational inertia about its center with the additional effect due to its displacement from the axis.

Step 3: Total moment of inertia
Adding the contributions of both disks about the given axis and simplifying, the total moment of inertia of the system is obtained.

Final Answer:
Total moment of inertia = 51 / 16 kg m²