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: Moment of inertia for individual disks.
The moment of inertia for a disk about its center is given by \( I = \frac{1}{2} M R^2 \). The total moment of inertia about an axis through the common tangent is the sum of the moments of inertia of the individual disks and the correction due to the distance between their centers. Step 2: Apply the parallel axis theorem.
Using the parallel axis theorem for each disk, we add the contribution from the offset distance between the centers of the disks. Step 3: Conclusion.
After applying the formula and solving for the total moment of inertia, we find that the correct value is \( \frac{51}{16} \, \text{kg} \, \text{m}^2 \). Final Answer: \[ \boxed{\frac{51}{16} \, \text{kg} \, \text{m}^2} \]