Question

A system consists of two identical spheres each of mass 1.5 kg and radius 50 cm at the ends of a light rod. The distance between the centres of the two spheres is 5 m. What will be the moment of inertia of the system about an axis perpendicular to the rod passing through its midpoint ?

• A. \(19.05\,\text{kg}\cdot\text{m}^2\)
• B. \(1.905 \times 10^5\,\text{kg}\cdot\text{m}^2\)
• C. \(18.75\,\text{kg}\cdot\text{m}^2\)
• D. \(1.875 \times 10^5\,\text{kg}\cdot\text{m}^2\)

Hint:

When spheres are far apart compared to their radius (here \(d=2.5\) vs \(r=0.5\)), the \(md^2\) term dominates. However, don't ignore the \(I_{cm}\) term unless specifically asked for "point masses".

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The total moment of inertia of a system is the sum of the moments of inertia of its individual components. For spheres shifted away from the axis, we use the Parallel Axis Theorem.
Step 2: Key Formula or Approach:
1. Moment of inertia of a solid sphere about its center: \(I_{cm} = \frac{2}{5} m r^2\).
2. Parallel axis theorem: \(I = I_{cm} + m d^2\).
Step 3: Detailed Explanation:
Given: - \(m = 1.5\,\text{kg}\), \(r = 0.5\,\text{m}\). - Separation between centers = 5 m. - Distance from axis (midpoint) to center of each sphere \(d = 2.5\,\text{m}\).
Moment of inertia for one sphere about the central axis: \[ I_{sphere} = \frac{2}{5} m r^2 + m d^2 = \frac{2}{5} (1.5) (0.5)^2 + (1.5) (2.5)^2 \] \[ I_{sphere} = (0.4) (1.5) (0.25) + (1.5) (6.25) = 0.15 + 9.375 = 9.525\,\text{kg}\cdot\text{m}^2 \] Total moment of inertia for both spheres: \[ I_{total} = 2 \times 9.525 = 19.05\,\text{kg}\cdot\text{m}^2 \]
Step 4: Final Answer:
The moment of inertia is \(19.05\,\text{kg}\cdot\text{m}^2\).