Before You Begin
Question #43637
The Question
A uniform thin spherical shell of mass and radius is free to rotate about a fixed vertical axis passing through its center . A particle of mass is projected from infinity in the horizontal equatorial plane of the shell, with an initial speed and an impact parameter . The particle enters the shell through a small hole. Inside the shell, there is a rigid diametric spoke of negligible mass lying in the equatorial plane. The spoke is oriented such that the particle strikes it exactly at the particle's point of closest approach to the center , and gets embedded in it. Assuming the shell and spoke are initially at rest, and considering only the gravitational interaction between the particle and the shell, find the final angular velocity of the system.
Options
- A.
- B.
- C.
- D.
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Intuition_Breaker
Train your brain to resist "obvious" answers that are designed to deceive your intuition.
Concepts Behind This Question
Impact Parameter Geometry|Impulsive Torque and Angular Momentum|Shell Theorem
Why This Question Is Tricky
Common Mistakes
The question tests the student's ability to compartmentalize physical regimes using Newton's Shell Theorem.
The lethal trap lies in calculating the distance of closest approach.
Students who misapply the physical principles will assume the shell acts as a point mass throughout the entire trajectory, leading them to calculate the periapsis of a Keplerian hyperbola ().
This yields an incorrect final moment of inertia and angular velocity (Option B).
The correct approach recognizes that once the particle enters the shell, the gravitational field vanishes, transforming the trajectory from a hyperbola to a straight line, fundamentally altering the point of closest approach ().
Quick Facts About This Question
Question Information
| Subject | Physics |
| Chapter | Atoms - Class 12 |
| Concept | Impact Parameter Geometry |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |