InterviewSolution
Saved Bookmarks
| 1. |
Calculate the moment of inertia of a ring of mass `2kg` and radius `2cm` about an axis passing through its centre and perpendicular to its surface. |
|
Answer» `I=MR^(2)` `=(2xx2)/(100)xx(2)/(100)=8xx10^(-4)kg m^(2)` |
|