InterviewSolution
Saved Bookmarks
| 1. |
A thin uniform circular ring is rolling down an inclined plane of inclination 30° without slipping. Its linearacceleration along the inclined plane is2(3) 93 |
|
Answer» Acceleration on an inclined plane a= g sinθ/ (1 + I/MR²) for circular rings: I= MR² so by putting the value in above equation we get a= g sinθ/ 2 a = g sin30/2 a= g/4 |
|