InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A particle is moving in a circle of radius R in such a way that any insant the total acceleration makes an angle of `45^(@)` with radius. Initial speed of particle is `v_(0)`. The time taken to complete the first revolution is:A. `(2R)/(v_(0))(1-e^(-2pi))`B. `(R)/(v_(0))(1-e^(-2pi))`C. `R/(v_(0))`D. `(2R)/(v_(0))` |
|
Answer» Correct Answer - b Total acceleration makes an anlge of `45^(@)` with radius i.e., tangential acceleration =radial acceleration. `R alpha=R omega^(2)` or `alpha=omega^(2)` or `(d omega)/(dt)=omega^(2)` or `(d omega)/(omega^(2))=dt` or `int_(omega_(0))^(omega) (domega)/(omega^(2))=int_(0)^(t)dt` `omega=(omega_(0))/(1-omega_(0)t)` or `(d theta)/(dt)=(omega_(0))/(1-omega_(0)t)` `int_(0)^(2pi)d theta=int_(0)^(t) (omega_(0)dt)/(1-omega_(0)t)` or `t= 1/(omega_(0))(1-e^(-2pi))` `=R/(v_(0))(1-e^(-2pi))` |
|