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A particle moves along a circle if radius (20 //pi) m with constant tangential acceleration. If the velocity of the particle is ` 80 m//s` at the end of the second revolution after motion has begun the tangential acceleration is .A. `40 m//s^(2)`B. `20 m//s^(2)`C. `10 m//s^(2)`D. `5 m//s^(2)` |
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Answer» Correct Answer - A `r=20/(pi)m, a_(1)` =cosntant `n=2^(nd)`=revolution v=80 m/s `omega_(0)=0, omega_(f)=v/r=80/(20//pi)=4pi ` rad/sec `theta=2pixx2=4pi` from `3^(rd)` `omega^(2)=omega_(0)^(2)+2 alpha theta` `rArr (4pi^(2))=0^(2)+2xxalphaxx(4pi)` `alpha=2pi rad//s^(2)` `a_(1)=alphar=2pixx20/(pi)` `=40 m//s^(2)` |
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