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A particle starts moving along a circle of radius `(20//pi)m` with constant tangential acceleration. If the velocity of the parthcle is `50 m//s` at the end of the second revolution after motion has began, the tangential acceleration in `m//s^(2)` is :A. `1.6`B. 4C. `15.6`D. `31.2` |
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Answer» Correct Answer - D As `v=omegar` `rArr" "50=omega((20)/(pi))` `rArr" "omega=(50pi)/(20)=(5pi)/(2)` `rArr" "(2pi)/(T)=(5pi)/(2)rArrT=(4)/(5)s` Given, `t=2T=(8)/(5)s` `therefore" "omega=alphat` `rArr" "(5pi)/(2)=alphaxx(8)/(5)rArr=(25pi)/(16)rads^(-2)` `therefore" "a_(t)=r(alpha)=((20)/(pi))((25pi)/(16))=31.2rads^(-2)` |
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