A particle moves in a circular path of radius R with an angualr velocity `omega=a-bt`, where a and b are positive constants and t is time. The magnitude of the acceleration of the particle after time `(2a)/(b)` isA. `(a)/(R)`B. `a^(2)R`C. `R(a^(2)+b)`D. `Rsqrt(a^(4)+b^(2))`

A particle moves in a circular path of radius R with an angualr veloci

1.	A particle moves in a circular path of radius R with an angualr velocity `omega=a-bt`, where a and b are positive constants and t is time. The magnitude of the acceleration of the particle after time `(2a)/(b)` isA. `(a)/(R)`B. `a^(2)R`C. `R(a^(2)+b)`D. `Rsqrt(a^(4)+b^(2))`
Answer» Correct Answer - D `alpha=(domega)/(dt)=-b` `a_(t)=Ralpha=-Rb` At `t=(2a)/(b),omega=-a` `a_(n)=Romega^(2)=Ra^(2)` Now, `a=sqrt(a_(t)^(2)+a_(n)^(2))=Rsqrt(a^(4)+b^(2))`

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A particle moves in a circular path of radius R with an angualr velocity `omega=a-bt`, where a and b are positive constants and t is time. The magnitude of the acceleration of the particle after time `(2a)/(b)` isA. `(a)/(R)`B. `a^(2)R`C. `R(a^(2)+b)`D. `Rsqrt(a^(4)+b^(2))`

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