Question

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)` is
A. `(a)/(R)`
B. `a^(2)R`
C. `R(a^(2)+b)`
D. `Rsqrt(a^(4)+b^(2))`

Answer 1

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))`