A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t = 0 is 4 m s - 1 , the time taken to complete the first revolution will be 1 α 1 - e - 2 π s , where α = ______.

Step-by-step solution

Given: a → c = a → t ⇒ v 2 r = d v d t ⇒ ∫ 4 v d v v 2 = ∫ 0 t d t r ⇒ - 1 v 4 v = t r ⇒ - 1 v + 1 4 = 2 t ⇒ v = 4 1 - 8 t = d s d t ⇒ 4 ∫ 0 t d t 1 - 8 t = ∫ 0 s d s r = 0 . 5 m , distance covered s = 2 π r = π ⇒ 4 × [ ln ( 1 - 8 t ) ] 0 t - 8 = π ⇒ ℓ n ( 1 - 8 t ) = - 2 π ⇒ 1 - 8 t = e - 2 π ⇒ t = 1 - e - 2 π 1 8 s So, α = 8 .