Answer:
twice as great as before
Explanation:
recall that centripetal acceleration is given by [tex]a =\frac{v^{2} }{r}[/tex]
let the original radius of the circular motion = r₁ and the reduced radius be =r₂
acceleration for original radius, a₁ = v² / r₁ ==> v²=a₁ r₁
acceleration for reduced radius, a₂ = v² / r₂ ==> v²=a₂ r₂
given that the speed remains the same, we can equate the two expressions above
a₁ r₁ = a₂ r₂ (rearranging)
a₂ = (r₁ / r₂) a₁
we are given that r₂ = 1/2 r₁ (substitute this into equation)
a₂ = (r₁ / r₂) a₁
a₂ = [r₁ / (1/2)r₁ ] a₁
a₂ = [r₁ / (1/2)r₁ ] a₁
a₂ = 2a₁
hence we can see that the acceleration of the path with the smaller raidus has an acceleration that is twice the original
