How Does Halving the Radius Affect the Period in Centripetal Acceleration?

[smillphysics]

What happens to the period when you cut the radius in half for a centripetal acceleration problem?


a=V^2/r
T=2pi*r/v


I need some background on this question. I believe the answer is the period is decreased by a factor of 2. I am just slightly confused.

 

[RoyalCat]

Well, that would depend. Are you keeping the acceleration on the body the same?
If so, then the problem isn't quite as trivial:
a_initial = v²/r_initial
a_final = u²/r_final
T_initial = 2πr/v
T_final = 2π(½r)/u

Since we want the acceleration before we cut the radius in half to be the same as after, all that remains to find u and T_final is a simple equation.