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.

 

[nlightner]

I can confirm that the period will indeed decrease by a factor of 2 when the radius is cut in half for a centripetal acceleration problem. This can be seen by rearranging the equation for the period (T=2pi*r/v) to solve for radius (r=T*v/2pi) and plugging it into the equation for centripetal acceleration (a=V^2/r). When the radius is halved, the value in the denominator decreases, resulting in a larger value for acceleration. This means that the object will complete one full revolution in a shorter amount of time, thus decreasing the period by a factor of 2. This relationship can also be seen by examining the units for acceleration, which are meters per second squared (m/s^2), and the units for period, which are seconds (s). Since acceleration is a measure of how quickly velocity changes over time, a larger acceleration will result in a shorter period. I hope this explanation helps to clarify any confusion.