Suppose the circular track in the figure below has a radius of 106 m and the runner has a speed of 4.9 m/s.
(a) What is the period of the motion?
(b) If the radius of the track were reduced to 50 m and the runner maintained this speed, by what factor would the runner's centripetal acceleration change?

Question

Suppose the circular track in the figure below has a radius of 106 m and the runner has a speed of 4.9 m/s.
(a) What is the period of the motion?
(b) If the radius of the track were reduced to 50 m and the runner maintained this speed, by what factor would the runner's centripetal acceleration change?

anonymous · Answer

If  r = 106 m and v = 4.9 m/s

So 
2pi/T = v/r
T = 2pi r/v
T = (2 x 3.14 x 106)/4.9
T = 135.85 sec

anonymous · Answer

that's the period? D:
omg thanks so much :D

anonymous · Answer

yes..., ur welcome

anonymous · Answer

good luck @SuperPhatty101  for the number b

anonymous · Answer

thanks :D

anonymous · Answer

Hei @SuperPhatty101 use the formula 

a = v^2/r 

for the answer "b"

anonymous · Answer

yeah I got it, thanks! ^_^

anonymous · Answer

nice, ur welcome :)