Assuming the keys are moving in uniform circular motion!!!!![/B]
Keys with a combined mass of 0.100 kg are attached to a 0.25 m long string and swung in a circle in the vertical plane.
a)What is the slowest speed that the keys can swing and still maintain a circular path?
b)What is the tension in the string at the bottom of the circle?
Given: V =10m/s, R=25m, Fc = mv^2/r, Ff = UFn
a) Centripetal Force is Fg, so Fg = MV^2/r
So, mg = mV^2/r
0.1(9.8) = (0.1)(V^2/0.25)
Mass cancels out..
9.8 = V^2/0.25
So, (0.25)(9.8) = V^2
Therefore, V^2 = 2.45, so V = 1.56m/s
The slowest speed the keys can travel and still maintain circular motion is 1.56m/s.
b) b) Fnet = Ft - Fg or Ft = Fnet + Fg (assuming this is uniform circular motion!!)
Ft = mV^2/r + mg
Ft = [ (0.1)(1.56)^2 / 0.25 ] + (0.1)(9.8)
Ft = 0.97N
Therefore, assuming that the keys are moving in uniform circular motion, the Ft at the bottom of the circle is 0.97N.
Ok.. Really having a tough time with this one.
Been looking over it for too long and eyes are gone blurry.. any help is appreciated!!!