Question

I'm swinging a pocket watch on its chain in a circular fashion. Because i have a crappy watch, the chain breaks exactly when the watch is traveling directly upward. If the chain, before breaking was 25.5 centimeters long, and I was swinging it a constant rate of 4 revolution per second, how high in the air (above the ground) will the watch go? (My hand was 0.8 meters above the ground when the chain broke)

Answer 1

v = rw

w = 4 rev/s = 25.13 rad/s

r = 25.5 cm = 0.255 m

v = wr = 6.408

s =v^2/2g

s = 2.095 m

max height = 0.8 + 2.095

hmax = 2.895 m

Answer 2

SOLUTION :

Angular acceleration = 4 * 2pi = 25.1327 rad/s

Initial velocity at breaking of chain upward = r * omega = 25.5 * 25.1327 = 640.88 cm/s = 6.4088 m/s

acceleration =  g  = 9.8   m/s^2 (downward)

Maximum height attained 

= 0.8  +  v^2 / 2g  (K. E. converted into P.E)

= 0.8 + (6.4088)^2 / (2*9.8) 

= 2.8955 m 

= 2.90 m approx. (ANSWER)\