A 0.55- ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.5 on a frictionless horizontal surface.If the cord will break when the tension in it exceeds 95 , what is the maximum speed the ball can have?

Answer 1

T= m(v^2/r)
95 N = 0.55(v^2/1.5)
95 N = (0.55v^2)/1.5
(95 N)(1.5) = 0.55v^2
142.5 = 0.55v^2
divide both sides by 0.55
answer is : 259.09 = v^2
take the square root of 259.09
final answer is : v = 16.1 m/s = the maximum speed the ball can have