Question

A 0.40 kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.0 m on a frictionless horizontal surface. if the cord will break when the tension in it exceeds 64 n, what is the maximum speed the ball can have?

Answer 1

The centripetal force for such an arrangement can be found through the equation Fc = mv^2/r where m is the mass of the rotating object, v is that object's velocity, and r is the radius of rotation. In this case, we know that the maximum Fc that can be tolerated by the cord is 64N. Thus we set the equation up and solve for the value of v for which Fc = 64.
   64 = 0.4*(v^2)/1
 64/0.4 = 160 =
v^2 v = sqrt(160) = 12.65 m/s

   At any speed faster than 12.65 m/s, the cord will break.