Question

A 0.55-kg ball, attached to the end of a horizontal cord, is revolved in a circle of radius 1.3 m on a frictionless horizontal surface. if the cord will break when the tension in it exceeds 75 n, what is the maximum speed the ball can have?

Answer 1

The tension of the cord is the centripetal force that keeps the ball in circular motion:
$T=F_c = m \frac{v^2}{r}$
where T is the tension of the cord, $F_c$ is the centripetal force, $m=0.55 Kg$ is the mass of the ball, v its speed and $r=1.3 m$ is the radius of the circle.

The maximum allowed tension is T=75 N, before the cord breaks. Using this value inside the formula, we can find which is the maximum allowed value fot the speed v:
$v= \sqrt{ \frac{T r}{m} }= \sqrt{ \frac{(75 N)(1.3 m)}{(0.55 kg)} }=13.3 m/s$