Question

A 0.40 kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.5 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 63 N, what is the maximum speed the ball can have?

Answer 1

The maximum speed the ball can have is 15.37 m/s

Step-by-step explanation:

Given;

mass of the ball, m = 0.4 kg

radius of the chord, r = 1.5 m

maximum tension on the chord, T = 63 N

The maximum tension on the chord is given by;

`T_(max) = (mv_(max)^2)/(r) \\\\v_(max)^2 = (T_(max) *r)/(m)\\\\ v_(max) = \sqrt{(T_(max) *r)/(m)} \\\\ v_(max) =\sqrt{(63 *1.5)/(0.4)}\\\\v_(max) = 15.37 \ m/s`

Therefore, the maximum speed the ball can have is 15.37 m/s