Answer to Question #116393 in Mechanics | Relativity for Foibe Kambala

Question

3. A 0.45 kg ball attached to the end of horizontal cord is rotated in a circle of radius 1.3 m
on 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

When the ball is doing circular motion then it will have a centripetal acceleration which is equal to $\frac{v^2}{r}$

$T_{max}+mg=mv^2/r$

$75+0.45\times10=0.45\times v^2/1.3$

$0.45v^2=(75+4.5)1.3$

$v^2=229.67$

$v=15.15m/s$