A 0.60-kgkg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.3 mm on a frictionless horizontal surface. Part A If the cord will break when the tension in it exceeds 60 NN , what is the maximum speed the ball can have

Respuesta :

11.4 m/s

Explanation:

The cord will break when the centripetal force exerted on it meets or exceeds the maximum tension [tex]T_{max}[/tex] that it can handle.

[tex]T_{max} = m\dfrac{v_{max}^2}{r}[/tex]

Solving for [tex]v_{max},[/tex] we get

[tex]v_{max}^2 = \dfrac{rT_{max}}{m}[/tex]

or

[tex]v_{max} = \sqrt{\dfrac{rT_{max}}{m}} =\sqrt{\dfrac{(1.3\:\text{m})(60\:\text{N})}{(0.6\:\text{kg})}}[/tex]

[tex]\:\:\:\:\:= 11.4\:\text{m/s}[/tex]