Answer to Question #219581 in Mechanics | Relativity for Moyo

Question

A stone of mass 0.5kg attached to a string of length 50cm will break when the tension exceed 20N. The stone is whirled in a vertical circle, the axis of rotation at height 100cm above the ground. At what position and angular speed is the break most likely to occur.

Answer 1

$\vec T = \vec F_{c}+\vec F_{g}$

$\text{where:}$

$\vec{F}_{c}-\text{centrifugal force}$

$\vec{F}_{c}= m\vec a= m\omega^2r$

$\vec{F_g}- \text{gravity force}$

$\vec F_{g}= mg$

$\text{The break will happen when:}$

$\vec F_c+\vec F_g - \text{maximum }$

$\text{The maximum occurs when the forces are equally directed,}$

$\text{that is, at the lowest point of the trajectory.}$

$\text{We take into account the fact that gravity is directed}$

$\text{ vertically downward}$

$\text {Lowest point of trajectory:}$

$100cm - r = 100cm-50cm = 0.5m$

$\vec F_c+\vec F_g = 20$

$m\omega^2r +mg = 20$

$0.5*\omega^2*0.5+0.5*9.8 = 20$

$\omega=\sqrt\frac{20-4.9}{0.25}=7.77\frac{rad}{s}$

$\text{Answer: }h = 0.5m\text{ over the ground}$

$\omega=7.77\frac{rad}{s}$