Answer to Question #141544 in Mechanics | Relativity for Akinibom Sandra

Question

A stone of mass 500g is attached to a string of length 50cm which will break if the tension in it exceeds 20N. The stone is whirled in a vertical circle, the axis of rotation being at a height of 100cm above the ground. The angular speed is very slowly increased until the string breaks. In what position is the break must likely to occur, and at what angular speed? Where will the stone hit the ground?

Answer 1

The total force exerced on a stone is the sum of gravity force ($m \vec{g}$ , where $\vec{g}$ is directed down) and centrifugal force ($m\omega^2 \vec{r}$ , directed from the axe of rotation to the stone). Also the sum of these forces projected on $\vec{e}_r$ is equal to the tension in the string, as string is implicitly supposed to be not extensible (=of constant lenght). Therefore we have :

$T = (m\vec{g} + m\omega^2\vec{r})\cdot \vec{e}_r = m\vec{g}\cdot\vec{e}_r+m\omega^2 l$

The tension is maximal, when $\vec{g}$ and $\vec{e}_r$ are in the same direction which happens in the lowest point of the circle. So the answer to the first question - it is more likely to occur at the lowest point of the circle.

Now let's calculate $T$ in this point as a function of $\omega :$

$T=mg + m\omega^2 l$

So the minimal angular speed to achieve the critical value $T_0=20N$ is :

$\omega = \sqrt{\frac{T-mg}{ml}} = \sqrt{\frac{20-5}{0.5\cdot0.5}}=2\sqrt{15}$ rad/s.

Now let's find where the stone will hit the ground. If the string breaks at the lowest point of a circle, after this the stone will move with the constant acceleration downwards $\vec{g}$ , from initial height $H=100cm=1m$ and initial horizontal velocity (as in the lowest point of a circle the velocity is horizontal) $v_0 = \omega l$ . Therefore we find that it will land at some horizontal distance from it's initial position, and to find this distance we apply a pretty standart algorithm of solving such problems :

$h(t) = H - \frac{gt^2}{2}$

$t_{landing} = \sqrt{\frac{2H}{g}} \approx\frac{1}{\sqrt{5}} s$ (if we take $g\approx 10 m/s^2$ )

$d_{landing} = v_{horizntal}\cdot t_{landing} = 2\sqrt{15}\cdot 0.5\cdot$ $\frac{1}{\sqrt{5}}=\sqrt{3} m\approx1.73 m$