Question

A stone tied to an inextensible string of length $l=1m$ is kept horizontal. If it is released, find the angular speed of the stone when the string makes an angle $\theta ={30}^o$ with horizontal.

Answer 1

When the stone is released, it moves down executing a circular motion. At any instant, the stone accelerates towards (w.r.t.) the center of its revolution 0, that is, centripetal acceleration a, and simultaneously accelerates down with an acceleration, that is, gravitational acceleration g; ${\overrightarrow{a}}_r={\omega }^{2}r$. Resolving ${\overrightarrow{a}}_r$, and along the tangent, we obtain the tangential acceleration of the stone as $a_t=g\cos \theta$

Angular acceleration, $\alpha =\frac{a_t}{r}$

$\Rightarrow \alpha =\frac{g\cos \theta }{l}(sincer=l)$

Putting $\alpha =\frac{\omega d\omega }{d\theta }$, we obtain $\frac{\omega d\omega }{d\theta }=\frac{g\cos \theta }{r}$

$\Rightarrow \omega d\omega =\left(g/l\right)cos\theta d\theta .$

Integrating both the sides, we obtain

${\int }_0^{\omega }\omega d\theta =\frac{g}{l}{\int }_0^{\theta }\cos \theta d\theta$

$\Rightarrow \frac{{\omega }^2}{2}=\frac{g}{l}\sin \theta or\omega =\sqrt{\frac{2g}{l}\sin \theta }$

Putting $l=1m$ and $\theta ={30}^o$, we obtain

$\omega =\sqrt{\frac{2\left(10\right)\sin {30}^o}{\left(1\right)}}=\sqrt{10}\Rightarrow \omega \cong 3.1rad{s}^{-1}$