Question

A ring of mass $2\,\pi kg$ and of the radius $0.25\,m$ is making $300\,rpm$ about an axis through its centre perpendicular to its plane. The tension (in newtons) developed in the ring is:

Answer 1

As $2 \text{t} sin \frac{\theta }{2} = \text{dm} \omega ^{2} \text{r}$ (for small angle sin $\frac{\theta }{2} \rightarrow \frac{\theta }{2}$ )
but $\text{dm} = \frac{\text{m}}{\ell } \theta \text{r}$
As $\ell = 2 \pi \text{r}$
$\therefore \text{T} = \frac{\text{m} \omega ^{2} \text{r}}{2 \pi }$



Put $\text{m} = 2 \pi kg \text{, } \omega = 1 0 \pi $ $rads^{- 1}$ and $r=0.25m$
$\therefore \text{T} = 2 5 0 \text{N}$