Physics - Rotation Question with Solution | TestHub

As$2\text{t}sin\frac{\theta }{2}=\text{dm}{\omega }^2\text{r}$(for small angle sin$\frac{\theta }{2}\to \frac{\theta }{2}$)

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 =10\pi$$\mathrm{rad} {\mathrm{s}}^{-1}$and$\mathrm{r}=0.25 \mathrm{m}$

$\therefore \text{T}=25\mathrm{0 }\text{N}$