When the bob of mass ' $m$ ' moves in a horizontal circle of radius ' $r$ ' with uniform speed ' $v$ '…
- $\frac{\mathrm{mgL}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}$
- $\frac{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}{\mathrm{mgL}}$
- $\frac{\operatorname{mgr}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}$
- $\frac{\mathrm{mgr}}{\mathrm{L}^2-\mathrm{r}^2}$
Solution
On taking the ratio of equation (1) and (2):
Centripetal force $=\mathrm{mg} \tan \theta$
From geometry, $\tan \theta=\frac{\mathrm{r}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}$
$\therefore$ Centripetal force $=\frac{\mathrm{mgr}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}$Asked in: MHT CET 2022 (07 Aug Shift 1)