If the angular momentum of a particle of mass $m$ rotating along a circular path of radius $r$ with uniform…
- $\frac{\mathrm{L}^{2}}{m r^{3}}$
- $\frac{\mathrm{L}^{2}}{m \mathrm{r}}$
- $\frac{\mathrm{L}}{\mathrm{mr}}$
- $\frac{\mathrm{L}^{2} m}{r}$
Solution
Centripetal force acting on the particle $F=\frac{m v^{2}}{r}=\frac{m\left(\frac{L}{m r}\right)^{2}}{r}=\frac{L^{2}}{m r^{3}}$
Asked in: JEE Mains - Rotational Motion - Test 4