A particle of mass ' $m$ ' is rotating in a circular path of radius ' $r$ '. Its angular momentum is ' $L$ '…

A particle of mass ' $m$ ' is rotating in a circular path of radius ' $r$ '. Its angular momentum is ' $L$ '. The centripetal force acting on it is ' $F$ '. The relation between ' $F$ ', ' $L$ ', ' $r$ ' and ' $m$ ' is
  1. $\mathrm{F}=\frac{\mathrm{L}}{\mathrm{mr}^2}$
  2. $\mathrm{L}=\mathrm{m}^2 \mathrm{Fr}^2$
  3. $\frac{\mathrm{L}^2}{\mathrm{~m}}=\mathrm{Fr}^3$
  4. $\frac{\mathrm{F}}{\mathrm{L}^3}=\mathrm{mr}^2$

Solution

Angular momentum, $\mathrm{L}=\mathrm{rp} \sin \theta=\mathrm{rp} \quad$ for U.C.M. $\quad\left[\because \theta=.90^{\circ}\right]$ $\therefore \quad \frac{\mathrm{L}^2}{\mathrm{mr}^3}=\frac{\mathrm{r}^2 \mathrm{~m}^2 \mathrm{v}^2}{\mathrm{mr}^3}=\frac{\mathrm{mv}^2}{\mathrm{r}}$ Given, Centripetal force, $F=\frac{m^2}{r}$ $\frac{\mathrm{L}^2}{\mathrm{mr}^3}=\mathrm{F} \Rightarrow \frac{\mathrm{~L}^2}{\mathrm{~m}}=\mathrm{Fr}^3$

Asked in: MHT CET 2024 (09 May Shift 2)

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