Question:

If the angular momentum of a particle of mass $m$ rotating along a circular path of radius $r$ with uniform speed is L, the centripetal force acting on the particle is

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This formula is very common in planetary motion and atomic physics (Bohr's model). Memorizing the relationship $F_c = \frac{L^2}{mr^3}$ can save valuable time in entrance exams!
Updated On: May 6, 2026
  • $\frac{L^2}{mr^3}$
  • $\frac{L^2}{mr}$
  • $\frac{L}{mr^2}$
  • $\frac{L^2m}{r}$
  • $\frac{Lm}{r^2}$
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The Correct Option is A

Solution and Explanation

Concept: This question requires linking Circular Motion and Angular Momentum.
Centripetal Force ($F_c$): The inward force required for circular motion: $F_c = \frac{mv^2}{r}$.
Angular Momentum ($L$): For a particle in circular motion: $L = mvr$.

Step 1:
Express velocity ($v$) in terms of angular momentum ($L$). From $L = mvr$, we can solve for $v$: \[ v = \frac{L}{mr} \]

Step 2:
Substitute $v$ into the centripetal force formula. \[ F_c = \frac{m \cdot (\frac{L}{mr})^2}{r} \] \[ F_c = \frac{m \cdot \frac{L^2}{m^2r^2}}{r} \] \[ F_c = \frac{L^2}{m r^2 \cdot r} = \frac{L^2}{m r^3} \]
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