# Central forces, angular momentum and effective potential

> A central force is directed along the line joining the particle to a fixed centre and has magnitude depending only on radial distance in the usual model.

Course: [University mechanics and waves](https://studybrix.com/en/courses/university-mechanics/index.md)
Level: university
Prerequisites: Vectors, conservation laws and derivatives.

## Notes

### Central force

A central force is directed along the line joining the particle to a fixed centre and has magnitude depending only on radial distance in the usual model.

### Conservation

Its torque about the centre vanishes, so angular momentum about that centre is conserved. The trajectory lies in a plane when angular momentum is nonzero.

### Radial energy

For conserved angular momentum L, the effective radial potential is V(r)+L²/(2mr²). The second term describes the angular contribution.

### Stability

A circular orbit corresponds to a stationary point of effective potential. A local minimum gives stability against small radial displacements in this model.

## Practice questions

### 1. Why is torque about the centre zero for a central force?

A. Velocity must vanish
B. Force must have zero magnitude
C. Position and force are parallel or antiparallel
D. Mass is zero

Answer: Position and force are parallel or antiparallel

Explanation: The cross product r×F vanishes for collinear vectors.

### 2. A 2 kg particle has tangential speed 3 m/s at radius 4 m. Find angular-momentum magnitude.

Answer: 24 kg m²/s

Explanation: For tangential motion L=mrv=2×4×3.

### 3. Which quantity is conserved about the centre under a central force?

A. Angular momentum
B. Radial velocity always
C. Kinetic energy always
D. Radius always

Answer: Angular momentum

Explanation: Zero torque implies conservation of angular momentum, not constant radius.

### 4. For L=6 kg m²/s, m=2 kg and r=3 m, find L²/(2mr²).

Answer: 1 J

Explanation: The angular contribution is 36/(2×2×9).

### 5. Explain how a radial turning point is identified in an effective-potential diagram.

Sample answer / marking guidance: At a turning point radial speed is zero, so total energy equals effective potential. Allowed radial regions have total energy at least as large as effective potential.

Explanation: Credit equality at the turning point and the inequality for allowed motion.

### 6. Which local feature suggests a stable circular orbit?

A. Any local maximum
B. A local minimum of effective potential
C. Any point with negative potential
D. Any zero of radius

Answer: A local minimum of effective potential

Explanation: Near a minimum, small radial displacements produce restoring behaviour.

### 7. If angular momentum is constant and radius halves, by what factor does tangential speed change?

Answer: 2

Explanation: L=mrvθ, so vθ is inversely proportional to r at fixed L and m.

### 8. Why must one specify the centre when asserting angular-momentum conservation?

Sample answer / marking guidance: Torque depends on the chosen origin. A force central about one origin need not exert zero torque about another.

Explanation: Credit origin dependence and the zero-torque criterion.

## Further reading

- https://ocw.mit.edu/courses/8-09-classical-mechanics-iii-fall-2014/
- https://openstax.org/details/books/university-physics-volume-3

Original AI-assisted educational draft; not independently teacher-reviewed. References support further study, not copied exercises, affiliation or curriculum accreditation.