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Physics · lesson 3 of 4

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.

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Prerequisites: Vectors, conservation laws and derivatives.

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.

Worked examples

Example 1

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

The cross product r×F vanishes for collinear vectors.

Position and force are parallel or antiparallel

Example 2

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

For tangential motion L=mrv=2×4×3.

24 kg m²/s

Check your understanding

8 questions

Decimals or fractions are accepted for numeric answers. Proofs and written responses use sample answers and require human marking. This is practice, not a protected examination.

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

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Position and force are parallel or antiparallel

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.kg m²/s

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24 kg m²/s

For tangential motion L=mrv=2×4×3.

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

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Angular momentum

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).J

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1 J

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

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

Sample answer and 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.

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

6 Which local feature suggests a stable circular orbit?

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A local minimum of effective potential

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?

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2

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 and marking guidance

Torque depends on the chosen origin. A force central about one origin need not exert zero torque about another.

Credit origin dependence and the zero-torque criterion.

Further reading and review references

These are reference links, not copied exercises, institutional endorsement or proof of independent review.