The Angular Momentum of a System of Particles — Question 8

Section 11-8 · Physics pages 1706–1707

Question 8 Fragment spin belongs in the angular momentum budget An isolated object initially has angular momentum 8ˆz kg m2/s about its stationary mass center. It breaks into two extended fragments. Their center motions carry orbital angular momenta 3ˆz and −1ˆz in the same units. C⃗ S1⃗ S2 Each extended fragment carries center orbital angular momentum and its own intrinsic spin. (a) Find the sum of their intrinsic spin angular momenta. (b) If fragment 1 has spin 4ˆz, find fragment 2 spin and explain the error in treating both as structureless particles.
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Question 8 - Solution Fragment spin belongs in the angular momentum budget C⃗ S1⃗ S2 (a) Derivation and calculation Conservation includes both fragment-center motion and intrinsic spin:⃗ Li =⃗Lorb,1 +⃗Lorb,2 +⃗S1 +⃗S2. Thus 8ˆz = 3ˆz −ˆz +⃗S1 +⃗S2,⃗ S1 +⃗S2 = 6ˆz kg m2/s . (b) Result and interpretation If⃗S1 = 4ˆz, then⃗ S2 = (6 −4)ˆz = 2ˆz kg m2/s . Treating both fragments as structureless particles would retain only 3 −1 = 2 of the original 8 units. The missing 6 units reside in fragment spin, so that simplified model would omit essential angular momentum. Physical check The complete accounting is 3 −1 + 4 + 2 = 8, despite any changes in kinetic energy during breakup.

Question 8

Fragment spin belongs in the angular momentum budget

An isolated object initially has angular momentum 8ẑkgm2/s8\hat z\,\mathrm{kg\,m^2/s} about its stationary mass center. It breaks into two extended fragments. Their center motions carry orbital angular momenta 3ẑ3\hat z and −1ẑ-1\hat z in the same units.

Diagram shown in the original SVG worksheet.

  1. Find the sum of their intrinsic spin angular momenta.

  2. If fragment 1 has spin 4ẑ4\hat z, find fragment 2 spin and explain the error in treating both as structureless particles.

Question 8 - Solution

Fragment spin belongs in the angular momentum budget

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Conservation includes both fragment-center motion and intrinsic spin: L→i=L→orb,1+L→orb,2+S→1+S→2.\vec L_i=\vec L_{{\mathrm{orb}},1}+\vec L_{{\mathrm{orb}},2}+\vec S_1+\vec S_2. Thus 8ẑ=3ẑ−ẑ+S→1+S→2,8\hat z=3\hat z-\hat z+\vec S_1+\vec S_2, S→1+S→2=6ẑkgm2/s.\boxed{\vec S_1+\vec S_2=6\hat z\ \mathrm{kg\,m^2/s}}.

(b) Result and interpretation

If S→1=4ẑ\vec S_1=4\hat z, then S→2=(6−4)ẑ=2ẑkgm2/s.\boxed{\vec S_2=(6-4)\hat z=2\hat z\ \mathrm{kg\,m^2/s}}. Treating both fragments as structureless particles would retain only 3−1=23-1=2 of the original 8 units. The missing 6 units reside in fragment spin, so that simplified model would omit essential angular momentum.

Physical check

The complete accounting is 3−1+4+2=83-1+4+2=8, despite any changes in kinetic energy during breakup.