Angular Momentum in Rotation About a Fixed Axis

1. One Particle Inside the Rotating Body

The angular momentum of a particle at point P is $$\mathbf{l}=\mathbf{r}\times\mathbf{p}$$ with $$\mathbf{r}=\mathbf{OC}+\mathbf{CP}$$. Here CP is perpendicular to the axis and has length r⊥, so the speed is $$v=\omega r_{⊥}$$. The part of l along the axis (chosen as the z-axis) becomes $$l_z = m\omega r_{⊥}^2$$.:contentReference[oaicite:0]{index=0} Notice that l and ω need not point in the same direction for a single particle.

2. Whole Rigid Body

Add the axial contributions of all particles: $$L_z=\sum_i m_i r_{i⊥}^2\,\omega = I\,\omega\tag{6.42b}$$ with the moment of inertia $$I=\sum_i m_i r_{i⊥}^2$$.:contentReference[oaicite:1]{index=1}

If the body is symmetric about the axis, the sideways part $$L_{⊥}$$ cancels out, so $$\mathbf{L}=I\,\omega\,\hat{\mathbf{k}}$$ points straight along the axis.:contentReference[oaicite:2]{index=2} For an asymmetric body, only the axial component $$L_z$$ equals $$I\omega$$; the full vector L does not have to line up with the axis.:contentReference[oaicite:3]{index=3}

3. Torque and Angular Acceleration

The rate of change of axial angular momentum equals the external torque on that axis: $$\frac{d}{dt}\!\left(I\omega\right)=\tau\tag{6.43c}$$.:contentReference[oaicite:4]{index=4} If the shape (and therefore I) is fixed, this reduces to $$\tau = I\alpha\quad\text{with}\quad \alpha=\frac{d\omega}{dt}\tag{6.41}$$.:contentReference[oaicite:5]{index=5}

4. Conservation of Angular Momentum

When the net external torque about the axis is zero, $$L_z = I\omega = \text{constant}\tag{6.44}$$.:contentReference[oaicite:6]{index=6}

Everyday feel-for-it demo: Sit on a swivel chair, spin with arms folded, then stretch your arms. Your moment of inertia grows, so the spin slows. Pull your arms back in and you speed up again—I ω stayed constant all along.:contentReference[oaicite:7]{index=7} Skaters, acrobats, and dancers use the same trick on ice or stage.

5. Key Takeaways

  • Particle angular momentum: $$\mathbf{l}=\mathbf{r}\times\mathbf{p}$$.:contentReference[oaicite:8]{index=8}
  • Axial component for one particle: $$l_z=m\omega r_{⊥}^2$$.:contentReference[oaicite:9]{index=9}
  • Rigid-body axial momentum: $$L_z=I\omega$$.:contentReference[oaicite:10]{index=10}
  • Torque–acceleration link (fixed I): $$\tau=I\alpha$$.:contentReference[oaicite:11]{index=11}
  • No external torque ⇒ $$I\omega$$ stays constant (swivel-chair test).:contentReference[oaicite:12]{index=12}

High-Yield Ideas for NEET

  1. The shortcut $$L_z = I\omega$$ quickly solves many fixed-axis problems.:contentReference[oaicite:13]{index=13}
  2. Use $$\tau = I\alpha$$ to jump straight from torque to angular acceleration.:contentReference[oaicite:14]{index=14}
  3. Conservation of I ω explains why changing body shape changes spin rate.:contentReference[oaicite:15]{index=15}
  4. Moment of inertia is always $$I=\sum_i m_i r_{i⊥}^2$$—memorize this form.:contentReference[oaicite:16]{index=16}
  5. Symmetry matters: in symmetric bodies L aligns with the axis; in asymmetric ones only Lz matters.:contentReference[oaicite:17]{index=17}