Physics 1600 9 Angular momentum

• 9.1 Angular Momentum and Torque
  • Oh, this is the one SSO said was difficult.
  • $\vec{L}=\vec{r}\times \vec{p}$
    • The cross product is involved.
    • If $\vec{r}$ is in the same direction as $\vec{p}$, then $L=0$, which is similar to angular momentum.
    • The dimension is perpendicular to $\vec{r}$ and $\vec{p}$.
  • $|L|=|r||p|\sin \alpha$
    • $=\vec{r}\sin \alpha \cdot |p|$
      • Multiply $\vec{r}$ projected onto the direction perpendicular to $\vec{p}$ by $|p|$.
    • $=\vec{p}\sin \alpha \cdot |r|$
  • Definition
    • Cartesian
      • Files／Screenshot 2022-12-11 at 2.46.07 AM.png
    • Polar
      • Files／Screenshot 2022-12-11 at 2.45.59 AM.png
      • The radial component is missing.
  • Torque
    • Files／Screenshot 2022-12-11 at 2.50.37 AM.png
    • Using the chain rule, $\dot{\vec{r}}\times \vec{p}$ is derived, but it becomes $\vec{v}\times \vec{p}$, so it becomes 0.
    • The direction is important, it is the cross product between the $r$ direction and the force direction.
      • Therefore, if the force is centripetal, $\tau =0$.
      • Only the $\theta$ component of the force affects the torque.
      • If the force is centripetal, the angular momentum only changes due to $r$.
      • Naturally, this value depends on the origin.
    • The force is the derivative of momentum, so it is the sum of all forces.
      • This intuition is important for understanding the equation.
  • 9.1.1 Particle Moving in a Straight Line
    • Even in a straight line, there is angular momentum if the origin is not on the line.
• Conical Pendulum
  • The direction of the cross product vector is important.
  • “The symmetry of the two masses makes the angular momentum independent of the location of the origin along the z axis.”
• 9.2 Central Forces
  • Angular momentum conservation
    • I want to include the proof here.
      • The elements of L are only the tangential velocity, which is the cross product between r and p.
    • If the acceleration only has a radial component, the tangential velocity is constant, and the angular momentum is constant.
  • Torque
    • Files／Screenshot 2022-12-11 at 2.50.37 AM.png
    • Using the chain rule, $\dot{\vec{r}}\times \vec{p}$ is derived, but it becomes $\vec{v}\times \vec{p}$, so it becomes 0.
    • The direction is important, it is the cross product between the r direction and the force direction.
      • Therefore, if the force is centripetal, $\tau =0$.
      • Only the $\theta$ component of the force affects the torque.
      • If the force is centripetal, the angular momentum only changes due to r.
      • Naturally, this value depends on the origin.
    • The force is the derivative of momentum, so it is the sum of all forces.
      • This intuition is important for understanding the equation.
  • 9.1.1 Particle Moving in a Straight Line
    • Even in a straight line, there is angular momentum if the origin is not on the line.
• Conical Pendulum
  • The direction of the cross product vector is important.
  • “The symmetry of the two masses makes the angular momentum independent of the location of the origin along the z axis.”
• 9.2 Central Forces
  • Angular momentum conservation
    • I want to include the proof here.
  • Effective Energy
    • The square of the polar velocity becomes ${\dot{r}}^2+r^2{\dot{\theta }}^2$.
      • This is not obvious, is it okay to simply add them?
      • Is it because it is the dot product of velocity vectors?
    • Total - $E_{\text{eff}}=$ r-direction KE
    • Therefore, in the Energy Diagram, when Total = $E_{\text{eff}}$, the r-direction velocity is 0.
  • If we calculate the initial condition momentum and it is conserved, we can say it is constant.
    • This is because it is a central force.
    • Assuming it is constant, if r decreases, $\dot{\theta }$ increases by the same amount.
• 9.3
  • Deriving Lrot and τrot
• 9.4
  • Rigid body

  • 7382. These systems consist of particles whose positions with respect to the COM are fixed.