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Conservation of Angular Momentum: Understanding Dynamics without External Torques

This section explores the principle of conservation of angular momentum, stating that in the absence of external torques, the total angular momentum of a system remains constant. Using Newton's second law in rotational motion, we establish that changes in the distribution of mass lead to compensating changes in angular velocity. Examples such as an ice skater and a merry-go-round illustrate these concepts, demonstrating how angular momentum conservation applies in various scenarios. This principle is foundational in understanding rotational dynamics in physics.

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Conservation of Angular Momentum: Understanding Dynamics without External Torques

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  1. Sect. 11.4: Conservation of Angular Momentum

  2. Recall Ch. 9:Starting with Newton’s 2nd Law in Momentum form: ∑Fext = (dptot/dt) For an isolated system with no external forces: ∑Fext = 0 so (dptot/dt)= 0 ptot = constant or In the absence of external forces, the total momentum of the system is conserved. • Following the translational – rotational analogies we’ve been talking about all through Ch. 1: Consider the most general form of Newton’s 2nd Law for Rotations: • Consider the case with no external torques: ∑τext= 0 so (dLtot/dt)= 0 Ltot = constant or In the absence of external torques, the total angular momentum of the system is conserved.

  3. Conservation of Angular Momentum In the absence of external torques, the total angular momentum of the system is conserved. That is,Ltot = constant = Linitial = Lfinal(1) • If the mass of an isolated system is redistributed, the moment of inertia changes. In this case, conservation of angular momentum, Linitial = Lfinal requires a compensating change in the angular velocity. That is, (1) requires: Iiωi = Ifωf = constant • This holds for rotation about a fixed axis and for rotation about an axis through the center of mass of a moving system • Note that this requires that the net torque must be zero

  4. Example: Ice skater • Iiωi = Ifωf = constant But I =∑(mr2)  Iiωi = Ifωf (1) Between the left figure & the right one, he redistributed the distances of part of his mass from the axis of rotation.  I =∑(mr2) changes because the r’s change. (1) requires a compensating change inω. a b

  5. Example 11.8: Merry-Go-Round • Merry-Go-Round:Assume ideal disk. M = 100kg, R = 2 m, Im = (½)MR2 • Student:m = 60 kg.Is = mr2. As he walks to center, his r changes, soIschanges.Angular momentum is conserved, so angular speed ω must also change. • We have: Iiωi = Ifωf (1) Im + Is = (½)MR2 + mr2 • Initially, he is on the edge at r = R = 2 m. At that time, ωi = 2 rad/s. Finally, he is at r = 0.5 m. Find ωf. Use (1): [(½)MR2 + mR2]ωi = [(½)MR2 + mr2]ωf Put in numbers & solve: ωf = 4.1 rad/s

  6. Example: Diver Iiωi = Ifωf = constant Angular momentum is conserved throughout her dive.

  7. Conceptual Example Conservation of angular momentum!! Linitial = Lfinal L = - L + Lperson  Lperson = 2 L Demonstration!

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