复旦大学：《大学物理》课程教学资源（PPT课件，英文）Chapter 10 Angular momentum

Chapter 10 Angular momentum

Chapter 10 Angular momentum

10-1 Angular momentum of a particle 1. Definition Consider a particle of mass m and linear momentum p at a position relative to the origin o of an inertial frame we define the angular momentum"lof the particle with respect to the origin o to be =r×P (10-1)

10-1 Angular momentum of a particle 1. Definition Consider a particle of mass m and linear momentum at a position relative to the origin o of an inertial frame we define the “angular momentum” of the particle with respect to the origin o to be (10-1) → P → r → L → → → L = r P x z y m  → P → r

Its magnitude is L=rp sin 6 (10-2) where 0 is the smaller angle between r and p, we also can write it as Note that for convenience and p are in Xy plane

Its magnitude is (10-2) where is the smaller angle between and , we also can write it as Note that , for convenience and are in xy plane. L = rp sin   = ⊥ = ⊥ L pr rp → P → r → P → r

2.The relation between torque and angular momentum Differentiating Eq(10-1) we obtain l dr x P+rx ∑F=x(106) Here ar dr the×P=yXP=0 d p and dt ∑ Eq(10-6) states that "the net torque acting on a particle is equal to the time rate of change of its angular momentum

2. The relation between torque and angular momentum Differentiating Eq(10-1) we obtain (10-6) Here , the → → → → → → → → =  +  = rF =  dt d P P r dt d r dt d L → → = v dt d r  =  = 0 → → → → P v P dt d r and Eq(10-6) states that “the net torque acting on a particle is equal to the time rate of change of its angular momentum.  → → = F dt d P

Sample pf roblem 10-1 A particle of mass m is released from rest at point p (a)Find torque and angular momentum with respect to P origin o (b show that the relation dl ∑τ yield a correct result mg:e

Sample problem 10-1 A particle of mass m is released from rest at point p (a) Find torque and angular momentum with respect to origin o (b) Show that the relation yield a correct result . b o y P x → r m mg  dt d L → →  =

Solution (a)t=mgbzo (b is the moment arm) L=r×mv= bmgtz (b )dL bmg=t

Solution: (a) ( b is the moment arm) (b) →  = 0  mgb z → → →  =  = 0 L r m v bmgt z  → → = =  0 bmg z dt d L

10-2 Systems of particles 1. To calculate the total angular momentum L of a system of particle about a given point, we must add vectorially the angular momenta of all the individual particles about this point. L=L1+L2+…LN=∑ (10-8) As time goes on, L may change. That is dl d ∑

10-2 Systems of particles 1.To calculate the total angular momentum of a system of particle about a given point, we must add vectorially the angular momenta of all the individual particles about this point. (10-8) As time goes on, may change. That is → L = → → → → → = + +  = N n L L L LN Ln 1 1 2  → → → → = + L +  = n dt d L dt d dt d L  1 2 → L

Total internal torque is zero because the torque resulting from each internal action reaction force pair is zero. Thus ∑n=∑ →dL (10-9) That is: the net external torque acting on a system of particles is equal to the time rate of change of the total angular momentum of the system Note that: (1) the torque and the angular momentum must be calculated with respect to the same origin of an inertia reference frame (2)Eg(10-9) holds for any rigid body

Total internal torque is zero because the torque resulting from each internal action- reaction force pair is zero. Thus (10-9) That is: “the net external torque acting on a system of particles is equal to the time rate of change of the total angular momentum of the system.” Note that: (1) the torque and the angular momentum must be calculated with respect to the same origin of an inertia reference frame. dt d L n ext → → →  =  = (2) Eq(10-9) holds for any rigid body

dAndo dl ∥/ P△P Suppose a force F acts on a particle which moves with P+△P momentum p .We can △P resolve f into two components P as shown in Fig 10-3: Fig 10-3 The component Fu gives a change in momentum A pu which changes the magnitude of p, on the other hand the f gives an increment ap that changes the direction of p

2. and Suppose a force acts on a particle which moves with momentum . We can resolve into two components, as shown in Fig 10-3: dt d P F → →  = dt d L → →  = → F → P →  P// → P → P → F// → F⊥ →  P⊥ → ⊥ → P+  P → F Fig 10-3 The component gives a change in momentum , which changes the magnitude of ; on the other hand, the gives an increment that changes the direction of . → F// → →  p// → P F⊥ → →  P⊥ P

The same analysis holds for the action of a →△L torque t As as shown in Fig 10-4. In this case Ai must be parallel to t We once again resolve T into two components∥L andt⊥ L. The component ∥/ ∥/ changes thei in magnitude ∥ but not in direction(Fig 10 L+△L 4a). The component△ gives an increment△L1⊥L which changes the direction of L but not its magnitude Fig 10-4 (Fig10-4b)

(a) (b) → //  → ⊥  → L → L →  L// →  L⊥ → ⊥ → L+  L We once again resolve into two components and . The component changes the in magnitude but not in direction (Fig 10- 4a ). The component gives an increment , which changes the direction of but not its magnitude (Fig10-4b). → →  // // L → →  ⊥ ⊥ L → L → L →  ⊥  → →  L⊥ ⊥ L Fig 10-4 The same analysis holds for the action of a torque , as shown in Fig 10-4. In this case must be parallel to . t L   = → →  →  L →  →  → // 