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 → → → → → → → → = + = rF = 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 → → → //
