Chapter 11: Angular Momentum

Transcription

1 Chapter 11: ngular Momentum Statc Equlbrum In Chap. 4 we studed the equlbrum of pontobjects (mass m) wth the applcaton of Newton s aws F 0 F x y, 0 Therefore, no lnear (translatonal) acceleraton, a0

2 For rgd bodes (non-pont-lke objects), we can apply another condton whch descrbes the lack of rotatonal moton If the net of all the appled torques s zero, we have no rotatonal (angular) acceleraton, α0 (don t need to know moment of nerta) We can now use these three relatons to solve problems for rgd bodes n equlbrum (a0, α0) Example roblem τ 0 The wheels, axle, and handles of a wheelbarrow wegh 60.0 N. The load chamber and ts contents

3 wegh 525 N. It s well known that the wheelbarrow s much easer to use f the center of gravty of the load s placed drectly over the axle. Verfy ths fact by calculatng the vertcal lftng load requred to support the wheelbarrow for the two stuatons shown. F F F D F D 1 2 F w F w m, m, m

4 Frst, draw a FBD labelng forces and lengths from the axs of rotaton F F Choose a drecton for the D rotaton, CCW beng axs postve s the conventon 1 2 F w a) 0 τ + τ + τ 0 D W F F + F 0 D 1 W 2 3 F + F F D 1 W 2 3 (525 N)(0.400 m) + (60.0 N)(0.700 m) F m F 194 N τ 3

5 pply to case wth load over wheel F Torque due to drt s zero, snce lever arm s zero axs F D F w b) 0 τ + τ + τ 0 D W F F + F 0 D 1 W 2 3 F + F F D 1 W 2 3 (525 N)(0 m) + (60.0 N)(0.700 m) F m F 32.3 N τ 2

6 Who? What s carryng the balance of the load? Consder sum of forces n y-drecton a) b) F F F F N N N Fy 0 FD FW + FN 0 FD + FW F N N F N F D F w F We dd not consder the Normal Force when calculatng the torques snce ts lever arm s zero

7 Center of Gravty F D The pont at whch the weght of a rgd body can be consdered to act when determnng the torque due to ts weght Consder a unform rod of length. Its center of gravty (cg) corresponds to ts geometrc center, /2. /2 cg Each partcle whch makes up the rod creates a torque about cg, but the sum of all torques due

8 to each partcle s zero So, we treat the weght of an extended object as f t acts at one pont Consder a collecton of pont-partcles on a massless rod The sum of the torques m gx m gx M 3 x cg 3 m 2 m + m 1 2 m x 1 gx2 Mgx 1 cg + m + m m 1 g 3 2 x M 2 + Mg m x 3 m 2 g 3 x 1 m 3 g x x 2 cm x cg x 3

9 ngular Momentum In Chapter 2, we defned the lnear momentum mv p nalogously, we can defne ngular Momentum Iω Snce ω s a vector, s also a vector has unts of kg m 2 /s The lnear and angular momenta are related p t v ( ) v & # Iω mr 2 $ t rm t % r " rp t r

10 gves us another way to express the rotatonal moton of an object For lnear moton, f an external force was appled for some short tme duraton, a change n lnear momentum resulted Smlarly, f an external torque s appled to a rgd body for a short tme duraton, ts angular momentum wll change If τ ext 0 then Ths s the rncple of Conservaton of ngular Momentum τ F ext ext Δt Δt f p f f p

11 How to nterpret ths? Say the moment of nerta of an object can decrease. Then, ts angular speed must ncrease. I I f > I f, f I ω I ω ω ω > ω f f I Example roblem For a certan satellte wth an apogee dstance of r 1.30x10 7 m, the rato of the orbtal speed at pergee to the orbtal speed at apogee s Fnd the pergee dstance r. Not unform crcular moton f

12 Satelltes generally move n ellptcal orbts. (Kepler s 1st aw). lso, the tangental velocty s not constant. " v If the satellte s ``crclng the r r Earth, the furthest pont n ts orbt from the Earth s called the ``apogee. The closest pont the ``pergee. For the Earth crclng the sun, the two ponts are called the ``aphelon and ``perhelon. v

13 Gven: r 1.30x10 7 m, v /v Fnd: r? Method: pply Conservaton of ngular Momentum. The gravtatonal force due to the Earth keeps the satellte n orbt, but that force has a lne of acton through the center of the orbt, whch s the rotaton axs of the satellte. Therefore, the satellte experences no external torques. " # $ $ % & " # $ $ % & r mr r mr I I v v 2 2 ω ω m 1.08x10 )(1/1.20) (1.30x10 ) v / (v v v 7 7 r r r r

14 Summary Translatonal Rotatonal x dsplacement θ v velocty ω a acceleraton α F cause of moton τ m nerta I ΣFma 2 nd aw Σ τi α Fs work τ θ 1/2mv 2 KE 1/2I ω 2 pmv momentum I ω