From Newton s 2 nd Law: F ma d dm ( ) m dt dt F d dt The tme rate of change of the lnear momentum of a artcle s equal to the net force actng on the artcle.
Conseraton of Momentum +x The toy rocket n dee sace breaks nto equal masses. Fnd the fnal eloctes of the masses and show that the elocty of the CM s equal to the ntal elocty. Ignore graty and ar resstance. x 2 cos30 cos 60 xf f y m m1 + m2 1 2 Substtute: yf 0 m sn30 m sn60 1 2 sn 60 sn 30 sn 60 2 m m( 2 )cos30+ m2cos60 sn 30 2 2(50 / ) 2 50 / (sn 60 / tan 30 + cos60) m s (sn 60 / tan 30 + cos60) m s 1 2 sn 60 sn 30 sn 60 (50 m/ s) 86.6 m/ s sn 30 Velocty s NOT consered!
MOMENTUM IS CONSERVED! NOT VELOCITY! before after before after Ths s true ONLY for The CM!
Center of Mass The geometrc center or aerage locaton of the mass.
Rotatonal & Translatonal Moton Objects rotate about ther Center of Mass. The Center of Mass Translates as f t were a ont artcle. d CM r dt CM
Center of Mass: Stablty If the Center of Mass s aboe the base of suort the object wll be stable. If not, t toles oer.
Newton s 1 st Law for Rotaton If the sum of the torques s zero, the system s n rotatonal equlbrum. x CM mx M What s the CM of the system? What f you don t gnore the mass of the lank?
Center of Mass The geometrc center or aerage locaton of the mass. Extended Body: System of Partcles: r CM 1 r M dm mx xcm M
Examle Extended Body: 1 rcm r dm M You must generate an exresson for the densty and the mass dfferental, dm, from geometry and by analyzng a str of the sgn. We assume the sgn has unform densty. If M s the total mass then the total olume and densty s gen by: x CM V 1 M abt, ρ 2 1 abt 2 M 2My dm ρ dv ρ ( ytdx) ytdx dx 1 abt ab 2 1 xdm M Where a, b and t are the wdth, heght and thckness of the sgn, resectely. Then the mass element for the str shown s: a 3 a 1 2My 2 b 2 x 2 ( ) 2 3 3 0 0 x dx x x dx a M ab ab a a
Rotatonal & Translatonal Moton Objects rotate about ther Center of Mass. The Center of Mass Translates as f t were a ont artcle. d CM r dt CM
System of Partcles Center of Mass The Center of Mass Translates as f t were a ont artcle and, f no external forces act on the system, momentum s then consered. Ths means: EVEN f the bat EXPLODED nto a thousand eces, all the eces would moe so that the momentum of the CM s consered that s, the CM contnues n the arabolc trajectory!!!! THIS IS VERY VERY IMPORTANT!
System of Partcles Center of Mass A rojectle s fred nto the ar and suddenly exlodes Wth no exloson, the rojectle would follow the dotted lne After the exloson, the center of mass of the fragments stll follows the dotted lne, the same arabolc ath the rojectle would hae followed wth no exloson! If no external forces act on the system, then the elocty of the CM doesn t change!!
System of Partcles Center of Mass CM dr d( mr / M) m dt dt M CM M m CM tot CM 1 m M CM 1 M tot tot M CM Note: If the elocty of the CM s zero, the total momentum s zero! M tot,fnal tot,ntal CM
System of Partcles 1 86.6 m/ s +x The toy rocket breaks nto equal masses. Fnd the fnal eloctes of the masses and show that the elocty of the CM s equal to the ntal elocty. Ignore graty and ar resstance. Partcle Velocty s NOT consered! CM,f 1 m M + 1 2 But CM elocty s consered! CM, 50 m ˆ s 2 50 m/ s 1 ˆ+ ˆ + ˆ ˆ m ( 1cos30 1sn 30 ) ( 2cos 60 2sn 60 ) 2 m m j m m j 1 86.6 / (cos30ˆ sn 30 ˆ) 50 / (cos60ˆ sn 60 ˆ) 2 m s + j + m s j ˆ 50 m s CM,
YOU TRY: Fnd the CM of a mong System of Partcles 10 At an nstant when a artcle of mass 80 g has a elocty of 25 m/s n the oste y drecton, a 75-g artcle has a elocty of 20 m/s n the oste x drecton. What s the seed of the center of mass of ths two-artcle system at ths nstant? a. 16 m/s 1 b. 45 m/s CM m c. 23 m/s M d. 20 m/s e. 36 m/s
A 3.0-kg ball and a 1.0-kg ball are laced at ooste ends of a massless beam so that the system s n equlbrum as shown. The drawng s not drawn to scale. a) If a 0.5 m, what s b? b) In general, What s the rato of the lengths, b/a? τ 0 Wa Wb 0 1 2 W m g m b a a a W m g m 1 1 1 2 2 2
A 50 kg erson sttng at the left end of a 75kg boat s 6 m from shore. The erson then walks to the other end of the boat 3m away. What haens to the boat? To the CM of the system? The CM of the system (erson + boat) s at rest and no external forces act, therefore, the total momentum before s zero. The walkng s nternal so that the there s no change n momentum and the CM remans at the same oston, relate to the shore. Therefore, the boat must moe to the left, to kee the CM at the same oston. That s, Conseraton of Momentum, causes the boat to moe to kee the CM unchangng. Ths s VERY mortant!!!
A 50 kg erson sts at the left end of a 75kg boat whch s 6 m from shore. The erson then walks to the other end of the boat 3m away. A) What s the CM of the erson-boat system relate to the shore? B) After the erson walks to the other end, how far s she from shore? mx xcm M (50 kg)(6 m) + (75 kg)(7.5 m) 125kg 6.9m 6.9m (50 kg)( x) + (75 kg)( x 1.5) 125kg x 7.8m
HW Romeo (77.0 kg) entertans Julet (55.0 kg) by layng hs gutar from the rear of ther boat (80.0 kg) at rest n stll water, 2.70 m away from Julet who s n the front of the boat. After the serenade, Julet carefully moes to the rear of the boat (away from shore) to lant a kss on Romeoʹs cheek. What s the boat gong to do? What haens to the CM of the system?
Exlodng Systems before after
Charle and hs gun, m 80kg, shoots off a bullet, m.01kg, straght ahead at 300 m/s. What s Charle s (+gun) recol elocty? 1. Defne the System so net external forces s zero. 2. THEN momentum s consered ntal fnal + + charle, bullet, charle, f bullet, f 0 charle, f + gun, f charle, f bullet, f m m c cf b bf cf m m b c bf.01kg 300 m/ s 80kg.04 m/ s Frctonless surface
Newton s 3 rd Law: Rocket Thrust ntal fnal 0 Mrocket mgas rocket M rocket M rocket m gas Δ Δ rocket gas gas mv gas Rocket Pushes Gas Out. Gas Pushes Back on Rocket.
Rocket Proulson The ntal mass of the rocket lus all ts fuel s M + Δm at tme t and elocty The ntal momentum of the system s (M + Δm) Fnal elocty gen by Proulson equaton: M f eln M f The force exerted on the rocket by the exhaust s gen by the Thrust Equaton: d Thrust : M e dt dm dt
P9.49 The frst stage of a Saturn V sace ehcle consumed fuel and oxdzer at the rate of 1.50 10 4 kg/s, wth an exhaust seed of 2.60 10 3 m/s. Calculate the thrust roduced by these engnes. d Thrust : M e dt dm dt
Rocket Problem A rocket for use n dee sace s to be caable of boostng a total load (ayload lus rocket frame and engne) of 3.00 metrc tons to a seed of 10 000 m/s. It has an engne and fuel desgned to roduce an exhaust seed of 2 000 m/s. How much fuel lus oxdzer s requred? M f eln M f M e ln M f e M e M f M ( ) 5 3 5 e 3.00 10 kg 4.45 10 kg ( ) 3 Δ M M M 445 3.00 10 kg 442 metrc tons f