Chapter 09 Center Of Mass And Linear Momentum - University Of Toledo

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Chapter 09Center of Mass and Linear Momentum Center of mass: The center of mass of a body or asystem of bodies is the point that moves as if all of themass are concentrated there and all external forces areapplied there. Note that HRW uses “com” but I will use “c.m.”because “c.m.” is more standard notation for the“center of mass”.

System of ParticlesConsider two masses m1 at x x1 and m2 at x2.x c. m .m1x1 m 2 x 2 m1 m 2

Center of mass for a system of n particles:x c. m .n1 mi x iM i 1y c. m . M is just the total massof the system1 n mi yiM i 1z c.m .1 n mi ziM i 1nM mii 1v Using vectors, we have: rc.m. x c.m. î y c.m.ˆj z c.m. k̂ Therefore:1 nvvrc.m. m i riM i 1

For a solid body, we can treat it as a continuousdistribution of matter dmx c.m .1 x dmMy c.m. 1y dm Mz c. m .1 z dmM If the object has uniform density,dm (M/V)dVx c. m .1 x dVVy c.m .1 y dVVz c.m .1 z dVV If an object has a point, a line or a plane ofsymmetry, the center of mass of such an object thenlies at that point, on that line or in that plane.

Sample 9-2: the figure shows auniform metal plate P of radius 2Rfrom which a disk of radius R hasbeen stamped out (removed). Usingthe x-y coordinate system shown,locate the center of mass of theplate.Notice that the object is symmetricabout the x-axis, so yc.m. 0. We justneed to calculate xc.m. 0.

Technique: Use symmetry asmuch as possible.CThe center of mass of the largedisk must be at the center of thedisk by symmetry. The sameholds true for the small disk.The hole is as if the smaller diskhad “negative” mass tocounteract the solid larger massdisk.S

Superimpose the two disks. Theoverlap of the “positive” masswith the “negative” mass willresult mathematically in a hole.SCThe center of mass will besomewhere over here.center of mass (c.m. or com) of Plate C: xC 0center of mass (c.m. or com) of Disk S: xS Rx c. m .m C x C mS x S 0 mS ( R ) mS Rm C mSm C mS m C mS

Superimpose the two disks. Theoverlap of the “positive” masswith the “negative” mass willresult mathematically in a hole.SNow we need mC and mS. Bothdisks have the same uniform massdensity ρ (but with different“signs”). Thus,() (m C ρC VC ρ π(2R ) 2 ρ 4πR 2Likewise,()(m S ρS VS ρ π(R ) ρ πR22))C

Superimpose the two disks. Theoverlap of the “positive” masswith the “negative” mass willresult mathematically in a hole.S() ( ) ρ(π(R ) ) ρ(πR )m C ρC VC ρ π(2R ) ρ 4πR2m S ρS VSx c.m .C222m C x C mS x S 0 mS ( R ) mS Rm C mSm C mS m C mS 2 1 ρ(πR ) R R 22 3 ρ(4πR ) ρ(πR )

Newton’s 2nd law for a system of particlesWe knowrr1 nrc.m. m i riM i 1nrrM rc.m. Σ m i rii 1take derivative with respect to timerrrrM v c.m. m1v1 m 2 v 2 . m n v ntake derivative with respect to time againrrrrM a c.m. m1a 1 m 2 a 2 . m n a nr rr rrM a c.m. F1 F2 . Fn Fnet

Newton’s second law for a system of particlesFnet , x M a c.m., xFnet , y M a c.m., yvvFnet M a c.m. Fnet ,z M a c.m.,zFnet is the net force of all external forces that act onthe system.M is the total mass of the system.ac.m. is the acceleration of the center of the mass

Linear Momentum The linear momentum of a particle is a vectorvdefined asvp mv Newton’s second law in terms of momentumvvdp ddv v dmv vv (mv) m v ma Fnetdt dtdtdtvvdPFnet dtMost of the time the massdoesn’t change, so this termis zero. Exceptions arerockets (Monday)

The figure gives the linear momentum versus time for aparticle moving along an axis. A force directed along theaxis acts on the particle.(a) Rank the four regions indicated according to themagnitude of the force, greatest first(b) In which region is the particle slowing?

Linear momentum of a system of particlesnnvvvvP p i m i v i M v c .m .i 1i 1vvP Mv c . m . Newton’s 2nd law for a system of particlesvvvdv c . m .dPv M M a c.m. FnetdtdtvvdPFnet dt

Daily Quiz, February 16, 2005A helium atom and a hydrogen atom can bind to form themetastable molecule HeH (lifetime of about 1µs). Consider onesuch molecule at rest in the lab frame at the origin. This moleculethen dissociates with the hydrogen atom having momentum mpvalong the x axis. What happens to the helium atom?

Daily Quiz, February 16, 2005A helium atom and a hydrogen atom can bind to form themetastable molecule HeH (lifetime of about 1µs). Consider onesuch molecule at rest in the lab frame at the origin. This moleculethen dissociates with the hydrogen atom having momentum mpvalong the x axis. What happens to the helium atom?

Daily Quiz, February 16, 2005A helium atom and a hydrogen atom can bind to form themetastable molecule HeH (lifetime of about 1µs). Consider onesuch molecule at rest in the lab frame at the origin. This moleculethen dissociates with the hydrogen atom having momentum mpvalong the x axis. What happens to the helium (mHe 4mp)atom?1) stays at x 02) goes along x at speed v3) goes along x at speed v4) goes along x at speed v/45) none of the above

Daily Quiz, February 16, 2005A helium atom and a hydrogen atom can bind to form themetastable molecule HeH (lifetime of about 1µs). Consider onesuch molecule at rest in the lab frame at the origin. This moleculethen dissociates with the hydrogen atom having momentum mpvalong the x axis. What happens to the helium (mHe 4mp)atom?pinitial 0 means pfinal 0mpv mHevHe 0 mpv 4mpvHe 04) goes along x at speed v/4 vHe mpv/4

Collisions take time!Even something thatseems instantaneous to ustakes a finite amount oftime to happen.

Impulse and Change in Momentumvvtf vdpv vv vvFnet dp Fdt p f p i p FdttidtCall this change in momentum the “Impulse” andgive it the symbol J.tf vv vv vp f p i p J Fdtti

Actual force functionversus time.Average force versuscollision time.

Impulse and Change in MomentumCall this change in momentum the “Impulse” andgive it the symbol J.tf vv vv vp f p i p J FdttiThe instantaneous force is hard (or very difficult) toknow in a real collision, but we can use the averageforce Favg. Thus,J Favg t.

Daily Quiz, February 18, 2005TableFoamAn egg was dropped on the table broke, but the eggdropped on the foam pad didn’t break. Why didn’t thisegg break?

Daily Quiz, February 18, 2005FoamTableAn egg was dropped on the table broke, but the egg dropped onthe foam pad didn’t break. Why didn’t this egg break?1) Foam egg’s speed was less.2) Foam egg’s change in momentum was less.3) Foam egg’s collision time was greater.4) What do you mean, it did break! 0) none of the above

Daily Quiz, February 18, 2005FoamTableAn egg was dropped on the table broke, but the egg dropped onthe foam pad didn’t break. Why didn’t this egg break?1) Foam egg’s speed was less.The eggs were dropped from same height, so theirspeeds were the same. mgh 1/2mv2

Daily Quiz, February 18, 2005FoamTableAn egg was dropped on the table broke, but the egg dropped onthe foam pad didn’t break. Why didn’t this egg break?2) Foam egg’s change in momentum was less.The table egg’s momentum stopped at the table, but thefoam egg bounced up making its change in momentumgreater than the table egg!

Daily Quiz, February 18, 2005FoamTableAn egg was dropped on the table broke, but the egg dropped onthe foam pad didn’t break. Why didn’t this egg break?3) Foam egg’s collision time was greater.Since J constant, if t is large then Favg will be small.J Favg t.

Conservation of Linear Momentum For a system of particles, if it is both isolated (thenet external force acting on the system is zero) andclosed ( no particles leave ror enter the system ) .rdPIfthen 0ΣF 0dtThereforerP cons tan torr rPi P fthen the total linear momentum of the systemcannot change.Law of conservation of linear momentum

Conservation of linear momentum along a specificdirection:If Σ Fx 0Then Pi, x Pf, xIf Σ Fy 0Then Pi, y Pf, yIf the component of the net external force on aclosed system is zero along an axis, then thecomponent of the linear momentum of the systemalong that axis cannot change.

Linear Momentum The linear momentum of a particle is a vectorvdefined asvp mv Newton’s second law in terms of momentumvvdp ddv v dmv vv (mv) m v ma Fnetdt dtdtdtvvdPFnet dtMost of the time the massdoesn’t change, so this termis zero. Exceptions arerockets (Tuesday)

Conservation of Linear Momentum For a system of particles, if it is both isolated (thenet external force acting on the system is zero) andclosed ( no particles leave ror enter the system ) .rdPIfthen 0ΣF 0dtThereforerP cons tan torr rPi P fthen the total linear momentum of the systemcannot change.Law of conservation of linear momentum

Conservation of linear momentum along a specificdirection:If Σ Fx 0Then Pi, x Pf, xIf Σ Fy 0Then Pi, y Pf, yIf the component of the net external force on aclosed system is zero along an axis, then thecomponent of the linear momentum of the systemalong that axis cannot change.

CollisionsIn absence of external forces,Linear momentum is conserved.Mechanical energy may or may not be conserved.Elastic collisions: Mechanical energy is conserved.Inelastic collisions: Mechanical energy is NOT conserved.But, Linear momentum is always conserved.

Conservation of Linear Momentumvvpi pf vvp 1i p 2 i vvm 1 v 1i m 2 v 2 i vvp 1f p 2 fvvm 1 v 1f m 2 v 2 f

Center of Mass motion is constant

Center of Mass MotionvvvvvP p 1i p 2 i M v c .m . ( m 1 m 2 ) v c .m .vPv v c .m . m1 m 2

Inelastic CollisionsPerfectly inelastic collision: The two masses stick togethervvvvvm 1 v 1 i m 2 v 2 i m 1 v 1 f m 2 v 2 f (m 1 m 2 )V fvVf0 m1vv 1im1 m 2v( v c .m . )

Inelastic CollisionsWas the mechanical energy:conserved (Ei Ef);lost(Ei Ef); orgained(Ei Ef);in the collision?

Inelastic CollisionsHow much mechanical energy was lost in the collision?v2 11(m 1 m 2 )V f (m 1 m 2 ) m 122 m1 m 2E lost E i E f E lost11 m 1 v 12i 22 1 22 2 21 m 12 v 1 i v 1 i2 m1 m 2 2m 121 m 1m 2 2 v 1 i v 1 im1 m 2 2 m1 m 2 m 1m 2 m1 m 2 2 v 1 i

Elastic CollisionsPerfectly elastic collision: Mechanical energy is conservedvvvvvm 1 v 1 i m 2 v 2 i m 1 v 1 f m 2 v 2 f ( M v c . m . )11v2v2E i m 1 v 1i m 2 v 2 i22 11v2v2m 1 v 1f m 2 v 2 f E f22

Elastic CollisionsPerfectly elastic collision: Mechanical energy is conservedvvvvvm 1 v 1 i m 2 v 2 i m 1 v 1 f m 2 v 2 f ( M v c . m . )vv 1fvv 2f m1 m 2 vv 1i m1 m 22m 2 vv 2im1 m 2 2m1 vv 1i m1 m 2m 2 m1 vv 2im1 m 2

What about two (or more) dimensions?Simply break the momenta and velocities intotheir x-, y-, and z-components.

Inelastic collisions in two dimensionsr rPi PfPi x Pf xPi y Pf yFor the case shown here:m1v1i m1v1 f cos θ1 m2 v2 f cos θ20 m1v1 f sin θ1 m2 v2 f sin θ2

Conservation of Linear Momentum For a system of particles, if it is both isolated (thenet external force acting on the system is zero) andclosed ( no particles leave ror enter the system ) .rdPIfthen 0ΣF 0dtThereforerP cons tan torr rPi P fthen the total linear momentum of the systemcannot change.Law of conservation of linear momentum

Linear Momentum is conservedv vPf PiLet the mass M change (M dM), which in turn makes thevelocity change. Rocket exhausts dM in time dt at avelocity U relative to our inertial reference frame.Mv dM U (M dM )( v dv) Velocity of rocket Velocity of rocket Velocity of exhaust relative to ref . frame relative to exhaust relative to ref . frame ( v dv) v relU UUSAvrelv dv

Linear Momentum is conservedv vPf PiU ( v dv) v relSubstitute and divide by dtMv dM U (M dM )( v dv)dvdMv rel M dtdtUUSAvrelv dv

Linear Momentum is conservedv vPf PiU ( v dv) v relSubstitute and divide by dtMv dM U (M dM )( v dv)dMdv v rel M vfvidv v rel Mi v f v i v rel ln MfMfMidMM USAvrelv dv

Problem 09-05Find the center of mass of the ammonia molecule.Mass ratio: N/H 13.9H to triangle center: d 9.40x10-11mN to hydrogen: L 10.14x10-11m

Problem 09-05Find the center of mass of theammonia molecule.

Problem 09-05Find the center of mass of theammonia molecule.

Problem 09-54Find the distance the spring iscompressed. m1 2.0kg, m2 1.0kg.

Problem 09-54Find the distance the spring iscompressed. m1 2.0kg, m2 1.0kg.

Problem 09-56Find the final velocity of A and is the collisionelastic?Mass A: mA 1.6kg, vAi 5.5 m/sMass B: mB 2.4kg, vBi 2.5 m/s, vBf 4.9 m/s

Problem 09-56Find the final velocity of A and is the collisionelastic?Mass A: mA 1.6kg, vAi 5.5 m/sMass B: mB 2.4kg, vBi 2.5 m/s, vBf 4.9 m/s

Problem 09-56Find the final velocity of A and is the collisionelastic?Mass A: mA 1.6kg, vAi 5.5 m/sMass B: mB 2.4kg, vBi 2.5 m/s, vBf 4.9 m/s

Problem 09-60Find the final speeds of the ball and block.Mass 1 (ball): m1 0.5kg, h 0.70mMass 2 (block): m2 2.5kg, v2i 0.0 m/s

Problem 09-60Find the final speeds of the ball and block.Mass 1 (ball): m1 0.5kg, h 0.70mMass 2 (block): m2 2.5kg, v2i 0.0 m/s

Problem 09-63Find the mass m to stop M and the final height of m.Mass 1 (baseball): m ?, hinitial 1.8mMass 2 (basketball): M 0.63kg, vMf 0.0 m/sElastic collisions: Mechanical energy is conserved.Basketball:1Mgh Mv 22 v 2ghrebounds upward with same speed -- only reversed direction

Problem 09-63Find the mass m to stop M and the final height of m.Mass 1 (baseball): m ?, hinitial 1.8mMass 2 (basketball): M 0.63kg, vMf 0.0 m/sBaseball:1mgh mv 22 v 2ghM m2mM m2mv Mf v Mi v mi 2gh 2ghM mM mM mM mM 3m 2gh 0M m m M / 3 0.21kg

Problem 09-63Find the mass m to stop M and final height of m.Mass 1 (ball): m1 0.5kg, h 0.70mMass 2 (block): m2 2.5kg, v2i 0.0 m/s

Problem 09-97Find the final speeds of the sleds.Mass 1(sleds): M 22.7kg,Mass 2 (cat): m 3.63kg, vi 3.05 m/s

Problem 09-97

Problem 09-97

Problem 09-130Ball 1 vo 10.0m/s at contact point of balls 2 and 3. All threeballs have mass m. Find the final velocities of all three balls.

Problem 09-130Ball 1 vo 10.0m/s at contact point of balls 2 and 3. All threeballs have mass m. Find the final velocities of all three balls.Pi mvo2ball1 balls 2 & 3Pf mV 2mv cosθ13θ 30o since all three balls are identical

Problem 09-130Ball 1 vo 10.0m/s at contact point of balls 2 and 3. All threeballs have mass m. Find the final velocities of all three balls.

Problem 09-130Ball 1 vo 10.0m/s at contact point of balls 2 and 3. All threeballs have mass m. Find the final velocities of all three balls.

Problem 09-130Ball 1 vo 10.0m/s at contact point of balls 2 and 3. All threeballs have mass m. Find the final velocities of all three balls.

A Quizype xCERNConsider a proton (mass mp and kinetic energy Ep) colliding headon with an electron (mass me) initially at rest. What is themaximum kinetic energy (Ee) that can be delivered to the electron?

A QuizConsider a proton (mass mp 1836me andkinetic energy Ep) colliding head on with anelectron (mass me) initially at rest. What isthe maximum kinetic energy (Ee) that can bedelivered to the electron?yCERNpe x1) Ee Ep 2) Ee Ep 3) Ee Ep4) not enough information

A QuizConservation of Momentumm p Vi m p Vf m p v f111222mVmVmv Conservation of Energyp ip fe f222112E p m p Vi and E e m e v f2224m e /m pEe Ep Ep2(1 me /m p )yConsider a proton (mass mp 1836me andCERNkinetic energy Ep) colliding head on with anpelectron (mass me) initially at rest. What is ethe maximum kinetic energy (Ee) that can bedelivered to the electron?x1) Ee Ep 2) Ee Ep 3) Ee Ep4) not enough information

Conservation of Linear Momentum For a system of particles, if it is both isolated (the net external force acting on the system is zero) and closed ( no particles leave or enter the system ) . If then Therefore or then the total linear momentum of the system cannot change. Law of conservation of linear momentum ΣF 0 r 0 dt dP r P .

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