Center of Mass and Linear Momentum Chapter 9 Center of Mass and Linear Momentum Copyright © 2014 John Wiley & Sons, Inc. All rights reserved. 1

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-1 Center of Mass Motion of rotating objects is complicated There is a special point for which motion is simple Center of mass of bat traces out a parabola, just as a tossed ball does All other points rotate around this point Figure 9-1 © 2014 John Wiley & Sons, Inc. All rights reserved. 2

Goals for Chapter 9 To learn the meaning of the momentum of a particle and how an impulse causes it to change To learn how to use the principle of conservation of momentum To learn how to solve problems involving collisions

Goals for Chapter 8 To learn the definition of the center of mass of a system and what determines how it moves To analyze situations, such as rocket propulsion, in which the mass of a moving body changes

Introduction In many situations, such as a bullet hitting a carrot, we cannot use Newton’s second law to solve problems because we know very little about the complicated forces involved. In this chapter, we shall introduce momentum and impulse, and the conservation of momentum, to solve such problems.

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-1 Center of Mass DEFINITION: center of mass (com) of a system of particles: © 2014 John Wiley & Sons, Inc. All rights reserved. 6

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-1 Center of Mass For two particles separated by a distance d, where the origin is chosen at the position of particle 1: © 2014 John Wiley & Sons, Inc. All rights reserved. 7

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-1 Center of Mass For two particles separated by a distance d, where the origin is chosen at the position of particle 1: For two particles, for an arbitrary choice of origin: © 2014 John Wiley & Sons, Inc. All rights reserved. 8

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-1 Center of Mass For many particles, we can generalize the equation, where M = m1 + m2 + . . . + mn: © 2014 John Wiley & Sons, Inc. All rights reserved. 9

9-1 Center of Mass In three dimensions, we find the center of mass along each axis separately: More concisely, we can write in terms of vectors: © 2014 John Wiley & Sons, Inc. All rights reserved. 10

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-1 Center of Mass For solid bodies, we take the limit of an infinite sum of infinitely small particles → integration! Coordinate-by-coordinate, we write: Here M is the mass of the object © 2014 John Wiley & Sons, Inc. All rights reserved. 11

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-1 Center of Mass If objects have uniform density = ρ (“rho”) Substituting, we find the center of mass simplifies: © 2014 John Wiley & Sons, Inc. All rights reserved. 12

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-1 Center of Mass The center of mass lies at a point of symmetry It lies on the line or plane of symmetry) It need not be on the object (consider a doughnut) © 2014 John Wiley & Sons, Inc. All rights reserved. 13

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-1 Center of Mass Answer: (a) at the origin (b) in Q4, along y=-x (c) along the -y axis (d) at the origin (e) in Q3, along y=x (f) at the origin © 2014 John Wiley & Sons, Inc. All rights reserved. 14

© 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-1 Center of Mass Example Subtracting Task: find COM of a disk with another disk taken out of it: Find the COM of the two individual COMs (one for each disk), treating the cutout as having negative mass Figure 9-4 © 2014 John Wiley & Sons, Inc. All rights reserved. 16

© 2014 John Wiley & Sons, Inc. All rights reserved.

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-1 Center of Mass Example Subtracting On the diagram, comC is the center of mass for Plate P and Disk S combined comP is the center of mass for the composite plate with Disk S removed Figure 9-4 © 2014 John Wiley & Sons, Inc. All rights reserved. 18

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-2 Newton's Second Law for a System of Particles Center of mass motion continues unaffected by forces internal to a system (collisions between billiard balls) Motion of a system's center of mass: Reminders: Fnet is the sum of all external forces M is the total, constant, mass of the closed system acom is the center of mass acceleration © 2014 John Wiley & Sons, Inc. All rights reserved. 19

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-2 Newton's Second Law for a System of Particles Examples Using the center of mass motion equation: Billiard collision: forces are only internal, F = 0 so a = 0 © 2014 John Wiley & Sons, Inc. All rights reserved. 20

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-2 Newton's Second Law for a System of Particles Examples Using the center of mass motion equation: Billiard collision: forces are only internal, F = 0 so a = 0 Baseball bat: a = g, so com follows gravitational trajectory Or does it… Figure 9-5 © 2014 John Wiley & Sons, Inc. All rights reserved. 21

© 2014 John Wiley & Sons, Inc. All rights reserved. 6-2 Newton's Second Law for a System of Particles Examples Using the center of mass motion equation: Exploding rocket: explosion forces are internal, so only the gravitational force acts on the system, and the COM follows a gravitational trajectory Figure 9-5 © 2014 John Wiley & Sons, Inc. All rights reserved. 22

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-2 Newton's Second Law for a System of Particles © 2014 John Wiley & Sons, Inc. All rights reserved. 23

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-2 Newton's Second Law for a System of Particles Answer: The system consists of Fred, Ethel and the pole. All forces are internal. Therefore the com will remain in the same place. Since the origin is the com, they will meet at the origin in all three cases! (Of course the origin where the com is located is closer to Fred than to Ethel.) © 2014 John Wiley & Sons, Inc. All rights reserved. 24

Momentum and Newton’s second law The momentum of a particle is the product of its mass and its velocity: What is the momentum of a 1000kg car going 25 m/s west?

Momentum and Newton’s second law The momentum of a particle is the product of its mass and its velocity: What is the momentum of a 1000kg car going 25 m/s west? p = mv = (1000 kg)(25 m/s) = 25,000 kgm/s west

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-3 Linear Momentum Momentum: Points in the same direction as the velocity Can only be changed by a net external force We can write Newton's second law thus: © 2014 John Wiley & Sons, Inc. All rights reserved. 27

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-3 Linear Momentum We can write Newton's second law thus: © 2014 John Wiley & Sons, Inc. All rights reserved. 28

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-3 Linear Momentum © 2014 John Wiley & Sons, Inc. All rights reserved. 29

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-3 Linear Momentum Answer: (a) 1, 3, 2 & 4 (b) region 3 Eq. (9-25) © 2014 John Wiley & Sons, Inc. All rights reserved. 30

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-3 Linear Momentum We can sum momenta for a system of particles to find: © 2014 John Wiley & Sons, Inc. All rights reserved. 31

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-3 Linear Momentum Taking time derivative write Newton's second law for system of particles as: Net external force on system changes linear momentum Without a net external force, the total linear momentum of a system of particles cannot change © 2014 John Wiley & Sons, Inc. All rights reserved. 32

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-5 Conservation of Linear Momentum Without a net external force, the total linear momentum of a system of particles cannot change This is called the law of conservation of linear momentum © 2014 John Wiley & Sons, Inc. All rights reserved. 33

Impulse and momentum Impulse of a force is product of force & time interval during which it acts. Impulse is a vector! On a graph of Fx versus time, impulse equals area under curve.

Note!! J , p, F are all VECTORS!) Impulse and momentum Impulse-momentum theorem: Impulse = Change in momentum J of particle during time interval equals net force acting on particle during interval J = Dp = pfinal – pinitial J = Net Force x time = (SF) x (Dt) so… Dp = (SF) x (Dt) SF = Dp/(Dt) Note!! J , p, F are all VECTORS!)

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-4 Collision and Impulse If F isn’t constant over time…. This means that the applied impulse is equal to the change in momentum of the object during the collision: © 2014 John Wiley & Sons, Inc. All rights reserved. 36

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-4 Collision and Impulse Given Favg and duration: We are integrating: we only need to know the area under the force curve © 2014 John Wiley & Sons, Inc. All rights reserved. 37

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-4 Collision and Impulse © 2014 John Wiley & Sons, Inc. All rights reserved. 38

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-4 Collision and Impulse Answer: (a) unchanged (b) unchanged (c) decreased © 2014 John Wiley & Sons, Inc. All rights reserved. 39

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-4 Collision and Impulse © 2014 John Wiley & Sons, Inc. All rights reserved. 40

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-4 Collision and Impulse Answer: (a) zero (b) positive (c) along the positive y-axis (normal force) © 2014 John Wiley & Sons, Inc. All rights reserved. 41

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-5 Conservation of Linear Momentum For an impulse of zero we find: Which is another way to say momentum is conserved! Law of conservation of linear momentum © 2014 John Wiley & Sons, Inc. All rights reserved. 42

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-5 Conservation of Linear Momentum Check components of net external force to determine if you should apply conservation of momentum © 2014 John Wiley & Sons, Inc. All rights reserved. 43

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-5 Conservation of Linear Momentum Internal forces can change momenta of parts of the system, but cannot change the linear momentum of the entire system © 2014 John Wiley & Sons, Inc. All rights reserved. 44

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-5 Conservation of Linear Momentum Internal forces can change momenta of parts of the system, but cannot change the linear momentum of the entire system Answer: (a) zero (b) no (c) the negative x direction © 2014 John Wiley & Sons, Inc. All rights reserved. 45

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-5 Conservation of Linear Momentum Do not confuse momentum and energy Change in KE => SPEED changes (+ or – directions) Change in P => direction may have changed, or velocity may have changed, or BOTH may have changed… © 2014 John Wiley & Sons, Inc. All rights reserved. 46

Compare momentum and kinetic energy Changes in momentum depend on time over which net force acts But… Changes in kinetic energy depend on the distance over which net force acts.

Ice boats again Two iceboats race on a frictionless lake; one with mass m & one with mass 2m. Wind exerts same force on both. Both boats start from rest, both travel same distance to finish line. Questions! Which crosses finish line with more KE? Which crosses with more Momentum?

Ice boats again Two iceboats race on a frictionless lake; one with mass m and one with mass 2m. The wind exerts the same force on both. Both boats start from rest, and both travel the same distance to the finish line? Which crosses the finish line with more KE? In terms of work done by wind? W = DKE = Force x distance = same; So ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22

Ice boats again Two iceboats race on a frictionless lake; one with mass m and one with mass 2m. The wind exerts the same force on both. Both boats start from rest, and both travel the same distance to the finish line? Which crosses the finish line with more KE? In terms of work done by wind? W = DKE = Force x distance = same; So ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22 Since m2 > m1, v2 < v1 so more massive boat loses

Ice boats again Two iceboats race on a frictionless lake; one with mass m and one with mass 2m. The wind exerts the same force on both. Both boats start from rest, and both travel the same distance to the finish line? Which crosses the finish line with more p? In terms of momentum? Force on the boats is the same for each, but TIME that force acts is different. The second boat accelerates slower, and takes a longer time. Since t2> t1, p2 > p1 so second boat has more momentum

p2 > p1 Ice boats again ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22 Check with equations! Force of wind = same; distance = same Work done on boats is the same, gain in KE same. ½ m1v12 = DKE1 = W = DKE2 = ½ m2v22 And m1v1 = p1; m2v2 = p2 ½ m1v12 = ½ (m1v1) v1 = ½ p1 v1 & same for ½ p2 v2 ½ p1 v1 = ½ p2 v2 Since V1 > V2, P2 must be greater than P1! p2 > p1

A ball hits a wall A 0.40 kg ball moves at 30 m/s to the left, then rebounds at 20 m/s to the right from a wall.

A ball hits a wall A 0.40 kg ball moves at 30 m/s to the left, then rebounds at 20 m/s to the right from a wall. What is the impulse of net force during the collision, and if it is in contact for 0.01 s, what is the average force acting from the wall on the ball?

Remember that momentum is a vector! When applying conservation of momentum, remember that momentum is a vector quantity!

Remember that momentum is a vector! When applying conservation of momentum, remember that momentum is a vector quantity! Use vector addition to add momenta in COMPONENTS!

Kicking a soccer ball – Example 8.3 Soccer ball 0.40 kg moving left at 20 m/s, then kicked up & to the right at 30 m/s at 45 degrees. If collision time is 0.01 seconds, what is impulse?

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-6 Types of Collisions Elastic collisions: Total kinetic energy is unchanged (conserved) A useful approximation for common situations In real collisions, some energy is always transferred Inelastic collisions: some energy is transferred Completely inelastic collisions: The objects stick together Greatest loss of kinetic energy © 2014 John Wiley & Sons, Inc. All rights reserved. 58

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-6 Momentum and Kinetic Energy in Collisions For one dimension inelastic collision © 2014 John Wiley & Sons, Inc. All rights reserved. 59

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-6 Momentum and Kinetic Energy in Collisions Completely inelastic collision, for target at rest: © 2014 John Wiley & Sons, Inc. All rights reserved. 60

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-6 Momentum and Kinetic Energy in Collisions © 2014 John Wiley & Sons, Inc. All rights reserved. 61

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-6 Momentum and Kinetic Energy in Collisions Answer: (a) 10 kg m/s (b) 14 kg m/s (c) 6 kg m/s © 2014 John Wiley & Sons, Inc. All rights reserved. 62

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-7 Elastic Collisions in One Dimension Total kinetic energy is conserved in elastic collisions © 2014 John Wiley & Sons, Inc. All rights reserved. 63

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-7 Elastic Collisions in One Dimension For a stationary target, conservation laws give: Figure 9-18 © 2014 John Wiley & Sons, Inc. All rights reserved. 64

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-7 Elastic Collisions in One Dimension With some algebra we get: Results Equal masses: v1f = 0, v2f = v1i: the first object stops Massive target, m2 >> m1: the first object just bounces back, speed mostly unchanged Massive projectile: v1f ≈ v1i, v2f ≈ 2v1i: the first object keeps going, the target flies forward at about twice its speed © 2014 John Wiley & Sons, Inc. All rights reserved. 65

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-7 Elastic Collisions in One Dimension For a target that is also moving, we get: © 2014 John Wiley & Sons, Inc. All rights reserved. 66

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-8 Collisions in Two Dimensions Apply conservation of momentum along each axis Apply conservation of energy for elastic collisions Example For a stationary target: Along x: Along y: Energy: © 2014 John Wiley & Sons, Inc. All rights reserved. 67

A two-dimensional collision Two robots collide and go off at different angles. You must break momenta into x & y components and deal with each direction separately

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-8 Collisions in Two Dimensions © 2014 John Wiley & Sons, Inc. All rights reserved. 69

© 2014 John Wiley & Sons, Inc. All rights reserved. 9-8 Collisions in Two Dimensions Answer: (a) 2 kg m/s (b) 3 kg m/s © 2014 John Wiley & Sons, Inc. All rights reserved. 70

Elastic collisions In elastic collision, total momentum of system in a direction is same after collision as before… if no external forces act in that direction: Pix = Pfx mavai +mbvbi =mavaf +mbvbf But wait – there’s more!

Objects colliding along a straight line Two gliders collide on an frictionless track. What are changes in velocity and momenta?

Elastic collisions In an elastic collision, the total kinetic energy of the system is the same after the collision as before. KEi = KEf ½ mavai2 + ½ mbvbi2 = ½ mavaf2 + ½ mbvbf2 =

Elastic collisions In an elastic collision, Difference in velocities initially = (-) difference in velocities finally (vai - vbi) = - (vaf – vbf)

Elastic collisions In an elastic collision, Difference in velocities initially = (-) difference in velocities finally (vai - vbi) = - (vaf – vbf) Solve generally for final velocities: vaf = (ma-mb)/(ma+mb)vai + 2mb/(ma+mb)vbi vbf = (mb-ma)/(ma+mb)vbi + 2ma/(ma+mb)vai

Elastic collisions  

Elastic collisions Behavior of colliding objects is greatly affected by relative masses.

Elastic collisions  

Elastic collisions Behavior of colliding objects is greatly affected by relative masses.

Elastic collisions  

Elastic collisions Behavior of colliding objects is greatly affected by relative masses. Stationary Bowling Ball!

Inelastic collisions In any collision where external forces can be neglected, total momentum conserved. Collision when bodies stick together is completely inelastic collision In inelastic collision, total kinetic energy after collision is less than before collision.

Some inelastic collisions Cars are intended to have inelastic collisions so the car absorbs as much energy as possible.

The ballistic pendulum Ballistic pendulums are used to measure bullet speeds

A 2-dimensional automobile collision Two cars traveling at right angles collide.

An elastic straight-line collision

Neutron collisions in a nuclear reactor

A two-dimensional elastic collision

Rocket propulsion Conservation of momentum holds for rockets, too! F = dp/dt = d(mv)/dt = m(dv/dt) + v(dm/dt) As a rocket burns fuel, its mass decreases (dm< 0!)

Rocket propulsion Initial Values m = initial mass of rocket v = initial velocity of rocket in x direction mv = initial x-momentum of rocket vexh = exhaust velocity from rocket motor (This will be constant!) (v - vexh) = relative velocity of exhaust gases

Rocket propulsion Changing values dm = decrease in mass of rocket from fuel dv = increase in velocity of rocket in x-direction dt = time interval over which dm and dt change

Rocket propulsion Final values m + dm = final mass of rocket less fuel ejected v + dv = increase in velocity of rocket (m + dm) (v + dv) = final x-momentum of rocket (+x direction) [- dm] (v - vexh) = final x-momentum of fuel (+x direction)

final momentum of the ejected mass of gas Rocket propulsion Final values mv = [(m + dm) (v + dv)] + [- dm] (v - vexh) initial momentum final momentum of the lighter rocket final momentum of the ejected mass of gas

Rocket propulsion Final values mv = [(m + dm) (v + dv)] + [- dm] (v - vexh) mv = mv + mdv + vdm + dmdv –dmv +dmvexh

Rocket propulsion Final values mv = [(m + dm) (v + dv)] + [- dm] (v - vexh) mv = mv + mdv + vdm + dmdv –dmv +dmvexh

Rocket propulsion Final values mv = mv + mdv + vdm + dmdv –dmv +dmvexh

Rocket propulsion Final values 0 = mdv + dmdv + dmvexh

Rocket propulsion Final values 0 = mdv + dmdv + dmvexh Neglect the assumed small term: m(dv) = -dmdv – (dm)vexh

Rocket propulsion Final values m(dv) = – (dm)vexh Gain in momentum of original rocket is related to rate of mass loss and exhaust velocity of gas!

Rocket propulsion Final values m(dv) = – (dm)vexh Gain in momentum of original rocket is related to rate of mass loss and exhaust velocity of gas! Differentiate both sides with respect to time!

Rocket propulsion Final values m(dv) = – (dm)vexh m(dv/dt) = – [(dm)/dt] vexh

Rocket propulsion Final values m(dv) = – (dm)vexh m(dv/dt) = – [(dm)/dt] vexh ma = Force (Thrust!) = – [(dm)/dt] (vexh)

Rocket propulsion Final values m(dv) = – (dm)vexh m(dv/dt) = – [(dm)/dt] vexh ma = Force (Thrust!) = – [(dm)/dt] (vexh) rate of change of mass Velocity of gas exhausted

Rocket propulsion Thrust = – [(dm)/dt] (vexh) Example: vexhaust = 1600 m/s Mass loss rate = 50 grams/second Thrust =?

Rocket propulsion Final values Thrust = – [(dm)/dt] (vexh) Example vexhaust = 1600 m/s Mass loss rate = 50 grams/second Thrust = -1600 m/s (-0.05 kg/1 sec) = +80N

Rocket propulsion Gain in speed? m(dv) = – (dm)vexh dv = – [(dm)/m] vexh integrate both sides

Rocket propulsion Gain in speed? m(dv) = – (dm)vexh dv = – [(dm)/m] vexh (integrate both sides) vf - vi = (vexh) ln(m0/m) m0 = initial mass m0 > m, so ln > 1 “mass ratio” gain in velocity Velocity of gas exhausted

Rocket propulsion – Gain in Speed vf - vi = (vexh) ln(m0/m) m0 = initial mass m0 > m, so ln > 1 “mass ratio” gain in velocity Velocity of gas exhausted Faster exhaust, and greater difference in mass, means a greater increase in speed!

Rocket propulsion Gain in speed? vf - vi = (vexh) ln(m0/m) Example: Rocket ejects gas at relative 2000 m/s. What fraction of initial mass is not fuel if final speed is 3000 m/s?

Rocket propulsion Gain in speed? vf - vi = (vexh) ln(m0/m) Example: Rocket ejects gas at relative 2000 m/s. What fraction of initial mass is not fuel if final speed is 3000 m/s? ln(mo/m) = (3000/2000) = 1.5 so m/mo = e-1.5 = .223

Rocket propulsion Gain in speed? vf - vi = (vexh) ln(m0/m) m0 = initial mass m0 > m, so ln > 1 “mass ratio” gain in velocity Velocity of gas exhausted STAGE rockets to throw away mass as they use up fuel, so that m0/m is even higher!