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

Center of Mass and Linear Momentum

Chapter 9

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

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.

9-1 Center of Mass

DEFINITION:

center of mass (com) of a system of particles:

For two particles separated by a distance d, where the origin is chosen at the position of particle 1:

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:

9-1 Center of Mass

For many particles, we can generalize the equation, where M = m

2 + . . . + m

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:

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

9-1 Center of Mass

If objects have uniform density = ρ (“rho”)

Substituting, we find the center of mass simplifies:

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)

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

9-1 Center of Mass

Example Subtractingo Task: find COM of a disk with another

disk taken out of it:

o Find the COM of the two individual COMs (one for each disk), treating the cutout as having negative mass

Figure 9-4

9-1 Center of Mass

Example Subtractingo On the diagram,

comC is the center

of mass for Plate P and Disk S combined

o comP is the center

of mass for the composite plate with Disk S removed

Figure 9-4

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:

1. Fnet

is the sum of all external forces

2. M is the total, constant, mass of the closed system

3. acom

is the center of mass acceleration

Examples Using the center of mass motion equation:o Billiard collision: forces are only internal, F = 0 so a = 0

Examples Using the center of mass motion equation:o Billiard collision: forces are only internal, F = 0 so a = 0o Baseball bat: a = g, so com follows gravitational trajectory

o Or does it…

Figure 9-5

Examples Using the center of mass motion equation:o 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

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.)

Momentum and Newton’s second law

The momentum of a particle is the product of its mass and its velocity:

m .p = v

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:

m .p = v

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

9-3 Linear Momentum

Momentum:o Points in the same direction as the velocity

o Can only be changed by a net external force

We can write Newton's second law thus:

9-3 Linear Momentum

We can write Newton's second law thus:

9-3 Linear Momentum

Answer: (a) 1, 3, 2 & 4 (b) region 3

Eq. (9-25)

9-3 Linear Momentum

We can sum momenta for a system of particles to find:

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

This is called the law of conservation of linear momentum

9-5 Conservation of Linear Momentum

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.

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)

Dp = (SF) x (Dt)

SF = Dp/(Dt)

Note!!

J , p, F

are all

VECTORS!)

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:

Given Favg

and duration:

We are integrating: we only need to know the area under the force curve

Answer: (a) unchanged (b) unchanged (c) decreased

Answer: (a) zero (b) positive (c) along the positive y-axis (normal force)

For an impulse of zero we find:

Which is another way to say momentum is conserved!

Law of conservation of linear momentum

Check components of net external force to determine if you should apply conservation of 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

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…

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

• 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

• 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

Ice boats again

• 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 m2v2

• 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?

9-6 Types of Collisions

Elastic collisions:o Total kinetic energy is unchanged (conserved)o A useful approximation for common situationso In real collisions, some energy is always transferred

Inelastic collisions: some energy is transferred

Completely inelastic collisions:o The objects stick togethero Greatest loss of kinetic energy

For one dimension inelastic collision

9-6 Momentum and Kinetic Energy in Collisions

Completely inelastic collision, for target at rest:

Answer: (a) 10 kg m/s (b) 14 kg m/s (c) 6 kg m/s

9-7 Elastic Collisions in One Dimension

Total kinetic energy is conserved in elastic collisions

For a stationary target, conservation laws give:

Figure 9-18

With some algebra we get:

Results

o Equal masses: v1f = 0, v

2f = v

1i: the first object stops

o Massive target, m2 >> m

1: the first object just bounces

back, speed mostly unchanged

o Massive projectile: v1f ≈ v

2f ≈ 2v

1i: the first object

keeps going, the target flies forward at about twice its speed

For a target that is also moving, we get:

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:o Along x:

o Along y:

o Energy:

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

Answer: (a) 2 kg m/s (b) 3 kg m/s

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 mbvbi

2 = ½ mavaf

2 + 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

Behavior of colliding objects is greatly affected by relative masses.

Elastic collisions

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

mv = initial x-momentum of rocket

vexh = exhaust velocity from rocket motor

(This will be constant!)

(v - vexh) = relative velocity of exhaust gases

m = initial mass of rocket

v = initial velocity of rocket

in x direction

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)

Rocket propulsion• Final values

mv = [(m + dm) (v + dv)] + [- dm] (v - vexh)

initial momentumfinal momentum of the lighter

rocket

final momentum of the ejected

mass of gas

Rocket propulsion

• Final values

mv = mv + mdv + vdm + dmdv –dmv +dmvexh

Rocket propulsion

• Final values

Rocket propulsion

• Final values

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

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/dt) = – [(dm)/dt] vexh

Rocket propulsion

• Final values

ma = Force (Thrust!) = – [(dm)/dt] (vexh)

Rocket propulsion

• Final values

ma = Force (Thrust!) = – [(dm)/dt] (vexh)

rate of change of

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?

dv = – [(dm)/m] vexh

integrate both sides

Rocket propulsion

• Gain in speed?

dv = – [(dm)/m] vexh (integrate both sides)

vf - vi = (vexh) ln(m0/m)

m0 = initial mass

m0 > m, so ln > 1

“mass ratio”

Velocity of gas exhaustedgain in velocity

Rocket propulsion – Gain in Speed

m0 = initial mass

m0 > m, so ln > 1

“mass ratio”

Faster exhaust, and greater difference in mass, means a greater increase in speed!

Rocket propulsion

• Gain in speed?

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?

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?

m0 = initial mass

m0 > m, so ln > 1

“mass ratio”

STAGE rockets to throw away mass as they use up fuel, so that m0/m is even higher!

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