Center of Mass & Linear

MomentumPhysicsMontwood High SchoolR. Casao

Center of Mass We have discussed parabolic trajectories using a

“particle” as the model for objects. But clearly objects are not particles – they are

extended and may have complicated shapes & mass distributions.

So if we toss something like a baseball bat into the air (spinning and rotating in a complicated way), what can we really say about it’s trajectory?

There is one special location in every object that provides us with the basis for our earlier model of a point particle.

That special location is called the center of mass. The center of mass will follow a parabolic trajectory –

even if the rest of the bat’s motion is very complicated

Center of Mass To start, let’s suppose that we have two masses m1

and m2, separated by some distance d. We have also arbitrarily aligned the origin of our

coordinate system to be the center of mass m1. We define the center of mass for these two particles

to be:

dmm

mx

21

2com

Center of Mass From this we can see that if

m2 = 0, then xcom = 0.

Similarly, if m1 = 0, then xcom = d.

Finally, if m1 = m2, then xcom = ½·d.

So we can see that the center of mass in this case is constrained to be somewhere between x = 0 and x = d.

Center of Mass Now lets shift the origin of

the coordinate system a little.

We now need a more general definition of the center of mass.

The more general definition (for two particles) is:

21

2211com mm

xmxmx

Center of Mass Now let’s suppose that we have lots of particles –

all lined up nicely for us on the x axis The equation would now be:

where M = m1 + m2 + … + mn

The collection of terms in the numerator can be rewritten as a sum resulting in:

M

xmxmxmx nn

...2211

com

n

iii xm

Mx

1com

1

Center of Mass This result is only for

one dimension however, so the more generalized result for 3 dimensions is shown here:

n

iii

n

iii

n

iii

zmM

z

ymM

y

xmM

x

1com

1com

1com

1

1

1

Center of Mass You know by intuition that the center of mass of

a sphere is at the center of the sphere; for a rod it lies along the central axis of the rod; for a flat plate it lies in the plane of the plate.

Note however that the center of mass doesn’t necessarily have to lie within the object or have any mass at that point: The COM of a horseshoe is somewhere in the

middle along the axis of symmetry. The COM of a doughnut is at it’s ‘geographic’

center, but there is no mass there either.

Center of Mass

Going back to our baseball bat, the COM will lie along the central axis (the axis of symmetry).

And it is the COM that faithfully follows the line of a parabola.

Newton’s 2nd Law for a System of Particles We know from experience that if you roll the cue ball into

another billiard ball that is at rest, that the two ball system will continue on away from you after the impact.

You would be very surprised if one or the other of the balls came back to you (without putting ‘English’ on the ball).

What continued to move away from you was the COM of the two ball system!

Remember that the COM is a point that acts as though all of the mass in the system were located there.

So even though we may have a large number of particles – possibly of different masses, we can treat the assembly as having all of it’s mass at the point of it’s COM.

Newton’s 2nd Law for a System of Particles So we can assign that point a position, a

velocity and an acceleration. And it turns out that Newton’s 2nd law holds

for that point:

where Fnet is the sum of all the external forces acting on the mass; M is the total mass; and acom is the acceleration of the center of mass.

comnet aMF

Newton’s 2nd Law for a System of Particles Circling back around to where we started from,

we can break the one vector equation down into an equation for each dimension:

z com,z net,

y com,y net,

xcom, xnet,

aMF

aMF

aMF

Newton’s 2nd Law for a System of Particles Now let’s re-examine what is going on when we have

the collision between the two billiard balls: Once the cue ball has been set in motion, there are no external

forces acting on the system. Thus, Fnet = 0 and as a consequence, acom = 0. This means that the velocity of the COM of the system must be

constant. So when the two balls collide, the forces must be

internal to the system (the balls exert forces on each other; no external forces act on the balls) – which doesn’t affect Fnet.

Thus the COM of the system continues to move forward unchanged by the collision.

Newton’s 2nd Law for a System of Particles When a fireworks rocket

explodes, the COM of the system does not change; while the fragments all fan out, their COM continues to move along the original path of the rocket.

This is also how a ballerina seems to defy the laws of physics and float across the stage when doing a grand jeté.

Suppose a cannon shell traveling in a parabolic trajectory (neglecting air friction) explodes in flight, splitting into two fragments of equal mass.

The fragments follow new parabolic paths, but the center of mass continues on the original parabolic path as if all the mass were still concentrated at that point.

Linear Momentum Linear momentum of an object is the mass of the

object multiplied by its velocity. Momentum: p = m·vcom Unit: kg·m/s or N·s Newton expressed his 2nd law of motion in terms of

momentum: The time rate of change of the momentum of a

particle is equal to the net force acting on the particle and is in the direction of that force.

Both momentum and kinetic energy describe the motion of an object and any change in mass and/or velocity will change both the momentum and kinetic energy of the object.

dt

pdFnet

Linear Momentum Momentum refers to inertia in motion. Momentum is a measure of how difficult it is to

stop an object; a measure of “how much motion” an object has.

More force is needed to stop a baseball thrown at 95 mph than to stop a baseball thrown at 45 mph, even though they both have the same mass.

More force is needed to stop a train moving at 45 mph than to stop a car moving at 45 mph, even though they both have the same speed.

Both mass and velocity are important factors when considering the force needed to change the motion of an object.

Impulse Impulse (J) = force·time Equation: J = F·t Unit: N·s The impulse of a force is equal to the change

in momentum of the body to which the force is applied. This usually means a change in velocity.

F·t = m·v where v = vf - vi

The same change in momentum can be accomplished by a small force acting for a long time or by a large force acting for a short time.

Impulse

The area under the curve in a force vs. time graph represents the change in momentum (m·v).

If your car runs into a brick wall and you come to rest along with the car, there is a significant change in momentum. If you are wearing a seat belt or if the car has an air bag, your change in momentum occurs over a relatively long time interval. If you stop because you hit the dashboard, your change in momentum occurs over a very short time interval.

Impulse If a seat belt or air bag brings you to a stop over a time

interval that is five times as long as required to stop when you strike the dashboard, then the forces involved are reduced to one-fifth of the dashboard values. That is the purpose of seat belts, air bags, and padded dashboards. By extending the time during which you come to rest, these safety devices help reduce the forces exerted on you.

If you want to increase the momentum of an object as much as possible, you apply the greatest force you can for as long a time as possible.

A 1000 kg car moving at 30 m/s (p = 30,000 kg m/s) can be stopped by 30,000 N of

force acting for 1.0 s (a crash!)

or by 3000 N of force acting for 10.0 s (normal stop)or by 3000 N of force acting for 10.0 s (normal stop)

Impulse and Bouncing Impulses are greater

when bouncing takes place.

The impulse required to bring an object to a stop and then throw it back again is greater than the impulse required to bring an object to a stop.

Conservation of Linear Momentum

In a closed system of objects, linear momentum is conserved as the objects interact or collide. The total vector momentum of the system remains constant.

p before interaction = p after interaction

Collisions A collision is an isolated event in which

two or more bodies (the colliding bodies) exert relatively strong forces on each other for a relatively short time.

An interesting point to note is that the definition doesn’t necessarily require the bodies to actually make contact – a near miss (near enough so that there are “relatively strong forces” involved) will suffice…

Collisions To analyze a collision we

have to pay attention to the 3 parts of a collision: Before During After

Momentum & Kinetic Energy in Collisions If we have a system where two bodies collide, then there must be some kinetic energy (and therefore linear momentum) present.

During the collision, the kinetic energy and linear momentum of each body is changed by the impulse from the other body.

We define two different kinds of collisions: An elastic collision is one where the total kinetic

energy of the system is conserved. An inelastic collision is one where the total kinetic

energy is not conserved. A collision is not necessarily completely elastic

or completely inelastic – most are in fact somewhere in between…

Momentum & Kinetic Energy in Collisions For any kind of collision however, linear momentum must be preserved.

This is because in a closed, isolated system, the total linear moment cannot change without the presence of an external force (and the forces involved in a collision are internal to the system – not external).

This does not mean that the linear momentum of the various colliding bodies cannot change.

The linear momentum of each body involved in the collision may indeed change, but the total linear momentum of the system cannot change.

Perfectly Inelastic Collisions Perfectly inelastic

collisions are those in which the colliding objects stick together and move with the same velocity.

Kinetic energy is lost to other forms of energy in an inelastic collision.

Inelastic Collision Example Cart 1 and cart 2 collide and stick together Momentum equation:

Kinetic energy equation:

v1 and v2 = velocities before collisionv = velocity after collision

'v)mm(vmvm 212211

iflost

221f

222

211i

KKK

v')m(m0.5K

vm0.5vm0.5K

Directions for Velocity Momentum is a vector, so direction is

important. Velocities are positive or negative to indicate

direction. Example: bounce a ball off a wall

Inelastic Collisions

Kinetic energy is lost when the objects are deformed during the collision.

Momentum is conserved.

Elastic Collisions Momentum and

kinetic energy are conserved in an elastic collision.

The colliding objects rebound from each other with NO loss of kinetic energy.

Elastic Collision Example Example: mass 1 and mass 2 collide and

bounce off of each other Momentum equation:

Kinetic energy equation:

v1 and v2 = velocities before collisionv1 and v2 = velocities after collision Velocities are + or – to indicate directions.

'vm'vmvmvm 22112211

222

211

222

211 'vm5.0'vm5.0vm5.0vm5.0

Elastic Collisions Involving an Angle Momentum is conserved in both the x-

direction and in the y-direction. Before:

positivesinθvv

positivecosθvv

negativesinθvv

positivecosθvv

22y2

22x2

11y1

11x1

Elastic Collisions Involving an Angle

After:

negativeθsin'v'v

positiveθcos'v'v

positiveθsin'v'v

positiveθcos'v'v

42y2

42x2

31y1

31x1

Elastic Collisions Involving an Angle

Directions for the velocities before and after the collision must include the positive or negative sign.

The direction of the x-components for v1 and v2 do not change and therefore remain positive.

The directions of the y-components for v1 and v2 do change and therefore one velocity is positive and the other velocity is negative.

Elastic Collisions Involving an Angle

px before = px after

py before = py after

Velocity after collision:

'vm'vmvmvm x22x11x22x11

'vm'vmvmvm y22y11y22y11

2y2

2x22

2y1

2x11

'v'v'v

'v'v'v

Elastic Collisions

Perfectly elastic collisions do not have to be head-on.

Particles can divide or break apart.

Example: nuclear decay (nucleus of an element emits an alpha particle and becomes a different element with less mass)

Elastic Collisions

mn = mass of nucleus mp = mass of alpha particle vn = velocity of nucleus before event vn’ = velocity of nucleus after event vp = velocity of particle after event

ppnpnnn vm'vmmvm

Recoil Recoil is the term that describes the backward

movement of an object that has propelled another object forward. In the nuclear decay example, the vn’ would be the recoil velocity.

bulletbulletgungunbulletgun vmvmv)m(m

Head-on and Glancing Collisions

Head-on collisions occur when all of the motion, before and after the collision, is along one straight line.

Glancing collisions involve an angle. A vector diagram can be used to

represent the momentum for a glancing collision.

Vector Diagrams

Use the three vectors and construct a triangle.

Vector Diagrams

Use the appropriate expression to determine the unknown variable.

35sin'

30sin'

35sin'

115sin

30sin'

115sin

BBRR

BBBB

RRBB

vmvm

vmvm

vmvm

Vector Diagrams

Total vector momentum is conserved. You could break each momentum vector into an x and y component.

px before = px afterpy before = py after

You would use the x and y components to determine the resultant momentum for the object in question

Resultant momentum = 2y

2x pp

Vector Diagrams Right triangle trigonometry can be

used to solve this type of problem:

Vector Diagrams

Pythagorean theorem:

If the angle for the direction in which the cars go in after the collision is known, you can use sin, cos, or tan to determine the unknown quantity. Example: determine final velocity vT if the angle is 25°.

2Tba2

bb2

aa vmmvmvm

Tba

bb

Tba

aa

vmmvm

25cos

vmmvm

25sin

Vector Diagrams

To determine the angle at which the cars go off together after the impact:

bb

aa1

vmvm

tanθ

Special Condition

When a moving ball strikes a stationary ball of equal mass in a glancing collision, the two balls move away from each other at right angles.

ma = mb

va = 0 m/s

Special Condition

Use the three vectors to construct a triangle.

Special Condition

Use the appropriate expression to determine the unknown variable.

40sin'

50sin'

40sin'

90sin

50sin'

90sin

BBAA

BBBB

AABB

vmvm

vmvm

vmvm

Ballistic Pendulum

In the ballistic pendulum lab, a ball of known mass is shot into a pendulum arm. The arm swings upward and stops when its kinetic energy is exhausted.

From the measurement of the height of the swing, one can determine the initial speed of the ball.

This is an inelastic collision. As always, linear momentum is conserved.

Ballistic Pendulum

Ballistic Pendulum

Potential energy of ball in gun:

Ball embeds in pendulum:

Ballistic Pendulum

Pendulum rises to a maximum height:

Solving for the initial speed of the projectile we get:

hg2m

mmv

b

pbb

Series of Collisions What happens when there are multiple

collisions to consider? Consider an object that is securely bolted

down so it can’t move and is continuously pelted with a steady stream of projectiles.

Each projectile has a mass of m and is moving at velocity v along the x axis.

Series of Collisions The linear momentum of each projectile is

m·v. Suppose that n projectiles arrive in an

interval of Δt. As each projectile hits (and is absorbed by)

the mass, the change in the projectile’s linear momentum is Δp; thus the total change in linear momentum during the interval Δt is n·Δp.

Series of Collisions The total impulse on the target is the same

magnitude but opposite in direction to the change in linear momentum – thus:

But we also know that:

This leads us to:

J n p

avg

JF

t

avg

J n nF p m v

t t t

Series of Collisions

So now we have the average force in terms of the rate at which projectiles collide with the target (n/Δt) and each projectile’s change in velocity (Δv).

In the time interval Δt, we also know that an amount of mass Δm = n·m collides with the target.

Thus our equation for the average force finally turns out to be:

avg

mF v

t

Series of Collisions Assuming that each projectile stops upon

impact, we know that:

In this case the average force is:

vvvvv if 0

avg

mF v

t

Series of Collisions Suppose instead that each projectile

rebounds (bounces directly back) upon impact – in this case we have:

In this case the average force is:

2fiv v v v v v

avg 2m

F vt

Elastic Collision Example Example: mass 1 and mass 2 collide and

bounce off of each other Momentum equation:

Kinetic energy equation:

v1 and v2 = velocities before collisionv1 and v2 = velocities after collision

Velocities are + or – to indicate directions.

'vm'vmvmvm 22112211

222

211

222

211 'vm5.0'vm5.0vm5.0vm5.0

Elastic Collision Example

Working with kinetic energy:

0.5 cancels out.

222

211

222

211 'vm5.0'vm5.0vm5.0vm5.0

2

12

1

22

22

2

1

22

222

21

211

222

222

211

211

222

211

222

211

'vv

v'v

mm

v'vm'vvm

vm'vm'vmvm

'vm'vmvmvm

Elastic Collision Example

The velocity terms are perfect squares and can be factored:

a2-b2 = (a – b)·(a + b)

We will use this equation later.

'vv'vv

v'vv'v

mm

1111

2222

2

1

Elastic Collision Example

Momentum equation:

'vv

v'v

mm

v'vm'vvm

vm'vm'vmvm

'vm'vmvmvm

11

22

2

1

222111

22221111

22112211

Elastic Collision Example

Both the kinetic energy and momentum equations have been solved for the ratio of m1/m2.

Set m1/m2 for kinetic energy equal to m1/m2

for momentum:

'vv

v'v

'vv'vv

v'vv'v

11

22

1111

2222

Elastic Collision Example

Get all the v1 terms together and all the v2 terms together:

Cancel the like terms:

2 2 2 2 1 1 1 1

2 2 1 1

' ' ' '

' '

v v v v v v v v

v v v v

'vvv'v 1122

Elastic Collision Example Rearrange to get the initial and final

velocities back together on the same side of the equation:

This equation can be solved for one of the two unknowns (v1´ or v2´), then substituted back into the conservation of momentum equation.

'v'vvv 2112

Change in Momentum Example A 0.5 kg rubber ball is thrown towards a wall with a

velocity of 14 m/s. It hits the wall, causing it to deflect in the opposite direction with a speed of 9 m/s. What is the change in the ball’s momentum?

∆p = F·∆t∆p = m·vf – m·vi ∆p = (0.5 kg)·(-9 m/s) – (0.5 kg)·(14 m/s)∆p = (-4.5 kg·m/s) – (7 kg·m/s)∆p = -11.5 kg·m/s

The change in the ball’s momentum is 11.5 kg·m/s.