Physics 1ALecture 8A

"The sun rises. In that short phrase, in a single fact, is enough information to keep biology, physics, and

philosophy busy for the rest of time.”--Lyall Watson

Center of MassUp until now, we have had to deal with simple objects: balls, particles, etc.

Complicated objects such as boomerangs, people, etc. can have different parts moving in different directions at the same time.

We define a special point to deal with complicated objects, we call it the center of mass (COM).

The center of mass of a system of particles is the point that moves as though all of the system’s mass were concentrated there and all of the external forces were applied there.

Center of MassA boomerang, for instance, will move like a particle at its center of mass point. But one edge may move backwards while the other edge moves forward.

The center of mass is, thus, important so that we can relate previous results, such as Newton’s Laws, to complicated objects.

How do we find the center of mass for different objects?

For discrete objects, like particles, we merely have to take into account the position and mass of all the particles.

Center of MassFor n discrete particles in one dimension we can write:

where mi is the mass of ith particle and xi is the position of the ith particle and M is the total mass.

Thus, M = m1 + m2 + m3 +...

If we are to move to three dimensions we can rewrite as:

€

xCOM =1M

mixii=1

n

∑

€

r COM =1M

mi r i

i=1

n

∑ where:

€

r = xˆ i + yˆ j + z ˆ k

Center of MassNow, if we want to move to actual solid objects, we can sum up differential masses to get:

where the mass must be uniform to use these equations.

We can extend this to three dimensions to get:

€

xCOM =1M

xdm∫

€

yCOM =1M

ydm∫

€

zCOM =1M

zdm∫

€

R = r COM =

1M

r dm∫

Center of MassMany times we would like to take the integral over the volume instead of the mass.

In this case we look at density, ρ:

This makes xCOM to be: €

ρ =MV

=dmdV

€

dm = MVdV

€

xCOM =1M

xdm∫ = 1M xMVdV

⎛

⎝ ⎜

⎞

⎠ ⎟ ∫

€

xCOM =1V

xdV∫

Similarly,

€

yCOM =1V

ydV∫

€

zCOM =1V

zdV∫

Newton’s Laws COMSince the center of mass is now calculable, we can use this to describe the motion of complicated objects.

We will treat complicated objects as point particles with all of the forces acting at the location of the center of mass.

Fgravity

Newton’s 2nd Law can be written as:

€

F ∑ = M a COM

where M = total mass.ΣF are external forces to the system.aCOM is the acceleration of the center of mass.

Linear Momentum COMRecall from earlier in the quarter, we defined linear momentum, p, as a vector that measures your motion:

The units of momentum are: [kg m/s]

As we mentioned before, the correct form of Newton’s Second Law is:

If the mass is constant:

€

p = m v

€

F ∑ = d

p dt

€

F ∑ = d

p dt

=d m v ( )

dt= m d

v dt

= m a

MomentumIf we have a system of particles or a complicated body:

So, essentially, if there is no net external force present upon a system; then its momentum will remain constant.

€

p com = M v com

where M is the mass of the total system and vcom is the velocity of the center of mass.

We can extend Newton’s Second Law to complicated systems via:

€

F ext∑ =

d p comdt

MomentumMathematically we say:

This is the law of conservation of momentum.

Also, since perpendicular directions are independent of one another, momentum in the x-direction is completely independent of momentum in the y-direction.

Energy and momentum do not both have to be conserved at the same time!

So in a closed, isolated system:

€

F ext∑ =

d p comdt

= 0

€

p com = constant

€

p i = p f

MomentumExampleTwo cars (A and B) have the same mass and velocity but are headed in opposite directions. They collide and come to rest. a) Is mechanical energy conserved? b) Is momentum conserved?

AnswerFirst, you must define a coordinate system.Let’s choose the direction of car A as the positive x-direction.

A Bv v

MomentumLet’s also define A and B as a single system. Every force between them is now internal.

Initially, the mechanical energy of the system is:Emec = KEA + PEA + KEB + PEB

Emec = KEA + 0 + KEB + 0 = (1/2)mv2 + (1/2)mv2Emec = mv2

Finally, the mechanical energy of the system is:Emec = KEA + PEA + KEB + PEB

Emec = 0 + 0 + 0 + 0 = 0

Here, Emec initial ≠ Emec final

Answer

MomentumNext, calculate momentum.

Initially, the momentum of the system is:

Answer

Finally, the momentum of the system is:

Thus, momentum is conserved, since:

ImpulseWhen applying a force to an object to change its motion, the time in which the force is applied is important in determining the end results.

Impulse, I, is a vector that measures how much force has changed motion:

We can also represent it as how much the momentum has changed:

€

I = Δ p = p f −

p i

€

I =

F

t1

t2

∫ dt

This is the Impulse-momentum theorem, that impulse is equivalent to the change in momentum.

MomentumThe impulse-momentum theorem basically states that your momentum will change if you apply a net force over any time period.

If the force is not constant, then use the average force applied.

The impulse imparted by a force during Δt is equal to the area under the force-time graph.

OR the impulse is equal to the average force multiplied by the time interval:

€

Δ p =

F avg( )Δt

MomentumExampleA soccer player kicks a ball (initially at rest) with a velocity of 20m/s. The mass of the ball is 0.45kg. The duration of the impact of her foot with the ball is 0.05s. a) What is the change in momentum of the ball? b) What is the average force applied to the ball by her foot?

AnswerFirst, you must define a coordinate system.Let’s choose the direction of motion of the soccer ball as the positive x-direction.

MomentumLook at the change in momentum:Answer

Since the ball is initially at rest: vi = 0. Such that:

From the Impulse-momentum theorem:

Conceptual QuestionA 1kg ball is thrown horizontally towards a wall with a speed of 10m/s. The initial velocity is chosen to be the positive x-direction for this question. The ball horizontally rebounds back from the wall with a speed of 10m/s in the negative x-direction. What is the change in momentum, Δp, of the ball?

v

A) 0 kg(m/s) î.

B) 10 kg(m/s) î.

C) –10 kg(m/s) î.

D) 20 kg(m/s) î.

E) –20 kg(m/s) î.

ball+x wall

v ball

For Next Time (FNT)

Finish Reading Chapter 8.

Start HW for Chapter 8.