Lecture 7 chapter 9 systems of particles wednesday 7 16 03
This presentation is the property of its rightful owner.
Sponsored Links
1 / 24

Lecture 7 Chapter 9 Systems of Particles Wednesday 7-16-03 PowerPoint PPT Presentation


  • 90 Views
  • Uploaded on
  • Presentation posted in: General

Lecture 7 Chapter 9 Systems of Particles Wednesday 7-16-03. Warm-up problem. Puzzle Question: Cite two possible reasons why it appears that some basket ball players and dancers have a greater hang time. Physlets. Topics. Center of mass Newton’s 2nd law for a system of particles

Download Presentation

Lecture 7 Chapter 9 Systems of Particles Wednesday 7-16-03

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Lecture 7 chapter 9 systems of particles wednesday 7 16 03

Lecture 7 Chapter 9Systems of Particles Wednesday7-16-03

Warm-up problem

Puzzle Question:

Cite two possible reasons why it

appears that some basket ball

players and dancers have a greater

hang time.

Physlets


Topics

Topics

  • Center of mass

  • Newton’s 2nd law for a system of particles

  • Linear momentum, 2nd law in terms of P

  • Conservation of P


Center of mass

Center of mass

The center of a body or a system of bodies is the point that

moves as though all of the mass were concentrated there

and all external forces were applied there.


Center of mass1

Center of Mass

Why is it important? For any rigid body the motion of the body is given by

the motion of the cm and the motion of the body around the cm.

As an example find the center of mass of the following system

d

.

x

m

M

xcm

m xcm = M (d-xcm)

xcm = M/(m + M) d


Problem 9 3

Problem 9.3

2 dimensions

.

Find xcm and ycm

ycmM = m1 y1 + m2 y2 + m3 y3

xcmM = m1 x1 + m2 x2 + m3 x3

ycm 15 = 3*0 +4*1 + 8*2

xcm 15 = 3*0 + 4*2 + 8*1

ycm = 20/15 = 1.33 m

xcm = 16/15 = 1.1 m


Lecture 7 chapter 9 systems of particles wednesday 7 16 03

If the 8 kg mass increases, how does the cm change?

xcm =(m1 x1 + m2 x2 + m3 x3)/(m1+m2+m3)

When m3 gets very large, suppose we neglect the other masses.

xcm ~(m3 x3)/(m3) ~ x3

The center of mass moves towards the large object as it should


Center of mass for a system of particles

Center of Mass for a system of particles

xcm = (m1 x1 + m2 x2 + m3 x3)/M


Newton s second law for a system of particles f net m a cm

Newton’s Second Law for a System of particles: Fnet= Macm

take d/dt on both sides

take d/dt again

Identify ma as the force on each particle


Linear momentum form of newton s 2nd law

Linear Momentum form of Newton’s 2nd Law

Important because it is a vector quantity that is conserved in interactions.

Now take derivative d/dt of


Law of conservation of linear momentum

Law of Conservation of Linear Momentum

If Fnet = 0 on a closed system where no mass enters or leaves the system, then dP/dt = 0 or P = constant.

Pi = Pf for a closed isolated system

Also each component of the momentum Px,Py,Pz is also constant

since Fx, Fy, and Fz all = 0

If one component of the net force is not 0, then that component of momentum is not a constant. For example, consider the motion of a horizontally fired projectile. The y component of P changes while the horizontal component is fixed after the bullet is fired.


Air track examples

Air track examples

Two carts connected by a spring. Set them into oscillation by pulling them

apart and releasing them from rest. Note cm does not move.

Now repeat with spring between two carts. Analogous to exploding mass.

Again observers in two different inertial reference frames will measure

different values of momentum, but both will agree that momentum is conserved.

Note if the net force vanishes in one inertial frame it will vanish in all

inertial frames


Lecture 7 chapter 9 systems of particles wednesday 7 16 03

Change in momentum of ball on left or right.

F(t)

-F(t)


Lecture 7 chapter 9 systems of particles wednesday 7 16 03

Andy Rodick has been clocked at serving a tennis ball up to 149 mph(70 m/s).

The time that the ball is in contact with the racquet is about 4 ms. The mass of a

tennis ball as is about 300 grams.

What is the average force exerted by the racquet on the ball?

Favg = J/Dt = Dp/ Dt

Dp =70( 0.3) - 0 = 21 kg-m/s

Favg = 21/0.004 = 5000 Newtons

What is the acceleration of the ball?

a = Favg /m = 5000N/0.3 kg = 16,667 m/s2

What distance the racquet go through while the ball is still in contact?

V2f - V2i =2ax

x = V2f /2a = (70)2/2(16667) = 0.15 m


Bouncing ball

BOUNCING BALL

hi

-vf

vf

After

bounce

Before

bounce

Initial

E = PE = mgh

hf

E = KE = 1/2mv2


Measuring velocities and heights of balls bouncing from a infinitely massive hard floor

Measuring velocities and heights of balls bouncing from a infinitely massive hard floor

Almost elastic collision

C.O.R. = Vf/Vi

R.E./C.E.= Hi/Hf

Almost inelastic collision


Conservation of momentum

Conservation of Momentum

In a closed isolated system containing a collision, the linear momentum

of each colliding body may change but the total linear momentum P of the system

can not change, whether the collision is elastic or inelastic.

Fnet = 0, dP/dt = 0, Hence, P = constant .

Each x,y, and z component of the momentum is a constant.


One dimension collision

One Dimension Collision

m1v1i + m2v2i = m1v1f + m2v2f

1/2 m1v1i2 = 1/2 m1v1f2 + 1/2m2v2f2

Kinetic energy is conserved too.

Total momentum before = Total momentum after


Balls bouncing off massive floors we have m 2 m 1

m1

Balls bouncing off massive floors, we have m2 >>m1

m2


Lecture 8 chapter 10 collisions thursday 7 17 03

Lecture 8 Chapter 10Collisions Thursday 7-17-03


Colliding pool balls the excutive toy m 2 m 1

Colliding pool balls The excutive toy m2 = m1


Two moving colliding objects problem 45

-v

-v

-v

v

v1f

v

Two moving colliding objects: Problem 45

m1v1i + m2v2i = m1v1f + m2v2f

1/2 m1v1i2 + 1/2 m2v2i2 = 1/2 m1v1f2 + 1/2m2v2f2

Just after the little ball

bounced off the big ball

Just after the big ball bounced

Just before each hits the floor

m1

m2


Lecture 7 chapter 9 systems of particles wednesday 7 16 03

-v

v

V1f=2V

V2f =0

For m2 =3m1 ,

v1f = 8/4V =2V

superball has twice

as much speed.

How high does it go?

4 times higher


Lecture 7 chapter 9 systems of particles wednesday 7 16 03

-v

v

V1f=3V

V2f = -V

V1f=3V

For maximum height consider m2 >>m1

V2f = -V

How high does it go?

9 times higher


  • Login