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

Download Presentation

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

Loading in 2 Seconds...

- 112 Views
- Uploaded on
- Presentation posted in: General

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

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 - - - - - - - - - - - - - - - - - - - - - - - - - -

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

- Center of mass
- Newton’s 2nd law for a system of particles
- Linear momentum, 2nd law in terms of P
- Conservation of P

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.

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

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

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

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

take d/dt on both sides

take d/dt again

Identify ma as the force on each particle

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

Now take derivative d/dt of

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.

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

Change in momentum of ball on left or right.

F(t)

-F(t)

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

hi

-vf

vf

After

bounce

Before

bounce

Initial

E = PE = mgh

hf

E = KE = 1/2mv2

Almost elastic collision

C.O.R. = Vf/Vi

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

Almost inelastic collision

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.

m1v1i + m2v2i = m1v1f + m2v2f

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

Kinetic energy is conserved too.

Total momentum before = Total momentum after

m1

m2

-v

-v

-v

v

v1f

v

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

-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

-v

v

V1f=3V

V2f = -V

V1f=3V

For maximum height consider m2 >>m1

V2f = -V

How high does it go?

9 times higher