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 greaterhang time.

Physlets

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 massThe center of a body or a system of bodies is the point thatmoves as though all of the mass were concentrated there and all external forces were applied there.

Center of MassWhy 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

xm M.xcm

d

m xcm = M (d-xcm) xcm = M/(m + M) d

Problem 9.3

2 dimensions

Find xcm and ycm

xcm M = m1 x1 + m2 x2 + m3 x3ycm M = m1 y1 + m2 y2 + m3 y3

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

xcm = 16/15 = 1.1 m

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

ycm = 20/15 = 1.33 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

Center of Mass for a system of particles

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

€

rr cm =

1

Mmi

i=1

n

∑ r r i €

ycm =1

Mmi

i=1

n

∑ y i

€

zcm =1

Mmi

i=1

n

∑ zi

€

xcm =1

Mmi

i=1

n

∑ x i

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

€

rr cm =

1

Mmi

i=1

n

∑ r r i take d/dt on both sides

€

rv cm =

1

Mmi

i=1

n

∑ r v i take d/dt again

€

ra cm =

1

Mmi

i=1

n

∑ r a i

€

ra cm =

1

M

r F i

i=1

n

∑

Identify ma as the force on each particle

€

ra cm =

1

M

r F net

€

rF net = M

r a cm

Linear Momentum form of Newton’s 2nd Law

€

rv cm =

1

Mmi

i=1

n

∑ r v i

€

rp i = mi

r v i is the definition of momentum of i'th particle

r P cm = M

r v cm is the momentum of the cm

€

rP cm =

r p i

i=1

n

∑i

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

€

rP cm = M

r v cm Now take derivative d/dt of

€

dr P cm

dt= M

dr v cm

dt = M

r a cm

€

rF net =

dr P cm

dt

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

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

F(t)-F(t)

€

J = F(t)dtt i

t f

∫ =dp

dtdt

t i

t f

∫ = dpp i

p f

∫ = p f − pi

Change in momentum of ball on left or right.

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.

Favg = J/t = p/ t

p =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 is the average force exerted by the racquet on the ball?

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

V2f - V2

i =2axx = V2

f /2a = (70)2/2(16667) = 0.15 m

hi

Initial

-vf

Beforebounce

vf

Afterbounce

BOUNCING BALLE = PE = mgh

E = KE = 1/2mv2

hf

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

Type of Ball

Coefficient of Restitution

(C.O.R.)

Rebound Energy/

Collision Energy

(R.E./C.E.)

Superball 0.90 0.81

Racquet ball 0.85 0.72

Golf ball 0.82 0.67

Tennis ball 0.75 0.56

Steel ball bearing 0.65 0.42

Baseball 0.55 0.30

Foam rubber ball 0.30 0.09

Unhappy ball 0.10 0.01

Beanbag 0.05 0.002

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

Almost elastic collision

Almost inelastic collision

C.O.R. = Vf/Vi

Conservation of MomentumIn 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

m1v1i + m2v2i = m1v1f + m2v2f

Total momentum before = Total momentum after

1/2 m1v1i2 = 1/2 m1v1f

2 + 1/2m2v2f2

Kinetic energy is conserved too.

€

v1 f =m1 − m2

m1 + m2

v1i

€

v2 f =2m1

m1 + m2

v1i

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

€

≅−m2

m2

v1i ≅ −v1i

€

v2 f =2m1

m1 + m2

v1i

€

v1 f =m1 − m2

m1 + m2

v1i

€

≅2m1

m2

v1i

€

≅0

m1

m2

Lecture 8 Chapter 10

Collisions Thursday

7-17-03

Colliding pool balls The excutive toy

m2 = m1

€

v2 f =2m1

m1 + m2

v1i

€

v1 f =m1 − m2

m1 + m2

v1i

€

≅v1i€

≅0

Two moving colliding objects: Problem 45

m1v1i + m2v2i = m1v1f + m2v2f

1/2 m1v1i2 + 1/2 m2v2i

2 = 1/2 m1v1f2 + 1/2m2v2f

2

-v

-v-vv

v1f

v

m1

m2

€

v1 f =m1 − m2

m1 + m2

v1i +2m2

m1 + m2

v2i

Just before each hits the floor

Just after the big ball bounced

Just after the little ball bounced off the big ball

€

v1 f =m1 − m2

m1 + m2

v1i +2m2

m1 + m2

v2i -vv

€

v1 f =m1 − m2

m1 + m2

(−V ) +2m2

m1 + m2

(V )

For m2 =3m1 ,v1f = 8/4V =2Vsuperball has twice as much speed.

€

v1 f =3m2 − m1

m1 + m2

(V )

How high does it go?€

h =V1 f

2

2g

4 times higher

V1f=2V

V2f =0

€

v1 f =m1 − m2

m1 + m2

v1i +2m2

m1 + m2

v2i -vv

€

v1 f =m1 − m2

m1 + m2

(−V ) +2m2

m1 + m2

(V )

€

v1 f =3m2 − m1

m1 + m2

(V )

How high does it go?

€

h =V 2

2g

V1f=3V

V2f = -V

9 times higher

For maximum height consider m2 >>m1

V1f=3V

V2f = -V