Physics 151: Lecture 19, Pg 1 Physics 151: Lecture 19 Today’s Agenda l Topics çExample problem (conservation of momentum) çImpulseCh. 9.2 çCenter of MassCh

Physics 151: Lecture 19, Pg 1

Physics 151: Lecture 19Physics 151: Lecture 19

Today’s AgendaToday’s Agenda

TopicsExample problem (conservation of momentum)Impulse Ch. 9.2Center of Mass Ch. 9.6


See text: 9-2

See Figure 12-2

video

rocket sciencerocket science

finalv+v

Conservation of MomentumConservation of Momentum


See text: 9-2

See Figure 12-2

Example ProblemExample Problem :

A object of mass m=0.200 kg is dropped from l =30.0 cm height above a basket base (M=0.200 kg) which is attached to the ceiling with a spring. If the spring is stretched by x = 0.05 m before, find the maximum distance the spring will stretch ?

d = 0.182 m

dm

m

lM

x


Force and ImpulseForce and Impulse

F

t

ti tf

tdtFI

t

The diagram shows the force vs time for a typical collision. The impulse, II, of the force is a vector defined as the integral of the force during the collision.

Impulse I I = area under this curve !

See text: 9-2

See Figure 12-2

Impulse has units of Ns.



F

t

dtdP

F

t

Using

PPPP

PF

it

f

tt

d

dtdtd

dtI

I P

the impulse becomes:

impulse = change in momentum !

See Figure 12-2

See text: 9-2



Two different collisions can have the same impulse since II dependsonly on the change in momentum,not the nature of the collision.

t

F

t

F

tt

same area

t big, FF smallt small, FF big

See text: 9-2



t

F

t

F

tt

t big, FF smallt small, FF big

soft spring

stiff spring

See text: 9-2

Animation


Lecture 19, Lecture 19, ACT 1ACT 1Force & ImpulseForce & Impulse

Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second.Which box has the most momentum after the force acts ?

(a)(a) heavier (b)(b) lighter (c)(c) same

F F light heavy


Lecture 19, Lecture 19, ACT 2ACT 2Force & ImpulseForce & Impulse

What is the average force that wall exerts on ball (0.40kg) if duration of wall-ball contact is 0.01 s ?

vi = 30m/s

vf = 20m/s

before collision

after collision

a) 20 N b) 200 N c) 2,000 N d) 20,000 N


Lecture 19, Lecture 19, ACT 3ACT 3

A 0.20 kg stone you throw rises 20.3 m in the air. The magnitude of the impulse the stone received from your hand while being thrown is

a. 0.27 Ns. b . 2.7 Ns. c . 4.0 Ns. d . 9.6 Ns. e . 34.3 Ns.


System of Particles:System of Particles:

Until now, we have considered the behavior of very simple systems (one or two masses).

But real life is usually much more interesting ! For example, consider a simple rotating disk.

An extended solid object (like a disk) can be thought of as a collection of parts. The motion of each little part depends on where it is in the object!


System of Particles: Center of MassSystem of Particles: Center of Mass

How do we describe the “position” of a system made up of many parts ?

Define the Center of MassCenter of Mass (average position):For a collection of N individual pointlike particles whose

masses and positions we know:

M

mN

1iii

CM

rR

(In this case, N = 2)

y

x

rr2rr1

m1

m2

RRCM



If the system is made up of only two particles:

21

2211

N

1iii

CM mmmm

M

m

rr

rR

y

x

rr2rr1

m1

m2

RRCM

m m m

m m1 2 1 2 2 1

1 2

r r r

R r r rCMm

M 1

22 1

where M = m1 + m2

So:

(r1 - r2)



If the system is made up of only two particles:

y

x

rr2rr1

m1

m2

RRCM

rr2 - r r1

where M = m1 + m2

+

R r r rCMm

M 1

22 1

If m1 = m2

the CM is halfway between

the masses.

R r r rCM 1 2 112



The center of mass is where the system is balanced !

m1m2

+m1 m2

+


Example Calculation:Example Calculation:

Consider the following mass distribution:

(24,0)

12m4

24m12m20mM

xmX i ii

CM )()()(

6m4

0m12m20mM

ymY i ii

CM )()()(

(0,0)

(12,12)

m

2m

m

RCM = (12,6)

See text: 9-6


Rr r

CMdm

dm

dm

M


For a continuous solid, we have to do an integral.

y

x

dm

rrwhere dm is an infinitesimal

mass element.

See text: 9-6

See example 8-4, A Triangular Plate

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Example: Astronauts & RopeExample: Astronauts & Rope

A male astronaut and a female astronaut are at rest in outer space and 20 meters apart. The male has 1.5 times the mass of the female. The female is right by the ship and the male is out in space a bit. The male wants to get back to the ship but his jet pack is broken. Conveniently, there is a rope connected between the two. So the guy starts pulling in the rope.

Does he get back to the ship? Does he at least get to meet the woman?

M = 1.5mm:-(

:-)


Lecture 19, Lecture 19, ACT 4ACT 4Center of Mass MotionCenter of Mass Motion

A woman weighs exactly as much as her 20 foot long boat. Initially she stands in the center of the motionless boat, a distance of 20

feet from shore. Next she walks toward the shore until she gets to the end of the boat.What is her new distance from the shore.

(There is no horizontal force on the boat by the water).

20 ft

? ft

20 ft

(a)(a) 10 ft (b)(b) 15 ft

(c)(c) 16.7 ft

before

after


Center of Mass Motion: ReviewCenter of Mass Motion: Review

We have the following law for CM motion:

This has several interesting implications:

It tell us that the CM of an extended object behaves like a simple point mass under the influence of external forces: We can use it to relate FF and AA like we are used to doing.

It tells us that if FFEXT = 0, the total momentum of the system does not change.As the woman moved forward in the boat, the boat went

backward to keep the center of mass at the same place.

See text: 9.6

CMEXT MAF


Recap of today’s lectureRecap of today’s lecture

Chapter 9, Center of MassElastic CollisionsImpulse

Documents

Physics 151: Lecture 19, Pg 1 Physics 151: Lecture 19 Today’s Agenda l Topics çExample problem (conservation of momentum) çImpulseCh. 9.2 çCenter of MassCh