Physics 1501: Lecture 18, Pg 1 Physics 1501: Lecture 18 Today’s Agenda l Announcements çHW#7: due Oct. 21 l Midterm 1: average = l Topics çImpulse çCenter

Physics 1501: Lecture 18, Pg 1

Physics 1501: Lecture 18Physics 1501: Lecture 18

TodayToday’’s Agendas Agenda Announcements

HW#7: due Oct. 21

Midterm 1: average =

TopicsImpulseCenter of MassRotational Kinematics


Momentum ConservationMomentum Conservation

The concept of momentum conservation is one of the most fundamental principles in physics.

This is a component (vector) equation.We can apply it to any direction in which there is no

external force applied. You will see that we often have momentum conservation

even when (mechanical) energy is not conserved.


Elastic vs. Inelastic CollisionsElastic vs. Inelastic Collisions A collision is said to be elastic when energy as well as

momentum is conserved before and after the collision. Kbefore = Kafter

Carts colliding with a spring in between, billiard balls, etc.

vvi

A collision is said to be inelastic when energy is not conserved before and after the collision, but momentum is conserved. Kbefore Kafter

Car crashes, collisions where objects stick together, etc.


Force and ImpulseForce and Impulse

F

t

ti tf

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 !

Impulse has units of Ns.



F

tt

Using

the impulse becomes:

impulse = change in momentum !



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



t

F

t

F

tt

t big, FF smallt small, FF big

soft spring

stiff spring


Lecture 18, Lecture 18, 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


Average Force and ImpulseAverage Force and Impulse

t

F

t

F

tt

t big, FFavav smallt small, FFavav big

soft spring

stiff spring

FFav av

FFav av


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:

(In this case, N = 2)

y

x

rr2rr1

m1

m2

RRCM



If the system is made up of only two particles:

y

x

rr2rr1

m1

m2

RRCM

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

+

If m1 = m2

the CM is halfway between

the masses.

Active Figure



The center of mass is where the system is balanced !Building a mobile is an exercise in finding centers of

mass. (Calder was a great applied physicist!)

m1m2

+m1 m2

+


Example Calculation:Example Calculation: Consider the following mass distribution:

(24,0)(0,0)

(12,12)

m

2m

m

RCM = (12,6)


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.


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?


Example: Astronauts & RopeExample: Astronauts & Rope

1. Find position relative to the male astronaut where they meet.

Use Center of Mass

M = 1.5mm


Example: Astronauts & RopeExample: Astronauts & Rope2. Find xcm

Fext=0 so vcm is constant at zero, xcm is constant.

M = 1.5mm

x0 20m


Example: Astronauts & Rope...Example: Astronauts & Rope...

3. Solve symbolically,



4. Put in numbers,



5. They end up 12m away from the ship,

So he doesn’t get back to the ship, but he does get to the woman.


Lecture 18, Lecture 18, ACT 2ACT 2Center 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 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.

Active Figure


Chap. 10: RotationChap. 10: Rotation

Up until now we have gracefully avoided dealing with the rotation of objects. We have studied objects that slide, not roll.We have assumed wheels are massless.

Rotation is extremely important, however, and we need to understand it !

Most of the equations we will develop are simply rotational versions of the ones we have already learned when studying linear kinematics and dynamics.


Rotational VariablesRotational Variables

Rotation about a fixed axis:Consider a disk rotating about

an axis through its center:

First, recall what we learned aboutUniform Circular Motion:

(Analogous to )


Rotational Variables...Rotational Variables... Now suppose can change as a function of time: We define the

angular acceleration:

Consider the case when is constant. We can integrate this to find and as a function of time:

constant


Rotational Variables...Rotational Variables...

Recall also that for a point a distanceR away from the axis of rotation:x = Rv = R

And taking the derivative of this we finda = R

R

v

x

constant


Summary Summary (with comparison to 1-D kinematics)(with comparison to 1-D kinematics)

Angular Linear

And for a point at a distance R from the rotation axis:

x = Rv = Ra = R


Example: Wheel And RopeExample: Wheel And Rope A wheel with radius R = 0.4m rotates freely about a fixed

axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a = 4m/s2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2 radians)

aa

R


Wheel And Rope...Wheel And Rope... Use a = R to find :

= a / R = 4m/s2 / 0.4m = 10 rad/s2

Now use the equations we derived above just as you would use the kinematic equations from the beginning of the semester.

= 0 + 0(10) + (10)(10)2 = 500 rad

aa

R

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Physics 1501: Lecture 18, Pg 1 Physics 1501: Lecture 18 Today’s Agenda l Announcements çHW#7: due Oct. 21 l Midterm 1: average = l Topics çImpulse çCenter