Physics 151: Lecture 12, Pg 1 REVIEW-2 Phys-151 l Work & Energy l Momentum Conservation l Center of...

Physics 151: Lecture 12, Pg 1

REVIEW-2REVIEW-2Phys-151Phys-151

Work & Energy Momentum Conservation Center of Mass Torque, Angular Momentum,

Moment of Inertia


Definition of Work:Definition of Work:

Ingredients: Ingredients: Force ( FF ), displacement ( rr )

Work, W, of a constant force FF

acting through a displacement rr

is:

W = F F . rr = F rr cos = Fr rr

FF

rr

displace

ment

Fr

“Dot Product”

See text: 7-1

Total Energy is Conserved


xx

vo

m

to

F

ExampleExampleWork Kinetic-Energy TheoremWork Kinetic-Energy Theorem

How much will the spring compress to bring the object to a stop if the object is moving initially at a constant velocity (vo) on frictionless surface as shown below ?

spring compressed

spring at an equilibrium position

V=0

t m


Lecture 13 -Lecture 13 - ACT 3ACT 3Work of SpringsWork of Springs

I have a spring with k = 20 N/m. Its compressed 30 cm by a 200 gram mass. How much work is done by the spring in this process ?

A) 0.6 J B) 6 J C) 0.9 J D) –0.9 J


Lecture 13, Lecture 13, ACT 5ACT 5Work & EnergyWork & Energy

Two blocks having mass m1 and m2 where m1 > m2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e. > 0) which slows them down to a stop.Which one will go farther before stopping ?

(a)(a) m1 (b)(b) m2 (c)(c) they will go the same distance

m1

m2


Another ExampleAnother Example

A 3.0-kg block is dragged over a rough horizontal surface by a constant force of 16 N acting at an angle of 37° above the horizontal as shown. The speed of the block increases from 4.0 m/s to 6.0 m/s in a displacement of 5.0 m. What work was done by the friction force during this displacement?

a. –34 Jb. –64 Jc. –30 Jd. –94 Je. +64 J


Work & Power:Work & Power: Example 3 :Example 3 :

What is the power required for a car (m=1000 kg) to climb a hill (5%) at v=30m/s assuming the coefficient of friction = 0.03 ?

100

5Car5%

V = 30 m/s Ptot = Phorizontal + Pvertical

v=const. - > a = 0

Phorizontal = F v = mg vhorizontal

Pvertical = F v = mg vvertical = (1000kg) (10m/s2 ) (30m/s)(5/100)Pvertical ~ 15 kW

Phorizontal ~ 0.03 (1000kg) (10m/s2 ) (30m/s) ~ 10 kW

Ptot ~ 10 kW + 15 kW = 25 kW


Example - 2Example - 2

A 12-kg block on a horizontal frictionless surface is attached to a light spring (force constant = 0.80 kN/m). The block is initially at rest at its equilibrium position when a force (magnitude P = 80 N) acting parallel to the surface is applied to the block, as shown. What is the speed of the block when it is 13 cm from its equilibrium position?

x

Fm

kv1= 0

v2=? 0.78 m/s


Lecture 16, Lecture 16, ACT 1ACT 1The Roller CoasterThe Roller Coaster

I have built a Roller Coaster . A motor tugs the cars to the top and then they are let go and are in the hands of gravity. To make the following loop, how high do I have to let the release the car ??

h ?

R

Car has mass m

A) 2R B) 5R C) 5/2 R D) none of the above


MOMENTUMMOMENTUM

F dp

dtpp = mvv

Momentum ConservationMomentum Conservation

ddtP 0F

PEXT

ddt

FEXT 0

A collision is said to be elastic when energy as well as momentum is conserved before and after the collision.

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

Kbefore Kafter

Kbefore = Kafter


ACT 4ACT 4

The value of the momentum of a system is the same at a later time as at an earlier time if there are no

a. collisions between particles within the system.b. inelastic collisions between particles within the

system.c. changes of momentum of individual particles

within the system.d. internal forces acting between particles within the

system.e. external forces acting on particles of the system.


Lecture 17, Lecture 17, ACT 3ACT 3Momentum ConservationMomentum Conservation

Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface.

The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck.Which box ends up moving fastest ?

(a)(a) Box 1 (b)(b) Box 2 (c)(c) same

1 2


Inelastic collision in 1-D: Example 1Inelastic collision in 1-D: Example 1

A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M, and V :What is the initial speed of the bullet v ?What is the initial energy of the system ?What is the final energy of the system ?Is energy conserved ?

v V

before after

x

See example 12-6


Lecture 17 Lecture 17 ACT 4ACT 4

3.0-kg mass sliding on a frictionless surface has a velocity of 5.0 m/s east when it undergoes a one-dimensional inelastic collision with a 2.0-kg mass that has an initial velocity of 2.0 m/s west. After the collision the 3.0-kg mass has a velocity of 1.0 m/s east. How much kinetic energy does the two-mass system lose during the collision?a. 22 Jb. 24 Jc. 26 Jd. 20 Je. 28


Elastic Collision in 1-DElastic Collision in 1-D After some moderately tedious algebra, (see text book

Chapter 9, section3) we can derive the following equations for the final velocities,

See text: 9.4

BBAv

mmm

vmmmm

v2

21

2

1

21

21

1)

2()(

BBAv

mmmm

vmm

mv

2

21

12

1

21

1

2)()

2(

1) m1v1,b + m2v2,b = m1v1,a + m2v2,a

2) v1,b - v2,b = - (v1,a - v2,a)

In general:


Example - Elastic CollisionExample - Elastic Collision

Suppose I have 2 identical bumper cars. One is motionless and the other is approaching it with velocity v1. If they collide elastically, what is the final velocity of each car ?

Note that this means,

m1 = m2 = m

v2B = 0

See text: 9.4

Animation


2-D Elastic Collisions2-D Elastic CollisionsBilliardsBilliards

Consider the case where one ball is initially at rest.

ppa

ppb

FF PPa

before afterthe final direction of the

red ball will depend on

where the balls hit.

vvcm

See Figure 12-14

See text: Ex. 9.11

Animation


ACT ACT

A 5.0-g particle moving 60 m/s collides with a 2.0-g particle initially at rest. After the collision each of the particles has a velocity that is directed 30° from the original direction of motion of the 5.0-g particle. What is the speed of the 2.0-g particle after the collision?a. 72 m/sb. 87 m/sc. 79 m/sd. 94 m/se. 67 m/s


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.


Force and ImpulseForce and Impulse

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


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


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


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.

See text: 9.6

CMEXT MAF


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 ACT Center 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


RecallRecallKinematic of Circular Motion:Kinematic of Circular Motion:

RR

vv

st

(x,y)

= t

s = v t

s = R = Rt v = R

x = R cos()= R cos(t)y = R sin()= R sin(t)

= tan-1 (y/x)

avR

2

For uniformcircular motion:

is angular velocity

x

y

Animation


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

Angular Linear

constant

0 t

0 021

2t t

constanta

v v at 0

x x v t at 0 021

2

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

x = Rv = Ra = R

See text: 10.3


Example:Example:

The turntable of a record player has an angular velocity of 8.0 rad/s when it is turned off. The turntable comes to rest 2.5 s after being turned off. Through how many radians does the turntable rotate after being turned off ? Assume constant angular acceleration.

a. 12 radb. 8.0 radc. 10 radd. 16 rade. 6.8 rad

See text: 10.1


Calculating Moment of Inertia...Calculating Moment of Inertia...

For a single object, I clearly depends on the rotation axis !!

L

I = 2mL2I = mL2

mm

mm

I = 2mL2

See text: 10.5

See example 10.4 (similar)


Moments of InertiaMoments of Inertia

Solid disk or cylinder of mass M and radius R, about a perpendicular axis through its center.

Some examples of I for solid objects:

RL

I r2 dm

I 1

2MR2

rdr


Moments of Inertia...Moments of Inertia...

Some examples of I for solid objects:

Solid sphere of mass M and radius R,

about an axis through its center.

I 2

52MR

R

See text: 10.5

See Table 10.2, Moments of Inertia

Thin spherical shell of mass M and radius R, about an axis through its center.

R

I 2

3MR2


Kinetic Energy of Rotation and Kinetic Energy of Rotation and Moment of InertiaMoment of Inertia

K 12

2I I m ri ii

2

See text: 10.4

where

IPARALLEL = ICM + MR2

Parallel Axis TheoremParallel Axis Theorem


Rotational Dynamics:Rotational Dynamics:What makes it spin?What makes it spin?

I=tot

See text: 10.6 and 10.7

DIRECTION: =rr X FF

MAGNITUDE: =rr FF sin sin


How Much WORK is Done ?How Much WORK is Done ?

W = Analogue of W = F •r W will be negative if and have opposite sign !

See text: 10.8

Work & Kinetic Energy:Work & Kinetic Energy:

K Wf i net 12

2 2I


Lecture 21, Lecture 21, ACT 2ACT 2 A uniform rod of mass M = 1.2kg and length L = 0.80 m,

lying on a frictionless horizontal plane, is free to pivot about a vertical axis through one end, as shown. If a force (F = 5.0 N, = 40°) acts as shown, what is the resulting angular acceleration about the pivot point ?

a. 16 rad/s2

b. 12 rad/s2

c. 14 rad/s2

d. 10 rad/s2

e. 33 rad/s2


Angular Momentum:Angular Momentum:Definitions & DerivationsDefinitions & Derivations

e have shown that for a system of particles

Momentum is conserved if

Fp

EXTddt

FEXT 0

See text: 11.3

L r p

r F The rotational analogue of force F F is torque

Define the rotational analogue of momentum pp to be

angular momentum

p=mv

Animation


Definitions & Derivations...Definitions & Derivations...

First consider the rate of change of LL: ddt

ddt

Lr p

ddt

ddt

ddt

r pr

p rp

v vm

0

So ddt

ddt

Lr

p

See text: 11.3


Lecture 21, Lecture 21, ACT 2ACT 2 Strings are wrapped around the circumference of two

solid disks and pulled with identical forces for the same distance. Disk 1 has a bigger radius, but both are made of identical material (i.e. their density = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest.

Which disk has the biggest angular velocity after the pull ?

(a)(a) disk 1

(b)(b) disk 2

(c)(c) same

FF

1 2


Example 2Example 2 A rope is wrapped around the circumference of a solid

disk (R=0.2m) of mass M=10kg and an object of mass m=10 kg is attached to the end of the rope 10m above the ground, as shown in the figure.

M

m

h =10 m

T

a) How long will it take for the object to hit the ground ?

a) What will be the velocity of the object when it hits the ground ?

a) What is the tension on the cord ?

1.7 s

11m/s

32 N


Example: Rotating RoadExample: Rotating Road

A uniform rod of length L=0.5m and mass m=1 kg is free to rotate on a frictionless pin passing through one end as in the Figure. The rod is released from rest in the horizontal position. What is

a) angular speed when it reaches the lowest point ?

b) initial angular acceleration ?

c) initial linear acceleration of its free end ?

See example 10.14

See text: 10.8

L

m = 7.67 rad/sa)

= 30 rad/s2b)

c) a = 15 m/s2


Example:Example:

A mass m = 4.0 kg is connected, as shown, by a light cord to a mass M = 6.0 kg, which slides on a smooth horizontal surface. The pulley rotates about a frictionless axle and has a radius R = 0.12 m and a moment of inertia I = 0.090 kg m2. The cord does not slip on the pulley. What is the magnitude of the acceleration of m?

a. 2.4 m/s2

b. 2.8 m/s2

c. 3.2 m/s2

d. 4.2 m/s2

e. 1.7 m/s2

See text: 10.1


Example :Example : Rolling MotionRolling Motion

A cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ?

M

h

Mv ?

Cylinder has radius R

M M

M

M

M


How high do we have to start the ball ?

1

2

-> The rolling motion added an extra 2/10 R to the height)

h

h = 2.7 R = (2R + 1/2R) + 2/10 R

Physics 151: Lecture 12, Pg 1 REVIEW-2 Phys-151 l Work & Energy l Momentum Conservation l Center of...

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Transcript of Physics 151: Lecture 12, Pg 1 REVIEW-2 Phys-151 l Work & Energy l Momentum Conservation l Center of...