Physics, Page 1

Chapter 11.Chapter 11. Rotational motion & Rotational motion & Angular momentumAngular momentum

각운동량 (Angular momentum)

각운동량 보존

( )L r p m r v≡ × = ×

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m(r x v)

τ = Iα

(1/2)Iω2

I

α

ω

θ

회전운동

(angular motion)

N.s

N

J

kg

m/s2

m/s

m

N.mF = ma운동방정식 (Newton’s 2nd)

J.smv모멘텀 (momentum)

J(1/2)mv2운동에너지 (KE)

kg.m2m관성 (inertia)

1/s2a가속도 (acceleration)

1/sv속도 (velocity)

-x변위 (displacement)

병진운동

(linear motion)

Review of last lecture Review of last lecture

오늘의오늘의 핵심주제핵심주제

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Question A disk of mass M and radius R rotates around the z axis with

angular velocity ωi. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity ωf.

A) ωf = ωi B) ωf = ½ ωi C) ωf = ¼ ωi

ωi

z

ωf

z

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ω1 ω2

I2I1

L L

What happens to your angular velocity as you pull in your arms? 1. it increases2. it decreases 3. it stays the same

Question

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s Rθ=

CMds dv R Rdt dt

θ ω⎛ ⎞= = =⎜ ⎟⎝ ⎠

CMCM

dva Rdt

α= =

미끄러지지미끄러지지 않고않고 가속되어가속되어 구르는구르는 경우경우::

0 0CM sa & f= =

일정한일정한 속도로속도로 구르는구르는 경우경우: :

CMs f M a=

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보기문제 11-2

(a) 수직 높이 h = 1.20 m 를 내려왔을 때 속력

(b) 쓸림힘의 방향과 크기

(a) 역학적 에너지 보존:2 21 1

2 2CM CMI Mv Mghω + =

225

CMCM

vI MR & =R

ω= 710CMv gh=

(b)

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요요

2

21 /

o CM CMo

CM

o

CM o

Mg T Ma

aR T I IR

IT aR

gaI MR

τ α

− =

⎛ ⎞= − = = ⎜ ⎟

⎝ ⎠⎛ ⎞

= −⎜ ⎟⎝ ⎠

=+

앞의 경사진 경우에서

θ = 90도 , 마찰력 장력 (T)

인 경우에 해당함.

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다시 보자! 돌림힘 (Torque)

r Fτ = ×

m3

m1m2

T1

T3

Since RT3 – RT1 = I2 α.

It takes force (torque) to accelerate the pulley.

Compare the tension T1 and T2 as block 3 falls

A) T1 < T3 B) T1 = T3 C) T1 > T3

Question

( )( ) ( ) ( )2

sin

t t

r F

rF r ma rm r mr

I

τ φ

α α

τ α

=

= = = =

∴ =

Physics, Page 10

Question You want to balance a hammer on the tip

of your finger, which way is easierA) Head upB) Head downC) Same

τ = I α

m g R sin(θ) = mR2 αmg

R

Torque increases with R

Inertia increases as R2

g sin(θ) / R = α

Angular acceleration decreases with R!, so large R is easier to balance.

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Question When a force is applied to the string, the spool will

1) Roll right 2) Roll Left 3) Depends on angle

F

Depends!

To solve, need to look at torque due to friction, and torque due to tension.

θ

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• There is another (slick) way to see this:• Consider the torque about the point of contact

between the spool and the ground. • We know the net torque about this point is zero.

– Since both Mg and f act through this point, they do not contribute to the net toque.

– Therefore the torque due to T mustalso be zero.

– Therefore T must actalong a line that passesthrough this point!

Consider all of the forces acting: tension T and friction f.

Using FNET = 0 in the x direction:

0fcosT =−θ θ= cosTf

aT bf− = 0 aT bf=

Using τNET = 0 about the CM axis: bacos =θ

ab

θ

T

f

Mg

θ

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( )l r p m r v≡ × = ×

각운동량 (Angular momentum)

sinl l mrv φ= =

2kg m /s⎡ ⎤⋅⎣ ⎦

( ) ( )( ) 2

sin tl mr v mr v

mr r mr

φ

ω ω

= =

= =

각속도로각속도로 표현하면표현하면,,

l Iω=

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( )d d dr dpl r p p rdt dt dt dt

= × = × + ×

0d r mv v mvdt

⎛ ⎞× = × =⎜ ⎟

⎝ ⎠

d dpl r r Fdt dt

τ∴ = × = × =

dldt

τ =

netd Ldt

τ =,1 1

n ni

net ii i

d L dldt dt

τ= =

= =∑ ∑1

n

ii

L l=

= ∑

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L Iω∴ =

( )sin 90i i i i i i i i il r p r p l r m vΔ= × = ⇒ =

( )( ) ( )sin siniz i i i i i i il l r m v r m vθ θ Δ Δ⊥= = =

221

2 2RLNote : K II

ω= =

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netd Ldt

τ= 0netτ =이므로 이면( ) 0d L t

dt=

( ) ( )i fL t L t=알짜알짜 돌림힘이돌림힘이 없다면없다면 총총 각운동량은각운동량은 항상항상 일정하다일정하다..

z i i f fL I Iω ω= = =일정

L r p= × =일정

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What happens to the angular momentum as you pull in your arms? 1. it increases2. it decreases 3. it stays the same

25

“no external forces gives constant angular momentum”

torques

Question

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Bonus Question!• There are No External forces acting on the

“student+stool” system.A) True B) False C) What!?

Key is no external torques!

The system has gravity and normal force.

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What happens to your angular velocity as you pull in your arms? 1. it increases2. it decreases 3. it stays the same

Question

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What happens to your kinetic energy as you pull in your arms? 1. it increases2. it decreases 3. it stays the same

The mass is closer so you have less inertia and you speed up. If you speed up then, your kinetic energy must increase as well.

K =12

2I ω =1

22 2

II ω =

12

2

IL (using L = Iω )

29

I think kinetic energy will decrease because as the bar stool spins faster and faster, rotational kinetic energy of the stool will increase. Maybe I just contradicted myself.

Question

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What about Energy Conservation?

A) Energy isn’t conserved here

B) Energy comes from weights

C) Gravitational energy is being converted to rotational kinetic energy

D) Energy comes from the stool

E ) I have no clue….

32

Physics, Page 22

z i i f fL I Iω ω= = =일정

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보기문제 11-7

( ), , , , ,

, ,2

wh i wh f b f wh i b f

b f wh i

L L L L L

L L

= + = − +

∴ =

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A student sits on a barstool holding a bike wheel. The wheel is initially spinning CCW in the horizontal plane (as viewed from above). She now turns the bike wheel over. What happens?

A. She starts to spin CCW.B. She starts to spin CW.C. Nothing

34

Start w/ angular momentum L pointing up from wheel. When wheel is flipped, no more angular momentum from it pointing up, so need to spin person/stool to conserve L!

Question

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R

m

F

v0

Example 10.10 A Revolving Ball on a Horizontal, Frictionless Surface

0=×=τ Fr ( )F//r

0==τ Ldtd constL =⇒

0 2 oRv v vr

= =

(a) R R/2 로 줄이면 각속도는?

(a) L1 = I1w1 = mR2w1

L2 = I2w2 = m(R/2)2w2

L1 = L2 => w2 = 4w1

(b) R R/2 로 줄이면 선속도는?

(b) L1 = L2 => mvoR = mvr

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ω++= IvRmvRmLT 21

Example 10.9 Two Connected Masses

m1

m2R O

v

v

I ( )RvIvRmm ++= 21

( )dtdv

RI

dtdvRmmL

dtd

TT ++==τ 21

aRIRmRm ⎟

⎠⎞

⎜⎝⎛ ++= 21

Text Rgm τ=⋅=τ 1

221

1

RImm

gma++

=∴

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Gyroscope 운동

• Suppose you have a spinning gyroscope in the configuration shown below:

• If the left support is removed, what will happen??

ωpivotsupport

g

48

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Gyroscopic Motion• The gyroscope does not fall down!• ... instead it precessesprecesses around its pivot axis !

ωpivot

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Gyroscope 운동

(a) 자전하지 않는 경우 :

아래로 떨어지면서 지지점 (O) 중심으로 회전

(b) 빠르게 자전하는 경우 :

처음에는 아래로 약간 처지다가 수평방향으로 회전

d Lr M gdt

τ = × =

( ) ( )sin 90r M g Mgr j Mgr jτ = × = =

// d L Lτ ⊥

따라서, t 에 의해 L 의 방향이 계속 바뀌는 옆돌기 운동

dL MgrdL Ld d dtL I

d Mgr dt I

φ φω

φΩω

= ⇒ = =

∴ = =

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Summary2 21 1

2 2CMK I Mvω= +

l r p= × l Iω=

dldt

τ =r Fτ = ×

netif =0τ( ) ( )i fL t L t=

ii

L l= ∑