¾ 각운동량 (Angular momentum)

Chapter 11.

Chapter 11. Rolling, Torque, and Rolling, Torque, and Angular momentum

Angular momentum

¾ 각운동량 (Angular momentum)

¾ 각운동량 보존

( )

L r r r ≡ × = r p m r v r r ×

병진운동 (linear motion)

회전운동

(angular motion)

변위 (displacement) x m θ

ω α

I (1/2)Iω²

τ = Iα

-

속도 (velocity) v

m(r x v)

rad/s 가속도 (acceleration) a

m/s m/s²

kg J N

rad/s²

관성 (inertia) m kg^.m²

운동에너지 (KE) (1/2)mv² J

운동방정식 (Newton’s 2^nd) F = ma N^.m

N^.s

모멘텀 (momentum) mv J^.s

Review of last lecture Review of last lecture

오늘의 오늘의 핵심주제 핵심주제

Question

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

angular velocity ω_i. 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 ω_f. A) ω_f = ω_i B) ω_f = ½ ω_i C) ω_f = ¼ ω_i

ω

z

ω

z Angular momentum will be conserved!

i

i MR

L ₁ ω₁ ²ω

2 0 1

I + =

=

f

f MR

L = I₁ω₁ + I₂ω₂ = ²ω

f

i MR

MR ²ω ²ω 2

1 =

f

i ω

ω = 2

1

ω₁ ω₂

I₂ I₁

L L

What happens to your angular velocity as you pull in your arms?

1. it increases 2. it decreases

3. it stays the same

CORRECT

Question

s = Rθ

CM

ds d

v R R

dt dt

θ ω

⎛ ⎞

= = ⎜ ⎟ =

⎝ ⎠

CM CM

a = dv =αR

미끄러지지

미끄러지지 않고않고 가속되어가속되어 구르는구르는 경우:경우:

s CM

f = M a

ur r

0 0

CM s

ar = & fur =

일정한일정한 속도로속도로 구르는구르는 경우: 경우:

보기문제 11-2

(a) 수직 높이 h = 1.20 m 를 내려왔을 때 속력 (b) 쓸림힘의 방향과 크기

(a) 역학적 에너지 보존: ^{1 2 1 2}

2 I^CMω + 2 Mv^CM = Mgh 2 2

5

CM CM

I MR & =v ω R

= 7

CM 10

v = gh

(b)

수직력

요요

2

1 / 2

o CM CM

o

CM o

CM o

Mg T Ma

R T I I a

R

T I a

R a g

I MR

τ α

− =

⎛ ⎞

= − = = ⎜ ⎟

⎝ ⎠

⎛ ⎞

= −⎜ ⎟

⎝ ⎠

= +

앞의 경사진 경우에서

θ = 90도 , 마찰력 Æ 장력 (T) 인 경우에 해당함.

다시 보자! 돌림힘 (Torque)

τ r r ur = × r F

m₃

m₁

m₂

T₁ T₃

Compare the tension T₁ and T₂ as block 3 falls

A) T₁ < T₃ B) T₁ = T₃ C) T₁ > T₃

Since RT₃ – RT₁ = I₂ α.

It takes force (torque) to accelerate the pulley.

Question

( )

( ) ( ) ( )

sin

t t

r F

rF r ma rm r mr I

τ φ

α α

τ α

=

= = = =

∴ =

Question

You want to balance a hammer on the tip of your finger, which way is easier

A) Head up

B) Head down C) Same

τ = I α

m g R sin(θ) = mR² α

mg 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.

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

• 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 must also be zero.

– Therefore T must act along a line that passes through this point!

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

Using F_NET = 0 in the x direction:

0 f

cos

T θ − = f = cosT θ

aT − bf = 0 aT = bf

Using τ_NET = 0 about the CM axis:

b cosθ = a

a b

θ T

f

Mg θ

( )

l ≡ × = r p m r v × r r r r r

각운동량 (Angular momentum)

sin l = =l mrv φ r

kg m /s2

⎡ ⋅ ⎤

⎣ ⎦

( ) ( )

( )

sin _t

l mr v mr v

mr r mr

φ

ω ω

= =

= =

각속도로각속도로 표현하면,표현하면,

ω

=

( )

d d dr dp

l r p p r

dt r = dt r r× = dtr × + ×r r dtr d r 0

mv v mv dt

⎛ ⎞

× = × =

⎜ ⎟

⎝ ⎠

r r r r

d dp

l r r F

dt dt τ

∴ r= ×r r = × =r r r

d l

τ = dt r r

net

d L

τ = dt r ur

,

1 1

n n

i

net i

i i

d L d l

dt dt τ

= =

= ∑ = ∑

ur r

r

1 n

i i

L l

=

=

∑

ur r

제2법칙

L I ω

∴ =

( )

sin 90

i i i i i i i i i

l = × r p = r p

⇒ = l r Δ m v r ur uur

( )( ) ( )

sin sin

iz i i i i i i i

l = l θ = r θ Δ m v = r

Δ m v

2

l

net

d L

dt = τ ur r

0 τ r

=

이므로 이면

( )

d L t 0

dt = ur

( ) ( )

i f

L t = L t

ur ur

알짜

돌림힘이

없다면

총

각운동량은

항상

일정하다

z i i f f

L = I ω = I ω = 일정 L = × = r p

ur r r 일정

What happens to the angular momentum as you pull in your arms?

1. it increases 2. it decreases

3. it stays the same CORRECT

“no external forces gives

constant angular momentum”

torques

Question

Bonus Question!

• There is No External force acting on the

“student+stool” system.

A) True B) False C) What!?

Key is no external torques!

The system has gravity and normal force.

What happens to your angular velocity as you pull in your arms?

1. it increases 2. it decreases

3. it stays the same

CORRECT

Question

What happens to your kinetic energy as you pull in your arms?

1. it increases 2. it decreases

3. it stays the same

CORRECT

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 = 1 2

I ω2 = 1 2

2 2

I I ω = 1 2

2

I L (using L = Iω )

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

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….

z i i f f

L = I ω = I ω = 일정

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

= + = − +

∴ =

ur ur ur ur ur

ur ur

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

CORRECT

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

R

m

F

v₀

= 0

×

=

τ r F

Example 10.10 A Revolving Ball on a Horizontal, Frictionless Surface

r r

r

(

)

= 0

=

τ L

dt

r d r ⇒ Lr = const

0 R 2 _o

v v v

= r =

(a) R Æ R/2 로 줄이면 각속도는?

(a) L₁ = I₁ω₁ = mR²ω₁ (b) L₁ = L₂ => mv_oR = mvr L₂ = I₂ω₂ = m(R/2)²ω₂

L = L => w = 4ω

(b) R Æ R/2 로 줄이면 선속도는?

ω + +

= m vR m vR I

L_{T 1 2}

Example 10.9 Two Connected Masses

m₁

m₂ R O

v

v

I

( )

R I v vR m

m + +

= _{1 2}

( )

dt dv R

I dt R dv m

m dt L

d

T

T = = + +

τ _{1 2}

R a R I

m R

m ⎟

⎠

⎜ ⎞

⎝

⎛ + +

= _{1 2}

T ext = m g⋅R = τ

τ ₁

2 2 1

1

I R m

m

g a m

+

= +

∴

a = ?

Gyroscope 운동

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

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

ω pivot

support

g

Gyroscopic Motion

• The gyroscope does not fall down!

• ... instead it precesses precesses around its pivot axis !

ω pivot

Gyroscope 운동

(a) 자전하지 않는 경우 :

아래로 떨어지면서 지지점 (O) 중심으로 회전

(b) 빠르게 자전하는 경우 :

처음에는 아래로 약간 처지다가 수평방향으로 회전

r M g d L τ = × = dt

r r ur ur

(

) ^{( )}

r M g Mgr j Mgr j

τr r= × ur = ^o ) = )

// d L L τr ur ⊥ ur

따라서, τ ^{에 의해 L} 의 방향이 계속 바뀌는 옆돌기 운동

dL Mgr

dL Ld d dt

L I

φ φ

ω φ

= ⇒ = =

Summary

2 2

1 1

2 ^CM 2

K = I ω + Mv

l = × r p r r r

l = I ω

d l τ = dt

r r

τ r r ur = × r F

if τ ^r

=0

( ) ( )

i f

L t = L t

ur ur

i i

L = ∑ l

ur r