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1 Rotational Motion 1. Remember how Newton’s Laws for translational motion were studied: 1. Kinematics (x = x 0 + v 0 t + ½ a t 2 ) 2. Dynamics (F = m a) 3. Momentum Conservation 2. Now, we repeat them again, but for rotational motion: 1. Kinematics (θ θ θ, ϖ ϖ ϖ, α α α) 2. Dynamics (τ τ τ = I α α α) 3. Angular Momentum Part III Part III Today Later Translational Motion and Rotational Motion Rotational Motion Angular Momentum Angular Momentum p p r L F F r v m p I L v m I × = × = τ τ τ ϖ ϖ ϖ ϖ ϖ ϖ

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### Transcript of Angular Momentum - Texas A&M...

1

Rotational Motion

1. Remember how Newton’s Laws for translationalmotion were studied:

1. Kinematics (x = x0 + v0 t + ½ a t2 )2. Dynamics (F = m a)3. Momentum Conservation

2. Now, we repeat them

again, but for rotationalmotion:

1. Kinematics (θθθθ, ωωωω, αααα)2. Dynamics (ττττ = I αααα)3. Angular Momentum

Part I I IPar t I I I

����������� �

Today Later

Translational Motion and Rotational Motion

àààà àààà

àààààààà

Rotational Motion

Angular MomentumAngular Momentum

pprL

FFr

v m pIL

v

mI

����

����

����

��

↔↔↔↔××××====↔↔↔↔××××====

≡≡≡≡↔↔↔↔≡≡≡≡↔↔↔↔↔↔↔↔

ττττωωωω

ωωωω

2

Rotational Motion

Rotational DynamicsRotational Dynamics

dtpd

amF�

��==

dtLd

I

��� ======== ααααττττ (torque)

mI

ααααax

No external force ààààMomentum conservation

No external torque ààààL conservation

F�

F�

r�

Rotational MotionAngular Momentum

Change in Change in Angular Momentum (I )Angular Momentum (I )

x

y

z

R.-H. Rule

x

y

z

R.-H. Rule

ωωωωfΙΙΙΙωωωωi àààà ΙΙΙΙωωωωfωωωωi

dtLd

I

��� ======== ααααττττ (torque)

F�

3

Rotational Motion

Example 1Example 1(a) What is the angular momentum of a 3.00-kg uniform

cylindr ical gr inding wheel of radius 20 cm and height 40 cm when rotating at 1500 rpm?

(b) How much torque is required to stop it in 10 s?

H0.4H

ωωωωif

if

tt

LL

tdLd

b

ILa

−−−−−−−−

========

====���

��

ττττ

ωωωω

)(

)(

Hints:

Rotational Motion

Rolling Motion w/o SlippingRolling Motion w/o Slipping

FasterFaster

Instantaneously restInstantaneously restàààà Instantaneous axisInstantaneous axis

ωωωω

ωωωω

Rolling Motion w/o SlippingRolling Motion w/o Slipping

FasterFaster

Instantaneously restInstantaneously restàààà Instantaneous axisInstantaneous axis

ωωωω

ωωωω

NewtonNewton’’ s 2s 2ndnd Law of MotionLaw of Motion

Translational Motion

Rotational Motion

)(integralnet tI

ωωωωτττταααα ��

� →→→→====

)(integralnet tvm

Fa

��

� →→→→====

Quick Question: All shapes have the same mass,but the speeds are different. Why?

Unlike a = F/m, even if F and m are given,we have to calculate ττττ and I.

4

Rotational Motion

Rotational Motion

ττττ = r � Fàààà àààààààà

θθθθ

àààà

F�

r�

ττττ�z

y

xF�

r�

z

Rotational Motion

Torque due to Gravity?Torque due to Gravity?

)(sin2

⊗⊗⊗⊗��������

������������

���� −−−−====

××××====

θθθθ

ττττ

mg xl

F r ���

ràààà

Fàààà

lCM

mass m

?====××××==== F r ���ττττ Rotational Motion

Vector Nature of Vector Nature of Angular QuantitiesAngular Quantities

Kinematical var iables to descr ibe the rotational motion:àààà Angular position, velocity and accelerationàààà c.w. or c.c.w. rotation (like +x or –x direction in 1D)àààà Vector natures!

)(

)(

)(

k

k

ˆˆ

ˆˆ

ωα

θω

θ

=

=

=

x

y

z

R.-H. Rule

>0 or <0

Rotational Motion

ParallelParallel--axis Theoremaxis Theorem

Icm = ½ MR02

d

2 MdII cm ++++====

Idisk = Icylinder

NewtonNewton’’ s 2s 2ndnd Law for RigidLaw for Rigid--body Rotationbody Rotation

ααααττττ ��I====net

Rotational Motion

2cm

2

2

1

2

1 ωIMMgH += v

Icm = Large àààà v = Small

Krot = (1/2) I ωωωω2 (àààà Ktrans = (1/2) m v2)

θθθθττττττττ sin Fr ====≡≡≡≡�

Rotational Motion

Ffr

1. F.B.D.1. F.B.D.

[x] Mg sinθθθθ – Ffr = M ax[y] FN – Mg cosθθθθ = 0

[⊗⊗⊗⊗] Ffr R0 = (2/5)MR02 (ax/R0)

We treat the box as a par ticle. Solid sphere (mass M, radius R0)

P is:(a) the point of contact with the

ground;(b) instantaneously rest relative

to the ground.

Thus:The motion of the wheel is a purerotation about the instantaneousaxis through P at the instant.

P is:(a) the point of contact with the

ground;(b) instantaneously rest relative

to the ground.

Thus:The motion of the wheel is a purerotation about the instantaneousaxis through P at the instant.

Ffr ≤≤≤≤ µµµµs FNFfr = µµµµk FN

Ffr

FNFN

Static fr iction: 0 < Ffr ≤≤≤≤ µµµµs FN

Ffr (static fr iction!)

2. Newton’s 2nd Law2. Newton’s 2nd Law

[x] mg sinθθθθ – µµµµk FN = m ax[y] FN −−−− mg cosθθθθ = 0

5

Rotational Motion

2cm

2

2

1

2

1 ωIMMgH += v

Icm = Large àààà v = SmallIcm = Large àààà v = Small

bbaa

ba

I I

L L

ωω =∴=

0ˆ )( Ng

�� ====++++====���� k���

i

Ia: big Ib: small

z

)(aωωωω� )(bωωωω�

Angular momentum conservationEnergy conservation

Rotational Motion

Recap: Example 1Recap: Example 1Calculate the torque on the 2.00-m longbeam due to a 50.0 N force (top) about (a) point C (= c.m.)(b) point P

Calculate the torque on the 2.00-m longbeam due to a 60.0 N force about (a) point C (= c.m.)(b) point P

Calculate the torque on the 2.00-m longbeam due to a 50.0 N force (bottom) about (a) point C (= c.m.)(b) point P

θθθθττττττττ sin Fr ====≡≡≡≡�

àààà

àààà

0ˆ )( ?]Why[ Ng

�� ====++++====���� k���

i

6

Rotational Motion

P.10-38, 10-65, 10-102 …

Assume the pulley bear ing

is fr ictionless …

à We can use conservation

of mechanical energy.

Avera

ge fr

iction

al toq

ue …

àW

e can

not use

conser

vatio

n

of mech

anica

l energ

y

because

F fris

non-co

nserva

tive

force.

Fr iction on the pulley (disk)by the bear ing (axis)Fr iction on the pulley (disk)by the bear ing (axis)

[⊗] FT2 R0 – FT1 R0 – ττττfr = (1/2)MR02 (ax/R0)

FT2FT1

Ffr

ax

α α α α [⊗]

x

Rotational Motion

7

Rotational Motion

Angular Momentum ConservationAngular Momentum Conservation

x

y

)()( 21 tF tF xx −−−−====

fx,fx,x,ix,i p ppp 2121 ++++====++++

bbaa

ba

I I

L L

ωω =∴=

Ia: big Ib: small

0ˆ )( Ng

�� ====++++====���� k���

i

z

)(aωωωω� )(bωωωω�

Rotational Motion

Application of Angular Momentum ConservationApplication of Angular Momentum Conservation

8

Rotational Motion

Example 2Example 2A person stands, hands at the side, on a platform that is rotating at a rate of 1.60 rev/s. I f the person now raises his arms to a hor izontal position, the speed of rotationdecreases to 0.800 rev/s.(a) Why does the speed of

rotation decrease? Explain using the twokey words;

external torqueangular momentum

(b) By what factor has the moment of iner tia of theperson changed?

(c) Compare Ki and KfΙΙΙΙiωωωωi ==== ΙΙΙΙfωωωωf

Rotational Motion

m2

m1

m1

ωωωωi

ωωωωf

Ii=MR2

If=MRf2

Rf

9

Rotational Motion

Cross Product: General

Rotational Motion

Cross Product: Cross Product: ττττττττ = = rr �� FF

x

z θθθθ

Direction and magnitude of torque?

Magnitude: ττττ = r F sinθθθθ

àààà àààà àààà

F�

ττττ�

r�

F�

r�

10

Rotational Motion

Cross Product: Cross Product: LL = = rr �� pp

θθθθ

Direction and magnitude of angular momentum?

Magnitude: L = r p sinθθθθ

x

z

àààà àààà àààà

r�

p�

vmp��

====prL���

××××====

r�

[General definition for any motion]

Rotational Motion

LL = = rr p p sinsinθθθθθθθθDirection and magnitude of angular momentum?à Magnitude: L = r p sinθθθθ

Lm = r (mv) sinθθθθ = R0 m v Lm = REarth m v sinθθθθ

θθθθ

ωωωωr�

v�

EarthRr ====�

r�

θθθθv�

m

11

Rotational Motion

Example 3Example 3

Suppose a meteor (m = 7.0 � 1010 kg) struck the Earth atthe equator with a speed v = 1.0 � 104 m/s and remained stuck. By what factor would this affect the rotationalfrequency of the Earth?

Basic concept: Lm + LE = Lm+E

“ By what factor ..”àààà R = ωωωω(after)/ωωωω(before)

Rotational Motion

Practice Problem 1Practice Problem 1

A uniform disk turns at 7.00 rev/s a r o u n d a f r i c t i o n l e s sspi ndl e. A non-r ot at i ng r od,of the same mass (m) as the disk and length (l) equal to the disk’s diameter , is dropped onto the disk. They then both turnaround the sp i n d l e w i t h t h ei r cen t er s superposed. There is no slipping between the rod and the disk. What is the moment of iner tia of the disk+rod system about the axis?

12

Rotational Motion

Example 4Example 4

A uniform disk turns at 7.00 rev/s around a fr ictionless spindle. A non-rotating rod, of the same mass (m) as the disk and length (l) equal to the disk’s diameter , is dropped onto the disk. They then both turnaround the spindle with their centers superposed. There is no slipping between the rod and the disk. What is the angular velocity (in rev/s) of the disk+rod system about the axis?

Rotational MotionAngular Momentum

Change in Change in Angular Momentum (I I )Angular Momentum (I I )

Only the direction of L changes;The magnitude of L does not change.

Out of page (+x axis)

F�

iL�

fL�

if LLL���

−−−−====∆∆∆∆

dtLd

I

��� ======== ααααττττ (torque)

13

Rotational MotionAngular Momentum and Torque

dtLd

I

��� == ατ (torque)

Rotational MotionAngular Momentum and Torque

z

y

x

Spinning

LSpinning Axisàààà

Ropeàààà

x

z

Rotation (in Rotation (in xyxy plane) ofplane) ofSpinning Bicycle WheelSpinning Bicycle Wheel