Ch 12 Rotation of a Rigid body

Ch 12 Rotation of a Rigid body 1ContentsRigid bodyCenter of mass: rCMRotational Energy Moment of inertia: IMathematics: cross productTorquePropertiesApplicationsRotation about a fixed axisStatic EquilibriumRolling MotionAngular MomentumExamples2Rigid Body (P340~P345)A rigid body is an extended object whose size and shape do not change as it moves.E.g.1st point: .2nd point: .3rd point: .Others: .We just need 6 parameters to describe a rigid body:Translational: rCM.Rotational: axis and angle .3xyz

xyz

Rotational MotionTop view from the tip of the axis of rotation:Right hand rule. Angular velocity:

Exercise: 4

Stop to think: =?

Center of Mass Benefits:

Exercise:Find the center of mass of this object:5

HomeworkStudent Workbook:P12-1~P12-26Rotational Energy & Moment of inertia (P346~P350)K= Mv2CM+Kmicro.For rotation and T0:

Note: This axis passes through the center of mass.Moment of inertia (in fact, it is a rank 2 tensor)

If , we get . 7

Parallel-axis Theorem (I=ICM+MR2)Total kinetic energies of these two cases:8

RK=?K=?

we getI=ICM+MR2.

Exercises ICM ofA cylinder.A thin rod.A sphere.Example 12.5A 1.0-m-long, 200 g rod is hinged at one end and connected to a wall. It is held out horizontally, then released. What is the speed of the tip of the rod as it hits the wall?9v=?HomeworkStudent Workbook12.712.812.1212.1310Cross Product (P368~P369)Geometrical definition: 11

Cross ProductProperties:Anti-commutative: AB=-BA.Distributive: A(aB+bC)=aAB+ bAC.Based on the right-hand rule, we get: Exercise: If A=(1, 0, 0),B=(1, 1 , 0).12

Torque (P370, P351~P356)Think of that, since Kinetic energy:

Can we define

If so, 13

Ft

TorqueDefinition:The torque contributed by Fi exerted on particle i is

It is dependent of the origin we chose. i = riFisinq=diFi=riFi,t.di: ()Moment arm.Lever arm. 14

Fi,tdiparticle model Particle model origin

TorqueImportant Properties: Based on Newtons 3rd Law of motion and assume thatGravitational torque: Owing to g=constant.When Fnet=0, tnet is independent of the choose of the origin.

Fictitious force = -mia.15

Action (torque)Purpose:Understand how to find the center of mass.Get the feeling of t=Ia=rFsinq.Objects:A coat hanger (lever)Actions:Find its center of mass.Exert different forces on this lever.Which point you chose as the origin. Why?16Stop to ThinkIs it possible for us to explain the equation r1F1=r2F2 for a balance by Newtons 2nd and 3rd Law directly?

Hint: acceleration constraint & 17r1r2F2F1HomeworkStudent Workbook:12.15, 12.17, 12.20, 12.22,12.2418Applications (P357~P367)Rotation about a fixed axis.Rotation Tangential (linear) motionStatic Equilibrium.Rolling Motion.19Rotation about a fixed axis (P357~P359)E.g. rope and pulley.Kinematics:

Dynamics: .

20

TranslationalRotationalvelocityvt=rwaccelerationat=ra.

rExerciseThe acceleration of box m1.Frictionless.Ignore drag.Massless string.The rope turns on the pulley without slipping.21Im1m2

Static Equilibrium (P360~P363)The condition for a rigid body to be in static equilibrium is both

Exercise: problem 12.62 (P381)A 3.0-m-long rigid beam with a mass of 100 kg is supported at each end. An 80 kg student stands 2.0 m from support I. How much upward force does each support exert on the beam?22

12Rolling Motion (P364~P367)Condition: vCM=Rw. aCM=Ra.Reason:The contact point P, which is instantaneously at rest.23PRw2RwInertial reference framePRwViewed from its center of massRwExerciseThe acceleration of this cylinder.I= MR2.No slipping.Friction?24qHomeworkStudent Workbook12.25, 12.29,12.3025Angular Momentum (P371~P375)Angular Momentum: Properties:

For a rigid bodyI is a tensor in general.26

riLi

ExercisesThe change of the divers angular velocity when he extend his legs and arms.

Rotating bikes wheel and rotating coins.Questions: The motion of this wheel or coin.Frictionless?Which point you chose to be as the origin, and why?27

ICM1ICM2v=Rw

HomeworkStudent WorkbookP12-11~P12-12Textbook 12.2012.5112.5412.65

28Summary29KinematicsSingle particleMany particlesRotationPositionrirCMSimiri/Mq=s/rMass (moment of inertia for rotation)miMSimiISimiri2.Velocityvi=dri/dtvCM=drCM/dtw=dq/dtMomentumpi=miviPCM=MvCMLi=ripiLtot=Iw (*) Accelerationai=dvi/dtaCM=dvCM/dta=dw/dtDynamicsForce (torque)Fnet on i Fnet on system =Fext Newtons 2nd Law Kinetic EnergyKi= mivi2.Ktot= MvCM2+Kmicro.Krot= Iw2.

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