lec15_Rotational inertia and torque

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1、PhysicsIClass15RotationalInertiaandTorque1Review:Definitions2Review:AnalogyBetweenLinearandRotationalVariablesReview:AngularVelocity•Righthandrule.•Curlyourfingersinthedirectionofrotation.•Thumbpointsinthedirectionangularvelocity.•Mathematically4AngularAcceler

2、ationdωMagnitude:α=dtDirection:Speedingup–inthedirectionofωωSlowingdown–oppositetodirectionofω5Review:LinearandAngularVariables6EnergyinRotationalCoordinates7RotationalInertia8Example:RotationalInertiaofaRodrL2Irod=∫rdmdLdm=MdL=drLLMML222Irod=∫rdm=∫rd

3、r=∫rdrLL00ML312Irod==MLL339PropertiesofRotationalInertia10RigidBodiesSphericalshapes:11TorqueSpecialcasewhenF⊥rRotationofmassattachedtoalightrodmFrAxisofrotation⊗dp2a2a2τ=rF=r=rma=rm=mr=mrαdtrr2mr≡Irotationalinertiaτ=IαMoregenerally:definit

4、ionoftorqueτ=r×F=IαProblemsoftheDay1.A32kgwheel(hoop)withradius1.2m,isrotatingat280rev/min.Itmustbebroughttoastopin15s.Howmuchworkmustbedonebyanoutsideforcetostopit?13SolutionA30kgwheel(hoop)withradius1.2m,isrotatingat100rad/sec.Itmustbebroughttoastopin15s.How

5、muchworkmustbedonebyanoutsideforcetostopit12W=K−K=0−Iωfi22I=∫rdmr=const=R.2222I=R∫dm=RM=1.2×30=43.2kg-mΔθω==100rad/secΔt12125W=−Iω=−43.2×100=−2.16×10J2214ProblemsoftheDay2.Calculatethetorqueandforcewhichneedtobeappliedattheedgeofthehoopforittobebroughttoastopi

6、n15s.Δω100τ=Iα=I=−43.2=92.8N-mΔt15τ92.8τ=FRF===77N(=17lb)R1.215