ReviewDirectionTheRightHandRule

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1、Physics 131:Rotational Motion IIIReview:Direction&The Right Hand RuleReview:Direction&The Right Hand RulelTo figure out in which direction the rotation vector points,curl the fingers of your right hand the same way the object turns,and your thumb will point in the direction of the rotation vector!lW

2、e normally pick the z-axis to be the rotation axis as shown.=z=z=zlFor simplicity we omit the subscripts unless explicitly needed.xyzxyzPhysics 131:Rotational Motion IIIReview:Torque and Angular AccelerationReview:Torque and Angular Acceleration NET=IlThis is the rotational analogue of FNET=mal lTor

3、que is the rotational analogue of force:Torque is the rotational analogue of force:The amount of“twist”provided by a force.l lMoment of inertiaMoment of inertia I I is the rotational analogue of massis the rotational analogue of massIf I is big,more torque is required to achieve a given angular acce

4、leration.Physics 131:Rotational Motion IIIRotationsRotationslTwo wheels can rotate freely about fixed axles through their centers.The wheels have the same mass,but one has twice the radius of the other.lForces F1 and F2 are applied as shown.What is F2/F1 if the angular acceleration of the wheels is

5、the same?(a)1(b)2 (c)4 F1F2Physics 131:Rotational Motion IIISolutionSolutionWe know butandsoF1F2Since R2=2 R1Physics 131:Rotational Motion IIIReview:Work&EnergyReview:Work&EnergylThe work done by a torque acting through a displacement is given by:lThe power provided by a constant torque is therefore

6、 given by:Physics 131:Rotational Motion IIIFalling weight&pulleyFalling weight&pulleylA mass m is hung by a string that is wrapped around a pulley of radius R attached to a heavy flywheel.The moment of inertia of the pulley+flywheel is I.The string does not slip on the pulley.Starting at rest,how lo

7、ng does it take for the mass to fall a distance L.I mRT mg aLPhysics 131:Rotational Motion IIIFalling weight&pulley.Falling weight&pulley.lFor the hanging mass use F=mamg-T=malFor the pulley+flywheel use =I =TR=I lRealize that a=RlNow solve for a using the above equations.I mRT mg aLPhysics 131:Rota

8、tional Motion IIIFalling weight&pulley.Falling weight&pulley.lUsing 1-D kinematics(Lecture 1)we can solve for the time required for the weight to fall a distance L:I mRT mg aLwherePhysics 131:Rotational Motion IIIRotation around a moving axis.Rotation around a moving axis.lA string is wound around a

9、 puck(disk)of mass M and radius R.The puck is initially lying at rest on a frictionless horizontal surface.The string is pulled with a force F and does not slip as it unwinds.What length of string L has unwound after the puck has moved a distance D?FRM Top viewPhysics 131:Rotational Motion IIIRotati

10、on around a moving axis.Rotation around a moving axis.lThe CM moves according to F=MAFM AlThe distance moved by the CM is thus RlThe disk will rotate about its CM according to =I lSo the angular displacement isPhysics 131:Rotational Motion IIIRotation around a moving axis.Rotation around a moving ax

11、is.lSo we know both the distance moved by the CM and the angle of rotation about the CM as a function of time:DFF Divide(b)by(a):(a)(b)LThe length of stringpulled out is L=R:Physics 131:Rotational Motion IIIComments on CM acceleration:Comments on CM acceleration:lWe just used =I for rotation about a

12、n axis through the CM even though the CM was accelerating!The CM is not an inertial reference frame!Is this OK?(After all,we can only use F=ma in an inertial reference frame).l lYES!YES!We can always write =I for an axis through the CM.This is true even if the CM is accelerating.We will prove this w

13、hen we discuss angular momentum!FRM A Physics 131:Rotational Motion IIIRollingRollinglAn object with mass M,radius R,and moment of inertia I rolls without slipping down a plane inclined at an angle with respect to horizontal.What is its acceleration?lConsider CM motion and rotation about the CM sepa

14、rately when solving this problem(like we did with the lastproblem).RI MPhysics 131:Rotational Motion IIIRolling.Rolling.lStatic friction f causes rolling.It is an unknown,so we must solve for it.lFirst consider the free body diagram of the object and use FNET=MACM:In the x direction Mg sin -f=MAlNow

15、 consider rotation about the CMand use =I realizing that =Rf and A=RRM fMgyxPhysics 131:Rotational Motion IIIRolling.Rolling.lWe have two equations:Mg sin -f=MAlWe can combine these to eliminate f:ARI MFor a sphere:Physics 131:Rotational Motion IIIRotationsRotationslTwo uniform cylinders are machine

16、d out of solid aluminum.One has twice the radius of the other.If both are placed at the top of the same ramp and released,which is moving faster at the bottom?(a)bigger one(b)smaller one(c)same Physics 131:Rotational Motion IIISolutionSolutionlConsider one of them.Say it has radius R,mass M and fall

17、s a height H.HEnergy conservation:-DU=DKbutandPhysics 131:Rotational Motion IIISolutionSolutionHSo:So,(c)does not depend on size,as long as the shape is the same!Physics 131:Rotational Motion IIISliding to RollingSliding to RollinglA bowling ball of mass M and radius R is thrown with initial velocit

18、y v0.It is initially not rotating.After sliding with kinetic friction along the lane for a distance D it finally rolls without slipping and has a new velocity vf.The coefficient of kinetic friction between the ball and the lane is .What is the final velocity,vf,of the ball?vf=Rf=Mgv0DPhysics 131:Rot

19、ational Motion IIISliding to Rolling.Sliding to Rolling.lWhile sliding,the force of friction will accelerate the ball in the-x direction:F=-Mg=Ma so a=-glThe speed of the ball is therefore v=v0-gt(a)lFriction also provides a torque about the CM of the ball.Using =I and remembering that I=2/5MR2 for

20、a solid sphere about an axis through its CM:Dxf=Mg(b)v f=Rv0Physics 131:Rotational Motion IIISliding to Rolling.Sliding to Rolling.lWe have two equations:lUsing(b)we can solve for t as a function of lPlugging this into(a)and using vf=R(the condition for rolling without slipping):Dx(a)(b)f=MgDoesnt d

21、epend on ,M,g!vf=Rv0Physics 131:Rotational Motion IIIRotationsRotationslA bowling ball(uniform solid sphere)rolls along the floor without slipping.What is the ratio of its rotational kinetic energy to its translational kinetic energy?Recall that for a solid sphere about an axis through its CM:(a)(b)

22、(c)Physics 131:Rotational Motion IIISolutionSolutionlThe total kinetic energy is partly due to rotation and partly due to translation(CM motion).rotationalKtranslationalKK=Physics 131:Rotational Motion IIISolutionSolutionSince it rolls without slipping:rotationalKTranslationalKK=Physics 131:Rotation

23、al Motion IIIAtwoods Machine with Massive Pulley:Atwoods Machine with Massive Pulley:lA pair of masses are hung over a massive disk-shaped pulley as shown.Find the acceleration of the blocks.m2m1RM yxm2gaT1 m1gaT2 lFor the hanging masses use F=ma -m1g+T1=-m1a -m2g+T2=m2a(Since for a disk)lFor the pu

24、lley use =I T1R-T2RPhysics 131:Rotational Motion IIIAtwoods Machine with Massive Pulley.Atwoods Machine with Massive Pulley.lWe have three equations and three unknowns(T1,T2,a).Solve for a.-m1g+T1=-m1a (1)-m2g+T2=m2a (2)T1-T2 (3)m2m1RM yx m2m1m2gaT1 m1gaT2 Physics 131:Rotational Motion IIIRecap of t

25、odays lectureRecap of todays lecturelReview lMany body dynamicslWeight and massive pulley lRolling and sliding examples lRotation around a moving axis:Puck on ice lRolling down an incline lBowling ball:sliding to rollinglAtwoods Machine with a massive pulley l lLook at textbook problems Look at textbook problems Chapter 11:#36,37,53,55,57