Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterPhysics 121, March 25, 2008.Rotational Motion and Angular Momentum.Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterPhysics 121.March 25, 2008.• Course Information• Topics to be discussed today:• Review of Rotational Motion• Rolling Motion• Angular Momentum• Conservation of Angular MomentumFrank L. H. Wolfs Department of Physics and Astronomy, University of RochesterPhysics 121.March 25, 2008.• Homework set # 7 is now available and is due on SaturdayApril 5 at 8.30 am.• There will be no workshops and office hours for the rest ofthe week. We will be busy grading exam # 2.• The grades for exam # 2 will be distributed via email onMonday March 31.• You should pick up your exam in workshop next week.Please check it carefully for any errors.Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterPhysics 121.Quiz lecture 17.• The quiz today will have 4 questions!Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterRotational variables.A quick review.• The variables that are used todescribe rotational motion are:• Angular position θ• Angular velocity ω = dθ/dt• Angular acceleration α = dω/dt• The rotational variables arerelated to the linear variables:• Linear position l = Rθ• Linear velocity v = Rω• Linear acceleration a = RαFrank L. H. Wolfs Department of Physics and Astronomy, University of RochesterRotational Kinetic Energy. A quick review.• Since the components of a rotating object have a non-zero(linear) velocity we can associate a kinetic energy with therotational motion:• The kinetic energy is proportional to square of the rotationalvelocity ω. Note: the equation is similar to the translationalkinetic energy (1/2 mv2) except that instead of beingproportional to the the mass m of the object, the rotationalkinetic energy is proportional to the moment of inertia I ofthe object:Note: units of I: kg m2K =12mivi2i!=12mi"ri( )2i!=12miri2i!#$%&'("2=12I"2I = miri2i!or I = r2dm"Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterTorque.• In general the torque associated witha force F is equal to• The arm of the force (also called themoment arm) is defined as rsinφ.The arm of the force is theperpendicular distance of the axis ofrotation from the line of action ofthe force.• If the arm of the force is 0, thetorque is 0, and there will be norotation.• The maximum torque is achievedwhen the angle φ is 90°. !!= rF sin"=!r #!FFrank L. H. Wolfs Department of Physics and Astronomy, University of RochesterTorque. A quick review.• The torque τ of the force F is relatedto the angular acceleration α:τ = Ια• This equation looks similar toNewton’s second law for linearmotion:F = ma• Note: linear rotational mass m moment I force F torque τArφFFrank L. H. Wolfs Department of Physics and Astronomy, University of RochesterRolling motion. A quick review.• Rolling motion is a combinationof translational and rotationalmotion.• The kinetic energy of rollingmotion has thus twocontributions:• Translational kinetic energy =(1/2) M vcm2.• Rotational kinetic energy =(1/2)!Icm!ω 2.• Assuming the wheel does notslip: ω = v / R.Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterHow different is a world with rotationalmotion? Sample problem.• Consider the loop-to-loop. Whatheight h is required to make it to thetop of the loop?• First consider the case withoutrotation:• Initial mechanical energy = mgh.• Minimum velocity at the top of theloop is determine by requiring thatmv2/R > mgorv2 > gR• The mechanical energy is thus equalto(1/2)mv2 + 2mgR > (5/2)mgR• Conservation of energy requiresh > (5/2)RhRFrank L. H. Wolfs Department of Physics and Astronomy, University of RochesterHow different is a world with rotationalmotion? Sample problem.• What changes when the objectrotates?• The minimum velocity at the topof the loop will not change.• The minimum translational kineticenergy at the top of the loop willnot change.• But in addition to translationalkinetic energy, there is now alsorotational kinetic energy.• The minimum mechanical energyis at the top of the loop has thusincreased.• The required minimum heightmust thus have increased.• OK, let’s now calculate by howmuch the minimum height hasincreased.hRFrank L. H. Wolfs Department of Physics and Astronomy, University of RochesterHow different is a world with rotationalmotion? Sample problem.• The total kinetic energy at the topof the loop is equal to• This expression can be rewrittenas• We now know the minimummechanical energy required toreach this point and thus theminimum height:hRNote: without rotation h ≥ 25/10 R !!!Kf=12I!2+12Mv2=12Ir2+ M"#$%&'v2Kf=1225M + M!"#$%&v2=710Mv2h !2710RFrank L. H. Wolfs Department of Physics and Astronomy, University of RochesterTorque and rotational motion.• Let’s test our understanding of the basic aspects of torqueand rotational motion by working on the following conceptproblems:• Q17.1• Q17.2• Q17.3• Q17.4• Q17.5Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterTorque.• The torque associated with aforce is a vector. It has amagnitude and a direction.• The direction of the torque can befound by using the right-handrule to evaluate r x F.• For extended objects, the totaltorque is equal to the vector sumof the torque associated with each“component” of this object.Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterAngular momentum.• We have seen many similaritiesbetween the way in which wedescribe linear and rotationalmotion.• Our treatment of these types ofmotion are similar if werecognize the followingequivalence: linear rotational mass m moment I force F torque τ = r x F• What is the equivalent to linearmomentum? Answer: angularmomentum.Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterAngular momentum.• The angular momentum isdefined as the vector productbetween the position vector andthe linear momentum.• Note:• Compare