Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterPhysics 121.March 20, 2008.Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterPhysics 121.March 20, 2008.• Course Information• Quiz• Topics to be discussed today:• Rotational Variables (Review)• Torque• Rolling Motion• Review for Exam 2Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterPhysics 121.March 20, 2008.• Homework set # 6 is now available on the WEB and will bedue on Saturday morning, March 22, at 8.30 am.• There will be no homework due on March 29.• Exam # 2 will take place on Tuesday March 25 at 8 am inHubbell. It will cover the material discussed in Chapters 7,8, and 9.• There will be no workshops or office hours on Tuesday -Friday next week.• Extra office hours will be scheduled for Sunday 3/23 andMonday 3/24.Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterPhysics 121.Quiz lecture 16.• The quiz today will have 3 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 variables.A quick review.• Things to consider when lookingat the rotation of rigid objectsaround a fixed axis:• Each part of the rigid object hasthe same angular velocity.• Only those parts that are locatedat the same distance from therotation axis have the same linearvelocity.• The linear velocity of parts of therigid object increases withincreasing distance from therotation axis.Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterRotational variables.A quick review.• Note: the acceleration at = rα isonly one of the two component ofthe acceleration of point P. Thetwo components of theacceleration of point P are:• The radial component: thiscomponent is always present sincepoint P carried out circular motionaround the axis of rotation.• The tangential component: thiscomponent is present only whenthe angular acceleration is notequal to 0 rad/s2.Frank L. H. Wolfs Department of Physics and Astronomy, University of RochesterRotational variables.A quick review.Angular velocity and acceleration are vectors! They have a magnitude and a direction. The direction of ω is found using the right-hand rule.The angular acceleration is parallel or anti-parallel to the angular velocity:If ω increases: parallelIf ω decreases: anti-parallelFrank 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.• Consider a force F applied to anobject that can only rotate.• The force F can be decomposedinto two two components:• A radial component directedalong the direction of the positionvector r. The magnitude of thiscomponent is Fcosφ. Thiscomponent will not produce anymotion.• A tangential component,perpendicular to the direction ofthe position vector r. Themagnitude of this component isFsinφ. This component will resultin rotational motion.ArφFFrank L. H. Wolfs Department of Physics and Astronomy, University of RochesterTorque.• If a mass m is located at the positionon which the force is acting (and weassume any other masses can beneglected), it will experience alinear acceleration equal to Fsinφ/m.• The corresponding angularacceleration is equal to Fsinφ/(mr).• Since in rotational motion themoment of inertia plays animportant role, we will rewrite theangular acceleration in terms of themoment of inertia:α = rFsinφ / (mr2) = rFsinφ / ΙArφFFrank L. H. Wolfs Department of Physics and Astronomy, University of RochesterTorque.• Consider rewriting the previousequation in the following way:rFsinφ = Ια• The left-hand-side of this equation iscalled the torque τ of the force F: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 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 RochesterRotational motion.Sample problem.• Consider a uniform disk withmass M and radius R. The disk ismounted on a fixed axle. A blockwith mass m hangs from a lightcord that is wrapped around therim of the disk. Find theacceleration of the falling block,the angular acceleration of thedisk, and the tension of the cord.• Expectations:• Linear acceleration shouldapproach g when M approaches 0kg.mM, RFrank L. H. Wolfs Department of Physics and Astronomy, University of RochesterRotational motion.Sample problem.• Start with considering the forcesand torques involved.• Define the sign convention to beused.• The block will move down andwe choose the positive and wechoose the positive y