Rotational Dynamics Moment Of Inertia.
Lecture 18Chapter 10Physics I11.18.2013Rotational DynamicsMoment of Inertia.Course website:http://faculty.uml.edu/Andriy Danylov/Teaching/PhysicsILecture 1fall.html95.141, Fall 2013, Lecture 18Department of Physics and Applied Physics
OutlineChapter 10 Moment of InertiaParallel Axis TheoremRotational kinetic energyRolling95.141, Fall 2013, Lecture 18Department of Physics and Applied Physics
Newton’s 2nd law of rotationForce causes linear acceleration: (N.2nd law): F maTorque causes angular acceleration: I I is the Moment of Inertia (rotational equivalent of mass)95.141, Fall 2013, Lecture 18Department of Physics and Applied Physics
Moment of inertia of a single particleA point mass is located at a distance R from an axis of rotation.A force is applied perpendicular to R.Let’s find a relation between torque and angular acceleration: F 90 RBy definition:N. 2nd law:Recall, last class:m RFSin RFF ma mR a R As a result, torque is: R(mR ) (mR 2 ) Moment of inertia of a single particle:Rotational N. 2nd law:95.141, Fall 2013, Lecture 18Department of Physics and Applied PhysicsI mR 2 I
Moment of inertia of many particleIf we have many point masses mi, located at distances Ri from anaxis of rotation. A force is applied perpendicular to R.Moment of inertia of N masses:m3m4R3R4I m1 R12 m2 R22 m3 R32 .m2NI mi RR2R1i 1m12iRotational N. 2nd law:N I ( mi R ) i 195.141, Fall 2013, Lecture 18Department of Physics and Applied Physics2i
Example: Moments of Inertia of two pointsTwo point masses connected to a massless rodI m1 R12 m2 R22I 5kg (2m) 2 7kg (2m) 2 48 kg m 2I m R m2 R21 122I 5kg (0.5m) 2 7kg (4.5m) 2 143 kg m 2The distribution of mass matters95.141, Fall 2013, Lecture 18Department of Physics and Applied Physics
Moment of inertia for extended objectsI R dm2 miRiThe rotational inertia of an object depends notonly on its mass distribution but also the locationof the axis of rotation—compare (f) and (g), for example.95.141, Fall 2013, Lecture 18Department of Physics and Applied Physics
Example: pulley and mass/PhysicsMAn object of mass m is hung from a cylindrical pulley of radius R andmass M and released from rest. What is the acceleration of the object?RWe have two objects in the system:- translational motion of m, describedby N.2nd law (1) F ma FT- rotational motion of M, described by the rotational N.2nd law(2) FT I - and there is a useful eq-n, which links these 2 eq-ns:(3) (1) F ma atan R amg FT ma1(2) I FT RSin90 I (3)atan R a a R95.141, Fall 2013, Lecture 18Department of Physics and Applied Physics yPhysics is over.Now, it is pure Algebra.3 eq-ns and 3 unknownsI ( see table10 20) mg
Example: pulley and mass/Algebra(1)mg FT maFT R I (3) a R R Multiply both sides of (1) by Rmg FT maa and add (1) and (2) up:FT R I(FT disappears)RamgR FT R FT R maR IRmgmRga a OrbetterIIm 2mR RRMR(2)I ( see table 10 20)sinceI 1MR 2295.141, Fall 2013, Lecture 18Department of Physics and Applied Physics FT FTymga Mm 2 mg
Torque due to gravity We often encounter systems in which there is a torque exertedby gravity. The torque on a body about any axis of rotation is RCM WSin RCM MgSin The proof RCMCM W Mg95.141, Fall 2013, Lecture 18Department of Physics and Applied Physics
Example Problem: Falling rodWhat is the angular acceleration of the rod shown below, if it is released fromrest, at the moment it is released? What is the linear acceleration of the tip?Rotational motion of the rod isdescribed by the rotational N.2nd law I Torque due to gravity (previous slide): RCM RCM WSin RCM MgSin MgRCM Sin90 Mg l 2lMg I 2MglMgl3g 1 2 2I2( 3 Ml ) 2lI 13 Ml2atan R 95.141, Fall 2013, Lecture 18Department of Physics and Applied PhysicsEach point on a rotating rigidbody has the same angularacceleration (previous class)!So we can apply it to the tip!!3g l (3g 2l ) 2
Parallel Axis Theorem (w/out proof)The moment of inertia about any axis parallel to that axis throughthe center of mass is given byII CMI I CM Mh2hCMI : moment of inertia about any parallel axisICM : moment of inertia about an axis through its center of massM : total massh : distance from a parallel axis to the center of mass.BTW: The moment of inertia of any object about an axis through its center ofmass is the minimum moment of inertia for an axis in that direction in space.95.141, Fall 2013, Lecture 18Department of Physics and Applied Physics
Parallel Axis Theorem: Example/SphereFor a uniform sphere of radius r0Moment of inertia for the sphere, rotating aboutan axis through its center of massI CM Mr2520Moment of inertia for the sphere about anaxis going through the edge of the sphere?Apply Parallel Axis Theorem:I I CM Mh 2 52 Mr02 Mr02 75 Mr02Through an axis a distance R r0 from the center?Very smallI I CM Mh 2 52 Mr02 MR 2 MR 2So, in this case we got aMoment of inertia of a single particle95.141, Fall 2013, Lecture 18Department of Physics and Applied PhysicsR r0
Rotational Kinetic Energy vi riSimple derivation: for pure rotationK rot mi vi mi12K rot mi ( ri ) 12 mi 2 ri2212K rot 122Since I thereforeK transvi ri 21 2 mv2 m r2i i2mr iiK rot I 122Similar to linear motion95.141, Fall 2013, Lecture 18Department of Physics and Applied PhysicsK rot1 2 I 2
Total kinetic energyRolling motion can be resolved into two motionsRotationalRollingTranslational Total Kinetic Energy KCM KrotKtot Mv2CM12Ktot MR I CM 222vCM R If there is no slipping12 I CM 1212295.141, Fall 2013, Lecture 18Department of Physics and Applied PhysicsKtot 12 (MR2 I CM ) 2
ConcepTest 1A force is applied to a dumbbellfor a certain period of time, firstas in (a) and then as in (b). Inwhich case does the dumbbellacquire the greatercenter-of-mass speed ?Because the same force acts for thesame time interval in both cases, thechange in momentum must be thesame, thus the CM velocity must bethe same.Dumbbell IA) case (a)B) case (b)C) no differenceD) it depends on the rotationalinertia of the dumbbellCMCM
ConcepTest 2A force is applied to a dumbbellfor a certain period of time, firstas in (a) and then as in (b). Inwhich case does the dumbbellacquire the greater energy ?Dumbbell IIA) case (a)B) case (b)C) no differenceD) it depends on the rotationalinertia of the dumbbellIf the CM velocities are the same, thetranslational kinetic energies mustbe the same. Because dumbbell (b)is also rotating, it has rotationalkinetic energy in addition.CMCM
Sphere rolling downan incline95.141, Fall 2013, Lecture 18Department of Physics and Applied Physics
Thank youSee you on Wednesday95.141, Fall 2013, Lecture 18Department of Physics and Applied Physics
Moment of inertiaThe distribution of mass matters here—these two objects have the samemass, but the one on the left has a greater rotational inertia, as so muchof its mass is far from the axis of rotation.object1object 2MMI1 I 295.141, Fall 2013, Lecture 18Department of Physics and Applied Physics
I : moment of inertia about any parallel axis I CM: moment of inertia about an axis through its center of mass M : total mass h : distance from a parallel axis to the center of mass. BTW: The moment of inertia of any object about an axis through its center of mass is the minimum moment of
31 Moment of Inertia by Integraion Monday, November 19, 2012 An Example ! The polar moment of inertia, J O, is the sum of the moments of inertia about the x and y axis y x yx2 4 1 2 4 yx 4m 4m 44 4 21.94 21.94 43.88 Ox y O O JII Jm m Jm 32 Moment of Inertia by Integraion Monday, N
3 Moment of Inertia - Composite Area Monday, November 26, 2012 Radius of Gyration ! The radius of gyration, k, is the square root of the ratio of the moment of inertia to the area I x x y y O xy O k A I k A J II k AA 4 Moment of Inertia - Composite Area Monday, November 26, 2012 Para
MOMENT OF INERTIA The moment of inertia, or the second moment of area, is defined as I y x2 dA I x y2 dA The polar moment of inertia J of an area about a point is equal to the sum of the moments of inertia of the area about any two point. I z J I y I x (x 2 y2) dA
Moment of Inertia is always positive. Units of Moment of Inertia are length raised to the fourth power, such as in4 or m4. Polar Moment of Inertia The second moment of Area A with respect to the pole O or the z-ax
dependent on the wheel inertia. So it is crucial that the inertia of the tire-wheel combination is simulated. The typical problem is that the dynamometer inertia is greater than the tire-wheel inertia. So the simulation must apply a force to the dyno inertia to compensate for the inertia difference
J J J N- m-s / rad B B B N- m-s/rad 2 T m L T m L Where: B T total viscous friction J T total rotational inertia B L load viscous friction B m motor viscous friction J m motor rotational inertia JL load rotational inertia With speed changer J1 N- m- s / rad N N J J B N m- s/rad N N 2 L 2 T m L 2 2 1 T m » ¼ º « ª » ¼ º .
Moment of inertia of an object is an indication of the level of force that has to be applied in order to set the object, or keep the object, in motion about a defined axis of rotation. Moment of inertia, which is a derivative of Newton’s second law, is sometimes referred to as the second moment
Counselling skills will be selected from the range below: Active listening Reflection Paraphrasing Summarising. Open questioning Body language 2 A record of feedback from others in relation to counselling skills Candidates must take account of the feedback from the teacher/lecturer. 3 A record of the candidate’s identified areas for improvement The Assessment Support Pack (ASP) for this Unit .
