Physics 207, Lecture 16, Oct. 27Connection with motion...Rotational Dynamics: What makes it spin?Torque is a vector quantityExercise Torque MagnitudeSlide 6For a point mass NET = m r2 a and inertiaCalculating Moment of InertiaCalculating Moment of Inertia...Moments of InertiaRotation & Kinetic EnergySlide 14Exercise Rotational Kinetic EnergyWork & Kinetic Energy:Angular Momentum:Angular momentum of a rigid body about a fixed axis:Example: Two DisksSlide 28Example: Throwing ball from stoolSlide 30Angular Momentum as a Fundamental QuantityFundamental Angular MomentumIntrinsic Angular MomentumAngular Momentum of a MoleculeAngular Momentum of a Molecule (It heats the water in a microwave over)Slide 36Physics 207: Lecture 16, Pg 1Physics 207, Lecture 16, Oct. 27Goals:Goals:•Chapter 12Chapter 12 Extend the particle model to rigid-bodies Understand the equilibrium of an extended object. Understand rotation about a fixed axis. Employ “conservation of angular momentum” conceptAssignment: HW7 due Oct. 29Wednesday: Review sessionPhysics 207: Lecture 16, Pg 2Connection with motion...So for a solid object which rotates about its center of mass and whose CM is moving: VCM2CM21RotationalTOTALVK MK MmNiii1CMrRPhysics 207: Lecture 16, Pg 3Rotational Dynamics: What makes it spin?NET = |r| |FTang| ≡ |r| |F| sin Only the tangential component of the force matters. With torque the position of the force mattersTorque is the rotational equivalent of forceTorque has units of kg m2/s2 = (kg m/s2) m = N mA constant torque gives constant angular acceleration iff the mass distribution and the axis of rotation remain constant. FTangential arFradial F A force applied at a distance from the rotation axis gives a torquePhysics 207: Lecture 16, Pg 4Torque is a vector quantityMagnitude is given by |r| |F| sin or, equivalently, by the |Ftangential | |r| or by |F| |rperpendicular to line of action | Direction is parallel to the axis of rotation with respect to the “right hand rule” And for a rigid object = I rF F cos(90°) = FTang. rFFradial F arF r90°r sin line of actionPhysics 207: Lecture 16, Pg 5Exercise Torque MagnitudeA. Case 1B. Case 2C. SameIn which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same.Remember torque requires F, r and sin or the tangential force component times perpendicular distanceLLF F axiscase 1 case 2Physics 207: Lecture 16, Pg 6Rotational Dynamics: What makes it spin?NET = |r| |FTang| ≡ |r| |F| sin Torque is the rotational equivalent of forceTorque is the rotational equivalent of forceTorque has units of kg m2/s2 = (kg m/s2) m = N mFTangential aarFradial FF A force applied at a distance from the rotation axisNET = r FTang = r m aTang = r m r = (m r2) For every little part of the wheelPhysics 207: Lecture 16, Pg 7For a point mass NET = m r2 and inertiaThis is the rotational version of FNET = ma Moment of inertia, Moment of inertia, I ≡ m r2 , (here (here II is just a point is just a point on the wheel) is the rotational equivalent of mass.on the wheel) is the rotational equivalent of mass. If I is big, more torque is required to achieve a given angular acceleration.FTangential aarFrandial FF The further a mass is away from this axis the greater the inertia (resistance) to rotationPhysics 207: Lecture 16, Pg 8Calculating Moment of Inertia where r is the distance from the mass to the axis of rotation.Niiirm12IExample: Calculate the moment of inertia of four point masses(m) on the corners of a square whose sides have length L, about a perpendicular axis through the center of the square:mmmmLPhysics 207: Lecture 16, Pg 9Calculating Moment of Inertia...For a single object, I depends on the rotation axis!Example: I1 = 4 m R2 = 4 m (21/2 L / 2)2LI = 2mL2I2 = mL2mmmmI1 = 2mL2Physics 207: Lecture 16, Pg 11Moments of Inertia Solid disk or cylinder of mass M and radius R, about perpendicular axis through its center.I = ½ M R2Some examples of I for solid objects:RLrdrdmr2IrdmFor a continuous solid object we have to add up the mr2 contribution for every infinitesimal mass element dm. An integral is required to find I :Use the table…Physics 207: Lecture 16, Pg 13Rotation & Kinetic EnergyConsider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods).The kinetic energy of this system will be the sum of the kinetic energy of each piece: K = ½m1v1+ ½m2v2+ ½m3v3+ ½m4v4rr1rr2rr3rr4m4m1m2m3 41221viiimKPhysics 207: Lecture 16, Pg 14Rotation & Kinetic EnergyNotice that v1 = r1 , v2 = r2 , v3 = r3 , v4 = r4 So we can rewrite the summation:We recognize the quantity, moment of inertia or I, and write:rr1rr2rr3rr4m4m1m2m3 24122141222141221][rrviiiiiiiiimmmK221RotationalIKPhysics 207: Lecture 16, Pg 15Exercise Rotational Kinetic EnergyA. ¼ B. ½C. 1D. 2E. 4We have two balls of the same mass. Ball 1 is attached to a 0.1 m long rope. It spins around at 2 revolutions per second. Ball 2 is on a 0.2 m long rope. It spins around at 2 revolutions per second. What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ?221IKiiirm2IBall 1Ball 2Physics 207: Lecture 16, Pg 19Work & Kinetic Energy:Recall the Work Kinetic-Energy Theorem: K = WNETThis applies to both rotational as well as linear motion.So for an object that rotates about a fixed axisFor an object which is rotating and translating NET2221 I WKif2CM212CM21VK MI Physics 207: Lecture 16, Pg 25Angular Momentum:0pdtdFWe have shown that for a system of particles, momentum is conserved if What is the rotational equivalent of this? angular momentumis conserved if vmpIL0dtLdPhysics 207: Lecture 16, Pg 26Angular momentum of a rigid bodyabout a fixed axis:Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin Is the sum of the angular momentum of each particle:Even if no
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