Monday, Nov. 4, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #14Monday, Nov. 4, 2002Dr. JaehoonYu1. Parallel Axis Theorem2. Torque3. Torque & Angular Acceleration 4. Work, Power and Energy in RotationToday’s homework is homework #14 due 12:00pm, Monday, Nov. 11!!Monday, Nov. 4, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu2Announcements• 2ndTerm Exam– Grading is completed• Maximum Score: 87• Numerical Average: 58.1• Four persons missed the exam without a prior approval– Can look at your exam after the class– All scores are relative based on the curve• One worst after the adjustment will be dropped– Exam constitutes only 50% of the total• Do your homework well• Come to the class and do well with quizzesMonday, Nov. 4, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu32ndTerm Exam DistributionsMean: 58Mean: 5114% ImprovementA lot narrower distribution. è Even improvementsBut as always, you could do better!!!Monday, Nov. 4, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu4Calculation of Moments of InertiaMoments of inertia for large objects can be computed, if we assume the object consists of small volume elements with mass, ∆mi.It is sometimes easier to compute moments of inertia in terms of volume of the elements rather than their massUsing the volume density, ρ, replace dmin the above equation with dV.The moment of inertia for the large rigid object isHow can we do this?∑∆=→∆iiimmrIi20lim∫=dmr2dVdm=ρThe moments of inertia becomes∫=dVrI2ρExample 10.5: Find the moment of inertia of a uniform hoop of mass M and radius R about an axis perpendicular to the plane of the hoop and passingthrough its center.xyROdmThe moment of inertia is∫=dmrI2What do you notice from this result?The moment of inertia for this object is the same as that of a point of mass M at the distance R.∫= dmR22MR=dVdmρ=Monday, Nov. 4, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu5xy(x,y)xCM(xCM,yCM)yCMCMParallel Axis TheoremMoments of inertia for highly symmetric object is easy to compute if the rotational axis is the same as the axis of symmetry. However ifthe axis of rotation does not coincide with axis of symmetry, the calculation can still be done in simple manner using parallel-axis theorem.2MDIICM+=yxrMoment of inertia is defined∫=dmrI2Since x and y arex’y’'xxxCM+=One can substitute x and y in Eq. 1 to obtain()()[]∫+++=dmyyxxICMCM22''Since the x’ and y’ are the distance from CM, by definition∫= 0'dmxDTherefore, the parallel-axis theoremCMIMD +=2What does this theorem tell you?Moment of inertia of any object about any arbitrary axis are thesame as the sum of moment of inertia for a rotation about the CM and that of the CM about the rotation axis.()(1) 22∫+= dmyx'yyyCM+=()()dmyxdmyydmxxdmyxCMCMCMCM∫∫∫∫+++++=2222'''2'2∫= 0'dmy()()dmyxdmyxICMCM∫∫+++=2222''Monday, Nov. 4, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu6Example 10.8Calculate the moment of inertia of a uniform rigid rod of lengthL and mass M about an axis that goes through one end of the rod, using parallel-axis theorem.The line density of the rod is Using the parallel axis theoremLM=λso the masslet is dxLMdxdm== λThe moment of inertia about the CM CMIMDIICM2+=The result is the same as using the definition of moment of inertia.Parallel-axis theorem is useful to compute moment of inertia of a rotation of a rigid object with complicated shape about an arbitrary axisxyLxdxCMMLML22212+=∫= dmr2dxLMxLL∫−=2/2/22/2/331LLxLM−=−−=33223LLLM124323MLLLM==3412222MLMLML=+=Monday, Nov. 4, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu7TorqueTorque is the tendency of a force to rotate an object about an axis. Torque, τ, is a vector quantity.FdrF=≡φτsinMagnitude of torque is defined as the product of the force exerted on the object to rotate it and the moment arm.FφdLine of ActionConsider an object pivoting about the point Pby the force F being exerted at a distance r. PrMoment armThe line that extends out of the tail of the force vector is called the line of action.The perpendicular distance from the pivoting point P to the line of action is called Moment arm.When there are more than one force being exerted on certain points of the object, one can sum up the torque generated by each force vectorially. The convention for sign of the torque is positive if rotation is in counter-clockwise and negative if clockwise. d2F221τττ +=∑22dFFd −=Monday, Nov. 4, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu8R1Example 10.9A one piece cylinder is shaped as in the figure with core section protruding from the larger drum. The cylinder is free to rotate around the central axis shown in the picture. A rope wrapped around the drum whose radius is R1exerts force F1to the right on the cylinder, and another force exerts F2on the core whose radius is R2downward on the cylinder. A) What is the net torque acting on the cylinder about the rotation axis?The torque due to F1 111FR−=τSuppose F1=5.0 N, R1=1.0 m, F2= 15.0 N, and R2=0.50 m. What is the net torque about the rotation axis and which way does the cylinder rotate from the rest?R2F1F2and due to F2 222FR=τUsing the above result221121FRFR+−=+=∑τττSo the total torque acting on the system by the forces ismNFRFR•=×+×−=+−=∑5.250.00.150.10.52211τThe cylinder rotates in counter-clockwise.Monday, Nov. 4, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu9Torque & Angular AccelerationLet’s consider a point object with mass mrotating on a circle.What does this mean?The tangential force Ftand radial force FrThe tangential force FtisWhat do you see from the above relationship?mrFtFrWhat forces do you see in this motion?ttmaF =The torque due to tangential force FtisrFt=τατI=Torque acting on a particle is proportional to the angular acceleration.What law do you see from this relationship?Analogs to Newton’s 2ndlaw of motion in rotation.How about a rigid object?rdFtdmOThe external tangential force dFtisαdmrdmadFtt==αατIdmr==∫∑2The torque due to tangential force FtisThe total torque is()ατdmrrdFdt2==What is the contribution due to radial force and why?Contribution from radial force is 0, because its line of action passes through the pivoting point, making the moment arm 0.αmr=rmat=α2mr=Monday, Nov. 4, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu10Example 10.10A uniform rod of length
View Full Document
Unlocking...