Wednesday, Nov. 10, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu11. Moment of Inertia2. Parallel Axis Theorem 3. Torque and Angular Acceleration4. Rotational Kinetic Energy5. Work, Power and Energy in Rotation 6. Angular Momentum & Its ConservationPHYS 1443 – Section 003Lecture #19Wednesday, Nov. 10, 2004Dr. Jaehoon YuToday’s homework is HW #10, due 1pm next Wednesday!!Wednesday, Nov. 10, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu2Moment of Inertia Rotational Inertia:What are the dimension and unit of Moment of Inertia?∑≡iiirmI22mkg⋅[]2MLMeasure of resistance of an object to changes in its rotational motion. Equivalent to mass in linear motion.Determining Moment of Inertia is extremely important for computing equilibrium of a rigid body, such as a building.dmrI∫≡2For a group of particlesFor a rigid bodyWednesday, Nov. 10, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu3Example for Moment of InertiaIn a system of four small spheres as shown in the figure, assuming the radii are negligible and the rods connecting the particles are massless, compute the moment of inertia and the rotational kinetic energy when the system rotates about the y-axis at angular speed ω.ISince the rotation is about y axis, the moment of inertia about y axis, Iy, isRKThus, the rotational kinetic energy is Find the moment of inertia and rotational kinetic energy when the system rotates on the x-y plane about the z-axis that goes through the origin O.xyThis is because the rotation is done about y axis, and the radii of the spheres are negligible.Why are some 0s?M MllmmbbOIRK2iiirm∑=2Ml=22Ml=221ωI=()22221ωMl=22ωMl=2iiirm∑=2Ml=()222 mbMl +=221ωI=()2222221ωmbMl +=()222ωmbMl +=2Ml+20m+⋅20m+⋅2Ml+2mb+2mb+Wednesday, Nov. 10, 2004 PHYS 1443-003, Fall 2004Dr. 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 dm in 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: 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 passing through 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=Wednesday, Nov. 10, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu5Example for Rigid Body Moment of InertiaCalculate the moment of inertia of a uniform rigid rod of length L and mass M about an axis perpendicular to the rod and passing through its center of mass.The line density of the rod is What is the moment of inertia when the rotational axis is at one end of the rod.xyLxdxLM=λso the masslet is dmThe moment of inertia is I∫= dmrI2Will this be the same as the above. Why or why not?Since the moment of inertia is resistance to motion, it makes perfect sense for it to be harder to move when it is rotating about the axis at one end.dxλ=dxLM=∫= dmr2dxLMxLL∫−=2/2/22/2/331LLxLM−⎥⎦⎤⎢⎣⎡=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛−−⎟⎠⎞⎜⎝⎛=33223LLLM⎟⎟⎠⎞⎜⎜⎝⎛=433LLM122ML=dxLMxL∫=02LxLM0331⎥⎦⎤⎢⎣⎡=()[]033−= LLM()33LLM=32ML=Wednesday, Nov. 10, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu6xy(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 if the 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 the same as the sum of moment of inertia for a rotation about the CMand that of the CM about the rotation axis.()22 (1)xydm=+∫'yyyCM+=()()dmyxdmyydmxxdmyxCMCMCMCM∫∫∫∫+++++=2222'''2'2∫= 0'dmy()()dmyxdmyxICMCM∫∫+++=2222''Wednesday, Nov. 10, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu7Example for Parallel Axis TheoremCalculate the moment of inertia of a uniform rigid rod of length L 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=+=Wednesday, Nov. 10, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu8Torque & Angular AccelerationLet’s consider a point object with mass m rotating 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 dFtis=tdF=∑τThe torque due to tangential force FtisThe total torque is=τdWhat 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=αI==tdmaαdmr=rdFt()αdmr2=∫dmr2ααIWednesday, Nov. 10, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu9Example for Torque and Angular AccelerationA uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to
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