1443-501 Spring 2002Lecture #13Dr. Jaehoon Yu1. Rotational Energy2. Computation of Moments of Inertia3. Parallel-axis Theorem4. Torque & Angular Acceleration5. Work, Power, & Energy of Rotational Motions Remember the mid-term exam on Wednesday, Mar. 13. Will cover Chapters 1-10. Today’s Homework Assignment is the Homework #5!!!Mar. 6, 20021443-501 Spring 2002Dr. J. Yu, Lecture #132Rotational EnergyWhat do you think the kinetic energy of a rigid object that is undergoing a circular motion is? Since a rigid body is a collection of masslets, the total kinetic energy of the rigid object isBy defining a new quantity called, Moment of Inertia, I, asKinetic energy of a masslet, mi, moving at a tangential speed, vi, isWhat are the dimension and unit of Moment of Inertia?rimiθOxyvi2== ω222121iiiiirmvmK22===∑∑∑ωωiiiiiiiiRrmrmKK222121∑=iiirmI22= ωIKR21The above expression is simplified as2mkg⋅[]2MLWhat similarity do you see between rotational and linear kinetic energies?What do you think the moment of inertia is?Measure of resistance of an object to changes in its rotational motion.Mass and speed in linear kinetic energy are replaced by moment of inertia and angular speed.Mar. 6, 20021443-501 Spring 2002Dr. J. Yu, Lecture #133Example 10.4In a system consists 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 ω.222222200 MlmmMlMlrmIiii=⋅+⋅++==∑Since the rotation is about y axis, the moment of inertia about y axis, Iy, is()2222222121ωωω MlMlIKR===Thus, 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 MllmmbbO()22222222 mbMlmbmbMlMlrmIiii+=+++==∑()()2222222222121ωωω mbMlmbMlIKR+=+==Mar. 6, 20021443-501 Spring 2002Dr. J. Yu, Lecture #134Calculation 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∫=dmr2dVmdVdmρρ == d ;The 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 passing through its center.xyROdmThe moment of inertia is222MRdmRdmrI ===∫∫What 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.Mar. 6, 20021443-501 Spring 2002Dr. J. Yu, Lecture #135Example 10.6Calculate 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 dxLMdxdm ==λThe moment of inertia is 12432233123332/2/32/2/22MLLLMLLLMxLMdxLMxdmrILLLL==−−====−−∫∫( )[ ]( )33033123303022MLLLMLLMxLMdxLMxdmrILL==−====∫∫Will 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.Mar. 6, 20021443-501 Spring 2002Dr. J. Yu, Lecture #136xy(x,y)xCM(xCM,yCM)yCMCMParallel Axis TheoremMoments of inertia for highly symmetric object is relatively easy 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()(1) 222∫∫+== dmyxdmrISince x and y arex’y’' ;' yyyxxxCMCM+=+=One can substitute x and y in Eq. 1 to obtain()()[]( ) ( )dmyxdmyydmxxdmyxdmyyxxICMCMCMCMCMCM∫∫∫∫∫+++++=+++=222222'''2'2''Since the x’ and y’ are the distance from CM, by definition∫∫==0'0'dmydmxDTherefore, the parallel-axis theorem2MDIICM+=What 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 CM and that of the CM about the rotation axis.Mar. 6, 20021443-501 Spring 2002Dr. J. Yu, Lecture #137Example 10.8Calculate 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 12432233123332/2/32/2/22MLLLMLLLMxLMdxLMxdmrILLLLCM==−−====−−∫∫MDIICM2+=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 axisxyLxdxCM341221222222MLMLMLMLML=+=+=Mar. 6, 20021443-501 Spring 2002Dr. J. Yu, Lecture #138TorqueTorque is the tendency of a force to rotate an object about some 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. d2F22221dFFd−=+=∑τττMar.
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