See example i for objects that are both rotating and

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(See Example I) For objects that are both rotating and moving linearly, you must include them twice; once as a linearly moving object (using m) and once more as a rotating object (using LEM). (See Example II) The LEM of a rotating mass is easily defined in terms of its moment of inertia, I. LEM = I/r 2 For example, using a standard table of Moments of Inertia, we can calculate the LEM of simple objects rotating on axes through their centers of mass: I LEM Cylindrical hoop mr 2 m Solid disk ½mr 2 ½m Hollow sphere 2 / 5 mr 2 2 / 5 m Solid sphere mr 2 m Example I A flywheel, M = 4.80 kg and r = 0.44 m, is wrapped with a string. A hanging mass, m, is attached to the end of the string. When the hanging mass is released, it accelerates downward at 1.00 m/s 2 . Find the hanging mass. To handle this problem using the linear form of Newton’s Second Law of Motion, all we have to do is use the LEM of the flywheel. We will assume, here, that it can be treated as a uniform solid disk. The only external force on this system is the weight of the hanging mass. The mass of the system consists of the hanging mass plus the linear equivalent mass of the fly-wheel. From Newton’s 2 nd Law we have F = ma, therefore, mg = [m + (LEM=½M)]a mg = [m + ½M] a (mg – ma) = ½M a m(g a) = ½Ma m = ½•M•a / (g a) m = ½• 4.8 • 1.00 / (9.81 1) m = 0.27 kg If a = g/2 = 4.905 m/s 2 , m = 2.4 kg If a = ¾g = 7.3575 m/s 2 , m = 7.2 kg Note, too, that we do not need to know the radius unless the angular acceleration of the fly-wheel is requested. If you need α , and you have r, then α = a/r. Example II Find the kinetic energy of a disk, m = 6.7 kg, that is moving at 3.2 m/s while rolling without slipping along a flat, horizontal surface. (I DISK = ½mr 2 ; LEM = ½m) The total kinetic energy consists of the linear kinetic energy, K L = ½mv 2 , plus the rotational kinetic energy, K R = ½(I)( ω) 2 = ½(I)(v/r ) 2 = ½(I/r 2 )v 2 = ½(LEM)v 2 . KE = ½mv 2 + ½•(LEM=½m)•v 2 KE = ½•6.7•3.2 2 + ½•(½•6.7)•3.2 2 KE = 34.304 + 17.152 = 51 J Final Note: This method of incorporating rotating objects into the linear equations of motion works in every situation I’ve tried; even very complex problems. Work your problem the classic way and this way to compare the two. Once you’ve verified that the LEM method works for a particular type of problem, you can confidently use it for solving any other problem of the same type.
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