Class Notes on Rotational Motion - Galileo and Einstein

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10Then just do an integral to add up the moments of inertia of all the nested hoops from 0 to R to find themoment of inertia of the disk:R3 1 4 1 2disk 22I 2 r dr R MR .0The same expression works for a cylinder, which for these purposes is just a stack of disks.What about a rod about one end? Assume its (constant) linear density is λ kg/m., so a length dx hasmass λdx. Then the moment of inertia isL2 3 2I x dx L / 3 ML / 3.0Suppose such a rod is hinged at one end, horizontal, and falls. What is the initial acceleration of its end?2 MgL / 2 I ( ML / 3) 3 gL / 2.So the end accelerates at Lα = 3g/2.If it is at first sight surprising that the end of the rod accelerates downwards faster then g, consider thefollowing system: a very light rod, hinged at one end, with a large weight attached to its middle.A very light rod hinged to a wall at one end has a mass atits center: how fast will the rod’s free end acceleratedownwards on release from a horizontal position?
11If let go from a horizontal position, the weight will completely dominate the situation, and accelerateddownwards very close to g. This means the far end of the light rod is forced downwards at close to 2g.Our example above is not that extreme, but the same idea.The Parallel Axis TheoremIf we know the moment of inertia of an object about a line through the center of mass, the moment ofinertia about any parallel line is easily found.We’ll prove this for a two-dimensional object (really a thin three-dimensional object):new axisrih m ir icenter of massThe moment of inertia of the two-dimensional object about any axisperpendicular to the object is simply related to that about theparallel axis through the center of mass.With the notation in the diagram,ICM2 miri, and recall mr i i0.iiNow I m r m ( r h)newaxisiCMCM2 2i i i ii I 2h m r m hIMh2.ii iii2This is the Parallel Axis Theorem:I I Mh2newaxisCM .where h is the perpendicular distance between the new axis and the parallel axis through the center ofgravity. It’s easy to extend this proof to a three-dimensional object: taking the parallel axes to be in thez-direction, replace r iwith (x i , y i ).
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Class Notes on Rotational Motion - Galileo and Einstein

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