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by: Mrs. Linda Wiegand


Mrs. Linda Wiegand
GPA 3.69


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Class Notes
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This 7 page Class Notes was uploaded by Mrs. Linda Wiegand on Friday September 18, 2015. The Class Notes belongs to PHY 4222 at University of Florida taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/206776/phy-4222-university-of-florida in Physics 2 at University of Florida.


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Date Created: 09/18/15
Speed of Lgm m veeuum of veeuum Linea dimensx em 5 Typical chemical enexgy ms 3 m sol 2 Linea dimensxon of nuclei Nucleax bmdmg enexgy 3 ms x 1012 c2 0 71quot m A 1 10 34 J s e 58 x 10 16 eV 5 9 c N71 72 o 510 keVc2 mp 65 x 1 2quot kg 928 MeVc2 e x 10quot m appxox sevexelxw lo m ep 1 p ox seveer ev pex ecom appxox 023 appxox w x 1019 at STP sevexalxloils m appxox Typically 8 MeV pex nucleon appxox Gxemeuonel eonecenc 667x 1011 N m2 kg 2 2 Ems 5975x 10241 2 85 x 108 225 ms 49 mxhx 150 cm p p 1abm760mmHg1x105N m 2 Lenth Speed Mess Foxce Pxessuxe Enexgy Powex Tmekneee of papa Mess of papexchp Lenth of dey Lenth of Fmgex at axm s lengcn Meunc ne Woxld pepulemn 1 m 7 29 25 m 1 km 7 1o3 m O 6214 mile m c 1 4 x 103 Wm at eaxbh s suxface Me 250 x 106 pxox e x 109 appxox Wobbling Disk Problem A thin7 uniform disk of mass m and radius a is attached to an axle which passes through its center a normal to the disk makes an angle 6 with the axle7 as shown in the gure The axle7 which can be considered to be massless7 is supported a distance d from the center of the disk by bearings which apply a force F to the axle The axle rotates with an angular frequency o What is the force required to maintain this uniform motion Ignore the weight of the disk ie7 dont worry about the gravitational force acting on the disk Side View of disk 2 and y are out of the page Solution Let7s rst de ne an appropriate set of xed and body coordinates7 and nd the relationship between them For the cced coordinates7 chose z7y7z as shown in the gure7 with 2 along the axle and z and y in the plane perpendicular to the axle let the z 7 y plane pass through the center of the disk In terms of these coordinates7 the angular velocity of the disk is w w ez 1 Let the body coordinates attached to the disk be 17 2737 with the origin at the center of the disk Let eg be normal to the disk with el and oz in the plane of the disk e2 lies along a line de ned by the intersection of the disk with z 7 y plane7 and el and eg lie in a plane which is perpendicular to oz and which rotates counterclockwise about the z axis with an angular frequency w With these conventions7 e2 always rotates in the z 7 y plane with no projection along ez The gure below shows the projection of the unit vectors e17e27e3 onto the z 7 y plane after a time t oz and eg make an angle wt with the x axis7 and e2 makes an angle wt with the y axis We are now in a position to write down the relationship X 39lt z aXis is out of the page between the two coordinate systems e1 cos 6cos wt em sin wt ey 7 sin 6 ez7 A D e2 7 sinwt em cos wt ey7 A 0 V e3 sin 6cos wt em sinwt ey cos 0oz A 4 V Its easy to verify that the body coordinates are orthonormal We also have em cos0coswte1 7 sinwteg sint9cos Wt637 A U V ey cos 6sinwte1 cos wteg sinl9sinwte37 A CT V ez 7sin0e1 cost9e3 A I Therefore7 from Eq 17 we see that the angular velocity in the body frame is w w7sin0e1 cos0e3 8 Notice that the angular velocity is also constant in the body frame This is what we should expect7 since the body frame is rotating with an angular velocity w with respect to the xed frame7 so that the time derivative of w is d w 6 wwgtltw dt 7 6t 6w E7 2 where we7ve used the fact that w gtlt w 07 and Sud6t denotes a time derivative in the body rotating frame Therefore7 since w is constant in the xed frame7 it is also constant in the body frame Now the body coordinates are also a set of principal axes for the disk7 so the inertia tensor is diagonal with In 22 maZZl7 and 33 ma22 Therefore7 in the body coordinates7 we have L1 Illwl iimazw sin 67 10 L2 12M 0 11 L3 Iggwg mazw cos 67 12 or in vector notation7 L iimazw sin6e1 mazw cos6e3 13 The angular momentum is time independent in the body frame of reference7 and is in the plane de ned by el and eg However7 L is time dependent in the the xed frame To see this7 rewrite Eq 13 in terms of the unit vectors for the xed frame 1 1 1 L Emazw sin 6cos 6 cos wt em Zmazw sin 6cos 6 sinwt ey Xma2w1 cos2 6 ez 14 In the xed frame7 lLl and L1 are constant7 but the projection of L into the z 7 y plane rotates around the z axis with an angular frequency w To nd the torque N which is needed to maintain this motion7 recall that the torque is the time rate of change of the angular momentum in the cced frame of reference Using Eq 147 we have dL 1 E Zmazwz sin6cos 6 7 sin wt em cos wt ey7 15 so that the torque is the z 7y plane7 and rotates about the z axis with an angular frequency w and a constant magnitude lNl 14ma2w2 sin6cos 6 We could also perform the calculation in the body frame of reference Recall that dL 6L 7E7EwgtltL7 16 where the FL6t denotes a time derivative taken in the body frame of reference Now in the body frame L is constant7 so that FL6t 0 The torque is then NwgtltL 1 1 7w sin 6 el w cos 6 e3 gtlt izmazw sin 6 el Emazw cos 6 e3 1 17711sz sin6cos6e27 17 so the torque has a constant magnitude and direction in the body frame Using Eq 3 to relate oz to the unit vectors in the xed frame7 we see that our two calculations in the xed and body frames agree The torque is supplied by the reaction of the bearings on the ends of the axle The forces are in the z 7y plane7 and are perpendicular to N the magnitude of the torque which they produce is 2Fd7 and this must equal 14ma2w2 sin0cos 07 so we nd mazwz F sin0cos 0 18 The force vanishes when 6 0 spinning the disk about the symmetry axis and t9 7r2 spinning the disk about a diameter In these situations the angular momentum is either parallel or antiparallel to the angular velocity7 and we have a state of dynamic balance Notes on kinematics in different coordinate systems January 15 2002 1 Cartesian rectangular coordinates yz In the Cartesian coordinate system the unit vectors are xed don7t change with time The position7 velocity7 and acceleration of a particle are then rt t em yt eyzt ez7 1 Vt rt x eany egJp ez7 2 at r39t iem1jey239ez7 3 and the kinetic energy is T gnu2 mo cz y 22 4 2 Cylindrical coordinates p gb z The rectangular coordinates in terms of the cylindrical coordinates are zpcosq 7 ypsinz57 22 5 The unit vectors are epcos emsinq ey7 Egg isin emcos ey7 ezez 6 This coordinate system is right handed and orthogonal however7 the direction of the unit vectors changes as the particle moves The time derivatives of the unit vectors are p 13 as ew z 0 7 We then have rpepzez7 8 vp epp 5e 2em 9 a p 7 Meg ltp 2p e 2e 10 T gnu2 mp 2 pzasz 22 11 3 Spherical coordinates T6gb The rectangular coordinates in terms of the spherical coordinates are x rsint9cos 7 y rsint9sinq57 z rcos6 The unit vectors are e sin0cos em sin6sin ey cost9ez7 e9 cos6cos gtem cos6sin ey 7 sint9ez7 e 7 sin em cos beg7 and their time derivatives are or Qee sint9ltg3e 7 e 79m cos 6 e 7 7sin0 3e7 7 cos aiee We then have r re7 Vre719 egrsint9 3e 7 a 7777192 7 rsin26 2eT r 2r9 7rsin6cos0 32e9 rsin6c 2sin6r 3 2rcos69c 6 1 i i T 5771092 r202 r2 sin2 19 12


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