Chapter5
             Kinetics of Rigid Bodies
             Question5–1
             Ahomogeneouscircularcylinderofmassmandradiusr isatrestatopathinsheet
             ofpaperasshowninFig.P5-1. Thepaperliesflatonahorizontalsurface. Suddenly,
             the paper is pulled with a very large velocity to the right and is removed from
             under the cylinder. Assuming that the removal of the paper takes place in a time
             t, that the coefficient of dynamic friction between all surfaces is µ, and that gravity
             acts downward,determine(a)thevelocityofthecenterofmassofthecylinderand
             (b) the angular velocity of the cylinder the instant that the paper is removed.
                                  r            g
                            ω     O
                                     x
                        Friction (µ)      Paper
                                Figure P5-1
             Solution to Question 5–1
             First, let F be the ground. Then, it is convenient to choose the following coordinate
             systemfixedinreferenceframeF:
                              Origin at O
                           Ex    ƒ    ToTheRight
                           Ez    ƒ      Into Page
                           Ey    ƒ       Ez ×Ex
                      182                                         Chapter5. KineticsofRigidBodies
                      Next, let D be the cylinder. Then, choose the following coordinate system fixed in
                      reference frame D:            Origin at O
                                                e        ƒ       Fixed in D
                                                 r
                                                e        ƒ                E
                                                 z                         z
                                                e        ƒ           ez ×er
                                                 θ
                      Now,inordertosolvethisproblem,weneedtoapplylinearimpulseandmomen-
                      tum to the center of mass of the cylinder and angular impulse and momentum
                      aboutthecenterofmassofthecylinder. Inordertoapplythesetwoprinciples,we
                      usethefreebodydiagramshowninFig.5-1where
                                     N     ƒ ReactionForceofPaper(Surface)onDisk
                                     mg ƒ ForceofGravity
                                     F     ƒ ForceofFriction
                                      f
                                                           O
                                                         mg
                                                              P
                                                                     F
                                                                      f
                                                           N
                                          Figure 5-1   Free Body DiagramofDisk
                         Nowfromthegeometrywehavethat
                                                   N    ƒ NEy
                                                   mg ƒ mgEy                                      (5.1)
                                                   F    ƒ −µkNk vrel
                                                    f              kv k
                                                                      rel
                      Nowweneedtodetermine vrel. Denoting the point of contact between the disk
                      andthepaperbyP (seeFig.5-1),wenotethat
                                                    vrel ƒ FvP −Fvpaper                           (5.2)
                      where Fvpaper is the velocity of the paper in reference frame F. Since the paper is
                      pulled in the positive E -direction, we have that
                                             x
                                                     Fv      ƒv     E                             (5.3)
                                                        paper   paper x
                                                                                                 183
                      Next, we need to determine FvP. From kinematics of rigid bodies we have that
                                              F     F      F R
                                               v − v ƒ ω ׄr −r …                               (5.4)
                                                 P     O             P    O
                                                                              F R
                      where R denotes the reference frame of the cylinder and  ω istheangular ve-
                      locity of reference frame R in reference frame F. From the geometry we have that
                                                       F R
                                                        ω ƒωEz                                  (5.5)
                      and
                                                      r −r ƒrE                                  (5.6)
                                                       P    O      y
                      Consequently,
                                            Fv −Fv ƒωE ×rE ƒ−rωE                                (5.7)
                                              P      O      z      y         x
                      Furthermore,
                                                        r  ƒxE                                  (5.8)
                                                         O      x
                      whichimpliesthat
                                                  F      Fd
                                                                    ˙
                                                   v ƒ        r   ƒxE                           (5.9)
                                                    O     dt „ O…      x
                      Wethenhavethat
                                            F      F               ˙
                                             v ƒ v −rωE ƒ„x−rω…E                               (5.10)
                                               P     O        x              x
                      Now,sincethepaperis“suddenly”pulledtotheright, it implies that the paper is
                      pulled such that its speed is extremely large. Therefore,
                                                              ˙
                                                     v     ≫x−rω                               (5.11)
                                                      paper
                      whichimpliesthat
                                                   ˙
                                                   x−rω−vpaper≪0                               (5.12)
                      Nowwehavethat
                                              ˙                         ˙
                          v ƒv −v          ƒ„x−rω…E −v           E ƒ„x−rω−v            …E      (5.13)
                           rel   P    paper             x    paper x               paper  x
                      ButfromEq.(5.12)weseethat
                                                        ˙
                                               v ƒ−|x−rω−v             |E                      (5.14)
                                                 rel               paper x
                      whichimpliesthat
                                                        vrel ƒ−E                               (5.15)
                                                       kv k       x
                                                         rel
                      Theforceoffriction is then given as
                                                      F ĵkNkE                                 (5.16)
                                                        f         x
                      Nowthatwehaveexpressionsforall of the forces, we move on to the application
                      of linear impulse and momentum and angular impulse and momentum
                   184                                  Chapter5. KineticsofRigidBodies
                   (a) Velocity of Center of Mass of Cylinder at Instant When Paper is Removed
                   Wehavethat
                                                  F¯     F
                                             Fƒm aƒm aO                           (5.17)
                   Differentiating Eq. (5.9) in reference frame F, we have
                                                F     ¨
                                                 a ƒxE                            (5.18)
                                                  O     x
                   Furthermore, using the result of the force resolution from above, we have
                    FƒN‚mg‚F ƒNE ‚mgE ‚µkNkE ƒ„N‚mg…E ‚µkNkE                      (5.19)
                                  f     y       y        x            y        x
                   EquatingFandmFaO,weobtain
                                                               ¨
                                       „N‚mg…E ‚µkNkE ƒmxE                        (5.20)
                                                y         x      x
                   whichyieldsthefollowingtwoscalarequations:
                                             N‚mg ƒ        0                      (5.21)
                                                           ¨
                                             µkNk     ƒ mx
                   Therefore,
                                                Nƒ−mg                             (5.22)
                   whichimpliesthat
                                               Nƒ−mgE                             (5.23)
                                                         y
                   whichfurtherimpliesthat
                                                kNkƒmg                            (5.24)
                   Wethenhavethat
                                                        ¨
                                                µmgƒmx                            (5.25)
                   This last equation can be integrated from 0 to ∆t to give
                                          Z∆t         Z∆t   ¨
                                           0 µmgdtƒ 0 mxdt                        (5.26)
                   Wethenhavethat
                                    ˙        ˙
                         µmg∆tƒmx„∆t…−mx„tƒ0…ƒmv „∆t…−mv „tƒ0…                    (5.27)
                                                          O         O
                   Notingthatthediskisinitially stationary, we have that
                                               v „t ƒ0…ƒ0                         (5.28)
                                                O
                   whichimpliesthat
                                             µmg∆tmv „∆t…                         (5.29)
                                                       O
                   Solving this last equation for v „∆t…, we obtain
                                            O
                                              v „∆t…ƒµg∆t                         (5.30)
                                               O
                   Therefore, the velocity of the center of mass at the instant that the paper is com-
                   pletely pulled out is
                                            Fv „∆t…ƒµg∆tE                         (5.31)
                                               O            x