44-50 Sprng 00 ecture # Dr. Jaehoon Yu. Rotatonal Energy. Computaton of oments of nerta. Parallel-as Theorem 4. Torque & Angular Acceleraton 5. Work, Power, & Energy of Rotatonal otons Remember the md-term eam on Wednesday, ar.. Wll cover Chapters -0. Today s Homework Assgnment s the Homework #5!!!
y r v θ m O Snce a rgd body s a collecton of masslets, the total knetc energy of the rgd object s By defnng a new quantty called, oment of nerta,, as Rotatonal Energy What do you thnk the knetc energy of a rgd object that s undergong a crcular moton s? Knetc energy of a masslet, m, movng at a tangental speed, v, s What are the dmenson and unt of oment of nerta? What do you thnk the moment of nerta s? What smlarty do you see between rotatonal and lnear knetc energes? mr K K m v mr ω ω R K mr The above epresson s smplfed as kg m [ ] easure of resstance of an object to changes n ts rotatonal moton. mr ω K R ω ass and speed n lnear knetc energy are replaced by moment of nerta and angular speed. ar. 6, 00 44-50 Sprng 00 Dr. J. Yu, ecture #
Eample 0.4 n a system conssts of four small spheres as shown n the fgure, assumng the rad are neglgble and the rods connectng the partcles are massless, compute the moment of nerta and the rotatonal knetc energy when the system rotates about the y-as at ω. l y O m m b b Thus, the rotatonal knetc energy s l mr ar. 6, 00 44-50 Sprng 00 Dr. J. Yu, ecture # Why are some 0s? Snce the rotaton s about y as, the moment of nerta about y as, y, s l + l + m 0 + m 0 l Ths s because the rotaton s done about y as, and the rad of the spheres are neglgble. ( l ) ω l ω K R ω Fnd the moment of nerta and rotatonal knetc energy when the system rotates on the -y plane about the z-as that goes through the orgn O. m r l + l + mb + mb ( l mb ) ( ) ( ) + K R ω l + mb ω l + mb ω
Calculaton of oments of nerta oments of nerta for large objects can be computed, f we assume the object conssts of small volume elements wth mass, m. The moment of nerta for the large rgd object s lm r m t s sometmes easer to compute moments of nerta n terms How can we do ths? of volume of the elements rather than ther mass Usng the volume densty, ρ, replace dm The moments of ρ ; dm ρdv ρ dm n the above equaton wth dv. dv nerta becomes Eample 0.5: Fnd the moment of nerta of a unform hoop of mass and radus R about an as perpendcular to the plane of the hoop and passng through ts center. y dm The moment of nerta s m r dm R dm 0 R r dm r dv O R What do you notce from ths result? The moment of nerta for ths object s the same as that of a pont of mass at the dstance R. ar. 6, 00 44-50 Sprng 00 Dr. J. Yu, ecture # 4
Eample 0.6 Calculate the moment of nerta of a unform rgd rod of length and mass about an as perpendcular to the rod and passng through ts center of mass. y d What s the moment of nerta when the rotatonal as s at one end of the rod. Wll ths be the same as the above. Why or why not? The lne densty of the rod s so the masslet s The moment of nerta s ar. 6, 00 44-50 Sprng 00 Dr. J. Yu, ecture # r dm / / r dm 0 λ dm λd d d [( ) 0] ( ) 4 0 d / / Snce the moment of nerta s resstance to moton, t makes perfect sense for t to be harder to move when t s rotatng about the as at one end. 5
y Parallel As Theorem oments of nerta for hghly symmetrc object s relatvely easy f the rotatonal as s the same as the as of symmetry. However f the as of rotaton does not concde wth as of symmetry, the calculaton can stll be done n smple manner usng parallel-as theorem. + C D y y y C r C D What does ths theorem tell you? (, y) C ( C, y C ) ar. 6, 00 44-50 Sprng 00 Dr. J. Yu, ecture # oment of nerta s defned r dm ( + y ) dm () Snce and y are C + ' ; y yc + One can substtute and y n Eq. to obtan [( + ) + ( y + y ) ] ' ' dm ( + y ) dm+ ' dm+ y y' dm+ ( ' + y' )dm C C C C C Snce the and y are the dstance from C, by defnton Therefore, the parallel-as theorem C y' ' dm y ' dm + C 0 0 D oment of nerta of any object about any arbtrary as are the same as the sum of moment of nerta for a rotaton about the C and that of 6 the C about the rotaton as.
Eample 0.8 Calculate the moment of nerta of a unform rgd rod of length and mass about an as that goes through one end of the rod, usng parallel-as theorem. y C d The moment of nerta about the C The lne densty of the rod s so the masslet s C r dm / / λ dm λd d 4 d / / Usng the parallel as theorem C + D + + 4 The result s the same as usng the defnton of moment of nerta. Parallel-as theorem s useful to compute moment of nerta of a rotaton of a rgd object wth complcated shape about an arbtrary as ar. 6, 00 44-50 Sprng 00 Dr. J. Yu, ecture # 7
P Torque Torque s the tendency of a force to rotate an object about some as. Torque, t, s a vector quantty. d r d oment arm F F φ ne of Acton agntude of torque s defned as the product of the force eerted on the object to rotate t and the moment arm. ar. 6, 00 44-50 Sprng 00 Dr. J. Yu, ecture # Consder an object pvotng about the pont P by the force F beng eerted at a dstance r. The lne that etends out of the tal of the force vector s called the lne of acton. The perpendcular dstance from the pvotng pont P to the lne of acton s called oment arm. When there are more than one force beng eerted on certan ponts of the object, one can sum up the torque generated by each force vectorally. The conventon for sgn of the torque s postve f rotaton s n counter-clockwse and negatve f clockwse. τ rf snφ Fd τ τ + τ Fd F d 8
Eample 0.9 A one pece cylnder s shaped as n the fgure wth core secton protrudng from the larger drum. The cylnder s free to rotate around the central as shown n the pcture. A rope wrapped around the drum whose radus s R eerts force F to the rght on the cylnder, and another force eerts F on the core whose radus s R downward on the cylnder. A) What s the net torque actng on the cylnder about the rotaton as? R F The torque due to F τ R F and due to F τ R F R So the total torque actng on the system by the forces s τ τ + τ R F + RF F Suppose F 5.0 N, R.0 m, F 5.0 N, and R 0.50 m. What s the net torque about the rotaton as and whch way does the cylnder rotate from the rest? Usng the above result τ R F + R F 5.0.0 + 5.0 0.50.5N m The cylnder rotates n counter-clockwse. ar. 6, 00 44-50 Sprng 00 Dr. J. Yu, ecture # 9
What does ths mean? r Torque & Angular Acceleraton F t F r m et s consder a pont object wth mass m rotatng n a crcle. What forces do you see n ths moton? The tangental force F t and radal force F r The tangental force F t s F ma The torque due to tangental force F t s What do you see from the above relatonshp? What law do you see from ths relatonshp? ar. 6, 00 44-50 Sprng 00 Dr. J. Yu, ecture # τ t t mrα Ft r mar t τ α mr α Torque actng on a partcle s proportonal to the angular acceleraton. How about a rgd object? The eternal tangental force df t s df t The torque due to tangental force F dm t s τ α dm α r r The total torque s O What s the contrbuton due to radal force and why? Analogs to Newton s nd law of moton n rotaton. dft dmat dmrα dτ df r ( r dm)α Contrbuton from radal force s 0, because ts lne of acton passes through the pvotng pont, makng the moment arm 0. 0 t
Eample 0.0 A unform rod of length and mass s attached at one end to a frctonless pvot and s free to rotate about the pvot n the vertcal plane. The rod s released from rest n the horzontal poston what s the ntal angular acceleraton of the rod and the ntal lnear acceleraton of ts rght end? / g Snce the moment of nerta of the rod when t rotates about one end We obtan α g g The only force generatng torque s the gravtatonal force g g τ Fd F g α ar. 6, 00 44-50 Sprng 00 Dr. J. Yu, ecture # r dm d 0 λ 0 Usng the relatonshp between tangental and angular acceleraton a t α g What does ths mean? The tp of the rod falls faster than an object undergong a free fall. 0
Work, Power, and Energy n Rotaton O dθr ds F φ et s consder a moton of a rgd body wth a sngle eternal force F eerted on the pont P, movng the object by ds. The work done by the force F as the object rotates through nfntesmal dstance dsrdθ n a tme s dw F d s ( F snφ ) rdθ What s Fsnφ? The tangental component of force F. What s the work done by radal component Fcosφ? Snce the magntude of torque s rfsnφ, The rate of work, or power becomes The rotatonal work done by an eternal force equals the change n rotatonal energy. The work put n by the eternal force then W ar. 6, 00 44-50 Sprng 00 Dr. J. Yu, ecture # Zero, because t s perpendcular to the dsplacement. dw τdθ P dw dw τ τdθ α τdθ θι θ ι τω dω ωdω τdθ ω ω ι f How was the power defned n lnear moton? ωdω dω dθ ω dθ ω f
Smlarty Between near and Rotatonal otons All physcal quanttes n lnear and rotatonal motons show strkng smlarty. Smlar Quantty ass ength of moton Speed Acceleraton Force Work Power omentum Knetc Energy ass Dstance Force Work Knetc near v a ar. 6, 00 44-50 Sprng 00 Dr. J. Yu, ecture # W P dr dv F f Fd ma oment of nerta Angle Torque Work Rotatonal Rotatonal r dm ω α d θ d ω τ F v P τω p m v mv (Radan) α K K R ω θ W ω θ θ f τdθ