Mechanics Cycle 3 Chapter 9++ Chapter 9++

Transcription

1 Chapter 9++ More on Knetc Energy and Potental Energy BACK TO THE FUTURE I++ More Predctons wth Energy Conservaton Revst: Knetc energy for rotaton Potental energy M total g y CM for a body n constant gravty To-Do: Knetc energy for rollng moton How to calculate power a qucke formula Appendces for Showng knetc energy has two parts More moments of nerta - spheres Rotaton About a Fxed Axs: Example of a Rod A unform rod or stck of mass m and length L has one end pvoted at a frctonless hnge. It can only rotate and ts rotaton axs s through P and perpendcular to the page. The rod has been ω gven an ntal angular velocty ω at the ntal angle θ wth the horzontal as shown. Then t s allowed to swng downward under the nfluence of gravty and, as a result, swngs down wth ncreasng angular velocty. How can we predct P g θ m L the new ω at the later angle θ, gven ω at θ? Answer: To have a shortcut to ths predcton we ask what s or s not conserved here? Lnear g ω momentum of the CM (or of the mass m) s OT m conserved because the pvot and gravty represent L a nonzero net external force durng the rotaton θ P (and they don t cancel each other n general); lnear momentum s not useful for pure rotatons anyway. Angular momentum of the CM (or of the ndvdual masses) around the pvot s OT conserved because gravty provdes a nonzero torque around that pvot. But the phrase wthout frcton s a bg hnt: knetc plus potental (total mechancal) energy IS conserved snce gravty s a conservatve force and the pvot s frctonless. So ths gets us a quck path to a predcton: Equate the total energy before and after. Recall that the entre knetc energy for fxed-axs rotaton s gven by K P = I P ω where I P s defned wth respect to the rotaton axs P. Thus I P here s the moment of 9-8

2 nerta for a rod around ts end: I P = I P (rod) = /3 ml. Recall also that the potental energy n unform gravty s as f all the mass m were concentrated at the CM: U=mgy CM, and, whle t s not new to us, we see the change n gravtatonal potental energy s found by followng the change n the rod s CM heght. The rod CM s orgnally at the heght L snθ above the horzontal poston. Its heght s L snθ at any other angle θ. Therefore potental energy change s mg tmes the change n heght, or ΔU = m g Δy = mg L (sn θ sn θ ) Snce the change n the rotatonal knetc energy s ΔK P = I P ω I ω = P I P (ω ω ) = 6 ml (ω ω ), conservaton of energy demands ΔK P = ΔU n gong from angular velocty ω to ω. Thus ΔK P = 6 ml (ω ω ) = ΔU = mg L (snθ snθ ) or ω ω = g 3 L (snθ snθ ) or, rearrangng the mnus sgns, ω = ω + 3 g L (sn θ sn θ) Snce θ decreases as the rod falls (and note that θ becomes negatve as t goes below the horzontal axs), we can verfy that ω > ω when we go from a hgher elevaton to a lower one. **************************************************** Problem 9- A unform thn beam of length L and mass M s n a vertcal poston wth ts lower end on a rough surface that prevents ths end from slppng. Suppose the beam s nudged so as to topple n the drecton shown. Fnd the angular velocty (as a vector : magntude and drecton) of the beam, about ts fxed end, just before mpact n terms of g and L. You mght lke to derve the answer by conservaton of energy, or you can use the result n the text for a quck escape and conservng your own energy! L **************************************************** 9-9

3 Rotatons About a Movng Axs: Rollng Knetc Energy (See the proof n the appendx at the end of ths chapter.) In partcular, the total knetc energy s where the energy assocated wth CM moton (.e., the translatonal moton) s and, for a rgd body, the nternal moton can always be wrtten n the rotatonal form where the rotaton s about the CM wth angular velocty Famous nclned plane example: usng energy conservaton to predct the speed of a ball or wheel rollng down an nclned plane. m, R Consder a cylnder startng from rest and rollng down a straght slope wthout slppng. otce the rotatonal axs s now movng (but stays perpendcular to the page). We use h conservaton of energy to predct how fast the cylnder s gong at the bottom. Includng the rotatonal KE, the decrease n PE s taken up by the ncrease n total KE: ΔK = - ΔU (remember ths s just K f - K = - (U f - U ), whch s equvalent to K + U = K f + U f, f you prefer!). Wth U = mgy, y f y = - h *, and K =, we fnd from the above two KE contrbutons: * Recall that the constant-gravty potental energy s easy to calculate! It s as f the total mass were concentrated at the CM pont (also, note n the fgure to the rght that the CM pont drops vertcally the same dstance h as the hll s hgh!). BUT how are and v CM related? Recall from Ch. : R h R Gettng back to the nclned plane, consder any knd of ball or cylnder, havng radus R, mass m, and moment of nerta about ts rollng axs (through ts CM). (As we dscuss n the appendx, the dmensonless constant would be equal to /5 for a sold 9-

4 unform sphere, and /3 for a hollow sphere, for examples.) Startng from rest and rollng all the way down the nclne, for a gven, the ball has a fnal speed v that can be predcted from the energy conservaton equaton: or Comments on ths result? Well, do ths problem! **************************************************************************************************** Problem 9- We should not be the only ones sufferng, er, havng fun here: (a) What happens when we change m for the above rollng object? Explan! (b) What happens when we change R for the above rollng object? Explan! (c) What happens when we change for the above rollng object? Explan! (c ) What happens when we keep on askng you more and more questons? Just gnore ths! (d) How does the fnal speed for ths rollng ball compare wth the fnal speed for any object that sldes wthout rotatng (and wthout frcton)? **************************************************************************************************** **************************************************************************************************** Problem 9- A homogeneous sphere of mass m and radus r has the CM moment of nerta gven by /5 mr, whch has been derved n the appendx. It starts from rest at the upper end of the track shown, and rolls wthout slppng untl t fles off the rght-hand end. (Ignore the possblty that the ball mght fly nto the ar before the end and not stay on the curvy track!) (a) For a rollng sphere wth no slppng, fnd the total knetc energy n terms of m and the speed v of ts center. HIT: (b) If H=6. m and h=. m and the track s horzontal at the rght-hand end, determne the dstance, to the rght of pont A, where the ball strkes the horzontal base lne. (otce how the mass cancels out agan n all of ths, as n the prevous examples.) *************************************************************************************************** 9-

5 Power and ts Calculaton What about the fact that some sources of work can do the job much faster than others? As usual, we want to talk about the (tme) rate of dong work and compare rates. We call the work rate or the rate of energy output or nput the power or power output or power nput. Average Power and Unts: P AV = work done n a tme nterval/tme nterval. The SI unt s the Watt (W) = Joule/second. As for the exctng Amercan unts we know and love, US HP = 55 ft lb/s = 746 W ( W =.738 ft lb/s). Instantaneous Power: dmenson: Recall that P = dw = F dx P = Fv D 3 dmensons: Recall dw = F dr cosθ P = dw = F dr P = Fvcosθ = F v 3D Example: A horse walkng along the shore pullng a barge through a rope wth tenson T n the rope θ v T barge P = Tv cosθ A lttle more nterestng example resdes n the followng problem: **************************************************************************************************** Problem 9-3 A bcyclst s coastng down a hll at constant speed v. The bcycle and woman have a total mass m. The hll s nclned at angle θ. a) What s the power generated by the force of gravty on the bcyclst? v b) What s the power generated by all the frcton forces operatng on the bcyclst? θ c) What s the power generated by the normal force on the bcycle due to the hll? d) Are your answers nstantaneous power or average power? *************************************************************************************************** 9-

6 Appendx A Knetc Energy when the CM s Movng We add translatonal moton to any rotatonal moton - the general theorem to be dscussed below s that any moton of a rgd body can be consdered as a combnaton of translatonal plus rotatonal moton. (Recall that earler we separated out the overall CM moton and we saw how t s related to the net external force.) The total moton can be decomposed nto the moton of the CM plus moton about the CM usng the postons and veloctes relatve to the CM, as follows. Recall we have used r as the poston of partcle n these knds of general dscussons. ow defne q as the poston of that partcle relatve to the CM poston: r = r CM + q or q = r - r CM whch corresponds to the vector addton trangle shown: Then we can talk about the velocty of the partcle n terms of ts velocty relatve to the CM moton, just by takng the dervatve of the above, v = d r = d r CM + d q but d r CM = v CM and d q u v = v CM + u Ths lets us derve the relaton between the total momentum and the CM velocty very easly: but p = m v = m v CM + m u = m v dq CM + m m dq = (changes n the ndvdual motons cannot change the CM remember the canoe problem as an example!) and concerned, there s no net extra nternal momentum: p = M v CM = p CM m = M. Thus, as far as momentum s Ths s somethng we already knew. (Remember that we defned p p total.) But there s an nternal contrbuton to knetc energy. See the next page. 9-3

7 We now show that the total knetc energy of a bunch of partcles reduces to a sum of "CM knetc energy" plus "nternal knetc energy. Rememberng our vector dot products stuff, we get K total = m v = m v v = m v ( CM + u ) v CM + u ( ) = m ( v + u CM v CM +u ) = ( CM m ) v + ( m CM u )v CM + m u As before, u = (the canoe dentty! ) and m advertsed earler, m = M. Therefore, as we K total = K CM + K nternal wth K CM = M v CM and K nternal = m u If the system s a rgd body, then K nt s due entrely to rotaton about the CM Therefore K nternal (rgd body) = m u v = ωr = ( m r ) ω K nternal (rgd body) = I CM ω where the axs goes through the CM and the drecton of the axs s determned by the stuaton (e.g., t s parallel to the surface for rollng). 9-4

8 Appendx B CM Moment of Inerta of a Unform Sold Sphere Answer: I CM = 5 M for a unform SOLID sphere, radus R, mass M, and R the axs through the center (.e., the CM moment of nerta) Proof (agan just for your enjoyment that s, enjoyng the fact that you don t have to master t BUT you mght be surprsed at your ablty to follow t): Thnk of as a stack of concentrc sold dscs, each of mass dm, as shown. Summng over the whole stack of lttle moments of nerta gves the ntegral as the lmt of dfferental dm (each wth dfferental di = ½ dm r ): I = Δm r dm r But dm = ρ dv for mass volume densty ρ and volume of the dsc dv = π r dh where the radus of the dsc s r and ts small heght s dh. Also, r = Rsnθ from the lttle golden trangle on the rght wth the sphere radus R as the hypotenuse. Contnung to relate everythng to θ, we have dh = Rdθsnθ from another golden trangle notng the hypotenus Rdθ s the lttle arc length subtended by dθ. So far we have gotten to I = dm r = ρπ (R sn θ) R dθ sn θ (R sn θ) = π ρπr5 sn 5 θ dθ Rdθ The mass densty for the unform sold sphere s ρ = M / 4 3 πr3 and now to do the ntegraton! Change varables from θ to x = cosθ (so dx = - snθ dθ) to obtan I = M 4 3 πr3 π πr 5 sn 5 θ dθ = 3 8 MR ( x ) ( dx) = MR ( x + x 4 )dx + = 3 8 MR ( ( / 3) + / 5) = 3 8 MR (6 / 5) = 5 MR (as promsed) CM Moment of Inerta of a Unform Hollow Sphere Answer: I CM = 3 M for a unform HOLLOW sphere, radus R, mass M, and R the axs through the center (.e., the CM moment of nerta) o proof shown someday when you ve got nothng to do and nowhere to go... try t! You can adapt the above knd of ntegraton to a hollow sphere (thnk of a stack of concentrc hoops). As we expect, t s bgger than for a sold sphere, snce more mass s farther out for a gven R. 9-5