1 Solutions fo Physics 1301 Couse Review (Poblems 10 though 18) 10) a) When the bicycle wheel comes into contact with the step, thee ae fou foces acting on it at that moment: its own weight, Mg ; the nomal foce upwad fom the gound, N ; the applied foce, F ; and the eaction foce fom the point of contact with the step, R. If a lage enough foce is applied (and in the ight place, as we shall see), the wheel will pivot about the edge of the step and lift up ove it. Fo the minimal amount of foce that would just make this happen, the wheel would be vey nealy in static equilibium. It would just begin to lift off the gound, causing the nomal foce N to dop to zeo. The eaction foce R acts at the contact point whee the wheel is pivoting, so the moment am along which it acts is zeo, hence the toque it poduces on the wheel is also zeo. So only two of the afoementioned foces poduce a toque on the wheel and only one of these is unknown, so the toque equation alone will suffice to detemine the value of F. The magnitude of a toque τ is given by τ = F = F sin θ, which is equivalent to the magnitude of the foce F times the pependicula (o closest) distance, sin θ, fom the pivot point to the line of action of the foce. Thus we find:
2 With the wheel just about to pivot aound the step, the net toque is exactly zeo, so the minimum applied foce needed to lift the wheel ove the step is given by + Mg h h F ( h ) = 0 F = Mg h h h. The fom of this esult appea easonable. When thee is no step ( h = 0 ), the hoizontal foce equied to lift the wheel becomes F = 0. Somewhat less tivially, we also find that it is impossible to lift the wheel, with a hoizontally applied foce, ove a step with a height equal to o geate than the wheel s adius ( F as h ). Notice that the moment of inetia, I, of the wheel does not ente into the calculation fo F. This indicates that the applied foce equied would be the same if the wheel wee a unifom solid disk of the same mass and adius. b) The foce F is applied at a height h above the floo along a vetical midline of one face of the cube (this is done so that the cube pivots up off the floo without tuning to one side o the othe). It is possible to push the cube had enough in this way so that it will pivot fowad on its leading edge, athe than simply emaining at est (unde static fiction) o sliding along the floo (with kinetic fiction). As the cube is about to pivot, the nomal foce fom the floo dops to zeo. Thee will then be two toques acting on the cube, since the diection of the fiction foce is paallel to its moment am, making the toque due to fiction zeo:
3 toque due to weight of cube τ W = + Mg ( 1 L ) sin 90o (the dashed aows toque due to applied foce epesent the lines of actions of each foce) τ F = F h sin 90 o So the cube is on the vege of tipping ove when the net toque fom these foce is just equal to zeo, o 1 Mg MgL Fh = 0 h = L. F If the foce is applied at a point highe above the floo than this, the cube will pitch fowad and tip ove, athe than slide along the floo (o possibly emain at est if the coefficient of static fiction, µ s, is lage enough). Note that as F gows vey lage, h becomes quite small: a vey stong applied foce deliveed vey low on the cube could still topple it. On the othe hand, since the foce cannot be applied any highe above the floo than h = L, we find that h = Mg L L Mg F F 1 F Mg ; this tells us that a foce less than ½Mg applied anywhee cannot cause the cube to tip ove. c) We can examine this situation fom eithe of two efeence fames, one which is stationay, fom which to watch the tuck acceleating, and the othe on the acceleating tuck. Both analyses should give the same esult. Fom the stationay viewpoint, it is the static fiction between the bottom of the stack of cubes and the tuck bed that causes the stack to acceleate along with the tuck. So long as the static fiction can hold, f s = Ma, M being the total mass of the cubes. The stack does not acceleate in the vetical diection, so N Mg = 0 N = Mg. We now look at the net toque about the cente of mass of the stack. The weight foce effectively acts fom this point, so the moment am fo this toque is zeo, making the toque due to the weight equal to zeo.
4 The othe two toques ae toque due to fiction τ f = + 1 (4L ) f s sin 90 o = + f s L = + MaL toque due to nomal foce τ N = x N sin 90 o = N x = Mg x (note that fo an acceleating object, the effective point at which the nomal foce acts is not diectly unde the cente of mass) The stack will not tip ove as long as the net toque is zeo, hence + MaL Mg x = 0 x = a L. g But this will only be possible as long as the effective point fom which the nomal foce acts is within the base of the stack. So the stack will not tip ove povided that x = a L 1 g L a 1 4 g.
5 Fom the point of view of a ide in the tuck bed, the stack of cubes is stationay, but it (as well as the passenge) expeiences a foce diected towad the ea of the tuck, in addition to the foce of gavity and the nomal foce. The ole of static fictional foce hee is to oppose this eawad foce and hold the stack of boxes in place. Since the stack is not obseved to acceleate, the equation fo the hoizontal foces on the stack is f s Ma = 0 and the vetical foce equation is N Mg = 0. We can look at the toques acting about the eawad end of the base of the stack of cubes. The static fictional foce poduces no toque because its diection of action is paallel to its moment am. The othe toques ae:
6 The stack will not topple ove as long as the net toque is zeo, so 1 + MaL + Mg x' 1 MgL = 0 x' = g L a L g = 1 a L g. The point of effective action fo the nomal foce must emain fowad of the tailing edge of the base of the stack; thus, x' = 1 a L 0 1 g a g 0 a 1 4 g, since L > 0 as we found fo the stationay efeence fame. f s max bed if We know that the magnitude of the static fiction foce cannot exceed = µ s N = µ s Mg = 0.3 Mg, so the stack will only stat to slide along the tuck Ma > f s max = 0.3 Mg a > 0.3 g. Hence, the stack of cubes will topple ove befoe it would stat to slide in the tuck bed. 11) Since we will be compaing all of the foces to the weight of the aluminum ball, we will find it useful to find the elative masses of the coppe and aluminum balls. If we call the mass of the coppe ball M and that of the aluminum ball m, we have M m = ρ Cu V Cu = ρ Cu 4π 3 3 Cu ρ Al V Al ρ Al 4π 3 = ρ 3 Cu 9000 kg. Cu = m ρ Al Al Al 700 kg. 3.0 cm..5 cm. = = o m. 3 notice that unit convesions ae unnecessay when woking with compaison atios
7 We can now analyze the foces on each ball with the aid of the two diagams above: the one to the ight povides geometical infomation which will allow us to descibe the angle θ above the hoizontal made by the line connecting the centes of the sphees. The foces on the coppe ball ae hoizontal: N c cos θ N = 0 ; vetical: N c sin θ Mg = 0, while the foces on the aluminum ball ae hoizontal: N l N c cos θ = 0 ; vetical: N v mg N c sin θ = 0, with mg being the weight of the aluminum ball. Fom these equations, we find that N = N c cos θ = N l, N c sin θ = Mg, N v = mg + N c sin θ. We ae now able to calculate the magnitudes of all the foces: N v = mg + N c sin θ = mg + Mg = mg mg = mg = 6.76 mg ; N c = Mg sin θ = mg ( 4 5.5) 6.47 mg ; and N = N l = N c cos θ = Mg cos θ = Mg cot θ = 144 sin θ 5 mg.5.94 mg. 4 1) a) In this situation, due to the absence of fiction, the two masses ae in a maginally stable static equilibium. We will call the downhill motion of m 1 positive, which automatically makes the uphill motion of m positive in tun. The net foce on m 1 acting paallel to the incline it is on is m 1 g sin 40º T = 0, while the net foce on m acting paallel to its incline is T m g sin θ = 0. When we solve each of these equations fo the tension T in the connecting cod and then equate the esults, we find m 1 g sin 40º = T = m g sin θ. Since both of the masses of the blocks ae known, we can solve fo the angle θ that is equied in ode fo this static equilibium to exist:
8 sin θ = m 1 sin 40 o m 1.8 kg θ 3.0 kg..7 o. b) With the inclusion of fiction, the static equilibium of the two masses on thei inclines becomes fa moe stable. In this potion of the Poblem, the angle fo the incline m ests upon is chosen to be θ = 3º. Thee ae now two cases to conside fo the limits of static equilibium: one in which the mass m 1 is just pevented fom sliding downhill, the othe in which it is just pevented fom being dawn uphill. The foces on each mass in the case whee m 1 esists sliding downhill ae foces pependicula to incline foces paallel to incline m 1 m N 1 m 1 g cos 40 o = m 1 a = 0 N m g cos 3 o = m a = 0 m 1 g sin 40 o T f s1max = m 1 a = 0 T m g sin 3 o f s max = m a = 0 Fom the equations fo the pependicula (nomal) foces on the blocks, we obtain N 1 = m 1 g cos 40º and N = m g cos 3º. Since the limits of static fiction on each block ae f s1max = µ s N 1 and f s max = µ s N, the equations fo the foces paallel to each incline can be solved fo the tension T in the cod, yielding m 1 g sin 40º µ s m 1 g cos 40º = T = m g sin 3º + µ s m g cos 3º. If we now solve this last equation fo m, we have m = sin 40o µ s cos 40 o sin 3 o + µ s cos 3 o m kg kg., which is the smallest mass m may have in ode to keep m 1 fom sliding downhill.
9 If the block m is massive enough, it will instead apply enough tension in the cod to pull block m 1 uphill. The limit of static fiction on m 1 in this diection is found by changing the equations fo the foces paallel to each incline to m 1 g sin 40 o T ' + f s1max = m 1 a = 0 and T ' m g sin 3 o + f s max = m a = 0 ; the equations fo the nomal foces ae not affected. Repeating the est of the calculations as we did befoe, we now have m = sin 40o + µ s cos 40 o sin 3 o µ s cos 3 o m kg kg., the geatest mass this block may have without ovecoming static fiction and pulling the othe block along afte it. c) We continue to use the Atwood machine descibed in pat (b), but now with a mass m = 9. kg. We have seen that this will be moe than sufficient mass to ovecome static fiction and set both blocks sliding along thei inclines, with m 1 moving uphill. The nomal foces acting on the blocks will still be as they wee in pat (b). Howeve, with the blocks in motion, thei acceleations will no longe be zeo; with kinetic fiction acting on the blocks, the equations fo the foces paallel to the inclines become m 1 g sin 40 o T '' + f k 1 = m 1 a and T '' m g sin 3 o + f k = m a, with f k = µ k N. We now solve each equation fo the new cod tension T and equate the esults to find m 1 g sin 40 o + µ k m 1 g cos 40 o m 1 a = T '' = m g sin 3 o µ k m g cos 3 o + m a (m 1 + m ) a = m 1 g sin 40 o + µ k m 1 g cos 40 o m g sin 3 o + µ k m g cos 3 o a = m 1 (sin 40 o + µ k cos 40 o ) m (sin 3 o µ k cos 3 o ) g m 1 + m 1.8 kg. ( ) 9. kg. ( ) (9.81 m kg. sec. ) 0.11 g o 1.19 m. sec.. The negative sign fo this acceleation agees with ou expectation that m 1 is being pulled uphill and it is m that is sliding downhill.
10 d) With the pulley now expeiencing fiction with the cod connecting the blocks, the tensions in the cod on eithe side of the pulley ae no longe equal. The equations fo the foces paallel to the inclines that we used in pat (c) must be modified to ead m 1 g sin 40 o T 1 + µ k m 1 g cos 40 o = m 1 a ' and T m g sin 3 o + µ k m g cos 3 o = m a '. Since block m is obseved to be sliding downhill, the acceleation of both blocks is a ' = 0.88 m. sec.. We can solve each of these foce equations fo the tensions, since they ae the only unknowns, to detemine [ ] T 1 = m 1 g(sin 40 o + µ k cos 40 o ) a ' (1.8 kg.) (9.81 m. m. ) ( ) ( 0.88 sec. sec. ) N. and T = m [ g(sin 3 o µ k cos 3 o ) + a '] (9. kg.) (9.81 m. m. ) ( ) + ( 0.88 sec. sec. ) N.
11 These unequal tensions apply a net toque to the pulley, so it will otate. The toques about the cente-of-mass of the pulley ae τ T1 = + T 1 sin 90 o = +T 1 and τ T = T sin 90 o = T, so the net toque on the pulley is ( T 1 T ) ( N ) 0.15 m N m. (the negative sign indicates that the otation is clockwise, as we would expect fom the diagams). Fo a igidly otating object (all points otate about the axis at the same angula speed) like the pulley, the net toque is τ net = I CM α, whee α is the angula acceleation. Since the edge of the pulley acceleates along with the cod, we have the same elationship as we do fo a olling object, α = a'. The moment of inetia of the pulley about its cente of mass is thus given by I CM = τ net α = τ net a' = τ net a' 0.15 m. ( 0.44 N m.) 0.88 m kg. m. sec. 13) a) This situation can be viewed fom eithe of two efeence fames; the Poblem is solved in much the same way in both cases. Fom the point of view of someone standing nea the conveyo belt, the gaphite block falls staight down, lands with zeo hoizontal velocity, and is acceleated up to the speed of the belt. Fon a viewpoint on the belt, the block falls along a paabolic ac and lands with a hoizontal speed of 3 m./sec., afte which the block is bought to est by kinetic fiction. In eithe efeence fame, we may apply the wok-kinetic enegy theoem: the magnitude of the obseved change in the kinetic enegy of the block is equal to the magnitude of the wok done on it by kinetic fiction. Since the conveyo belt is hoizontal, the nomal foce on the block is N = Mg, so the kinetic fictional foce on the block is f k = µ k N = µ k Mg. The fiction acts in the opposite diection to the path along which the block slides, so the fictional wok done on the block is = f k Δ x = µ k Mg L cos 180 o = µ k MgL, whee W f L is the distance the block slides along the belt. By the wok-kinetic enegy theoem then, we find W f = ΔK µ k MgL = 1 Mv L = v µ k g (3.0 m. sec. ) m..9 m. sec. b) Hee, thee is no fiction between any of the sufaces in contact, so we do not need to be concened with the nomal foces on the blocks. Since the only hoizontal foce on m 1 would be due to fiction, in this case the uppe block is not acceleated; thus, a 1 = 0. In the absence of fiction, the only hoizontal foce acting on m is the applied foce F = 6 N. ; hence, the acceleation of the lowe block is
12 a = m F = 6 N. 4.6 m. 1.3 kg. sec.. The lowe block will ultimately be pulled out fom unde the uppe one. c) Since the contact sufaces between the blocks and between the lowe block and the tabletop ae hoizontal, the nomal foces on the blocks ae simply N 1 = m 1 g and N = ( m 1 + m ) g. The static fictional foce acting between the blocks is then f s µ s N 1 = µ s m 1 g. So the two blocks will move togethe as a unit povided that this limit applies. (Since the tabletop is fictionless, N is not impotant to know and contact with the uppe block is the only souce of fiction.) The hoizontal foce equation fo the lowe block is F f s = m a, while that fo the uppe block is f s = m 1 a 1. If the two blocks ae to stay togethe, thei acceleations must match, implying that a 1 = f s m 1 = a = F f s the maximum possible value fo the static fictional foce the limit fo the applied foce is given by µ s m 1 g m 1 = F max µ s m 1 g F max = µ s (m m 1 + m ) g f smax m. When we apply = µ s N 1, we find that 0.45 ( kg.) (9.81 m. sec. ) 8.83 N. Since F max > 6 N., a hoizontal applied foce at this level still pemits the blocks to stay togethe. Because they move as a unit, thei mutual acceleation is F 6 N. a = m 1 + m = 3.00 m. ( kg.) sec.. Note that this means that the static fictional foce hee is f s = m 1 a = ( 0.7 kg. ) ( 3.00 m./sec. ) =.10 N., which is
13 cetainly less than check, we find that f smax = µ s m 1 g = 0.45 (0.7 kg.) (9.81 m. sec. ) 3.09 N. As a a = F f s N m. m 1.3 kg. sec. = a. Fo an applied hoizontal foce F = 10 N. > F max, the blocks will not move togethe and the kinetic fictional foce f k = µ k m 1 g must be used in ou analysis. Upon eplacing f s with f k in the hoizontal foce equations fo each block, we now have a 1 ' = f k = µ k m 1 g = µ k g 0.3 (9.81 m. m. ) 3.14 m 1 sec. sec. and m 1 a ' = F f k m 10.0 N. 0.3 (0.7 kg.) (9.81 m. sec. ) 1.3 kg N. 1.3 kg m. sec.. The lowe block will be pulled out fom unde the uppe one, but hee the uppe block will move fowad a bit fist, which was not the case in the situation we teated in pat (a) above. d) Nothing else about this system is changed if we apply the foce to the lowe block with a sping, instead of a cod. In ode fo the two blocks to continue moving togethe, the foce applied must not exceed F max 8.83 N. Hooke s Law tells us that the estoing foce of a sping is given by F sp = k Δx, whee Δx is the displacement of the sping fom equilibium. A maximum allowed foce of 8.83 N. equies that the maximum pemitted displacement fo this sping be (Δ x ) max = F max 8.83 N m. This k 7 N./m. will set the limit on the amplitude fo the pai of blocks oscillating at the end of this sping. e) Thee ae two phases to the motion of any of the pieces of tablewae: an object is fist caied along on the apidly acceleating tablecloth by kinetic fiction, then, once the fabic has been extacted fom undeneath it, the same object deceleates to est due to the kinetic fiction between the object and the tabletop. We will follow the motion of one such object, which is initially at est on the tablecloth. If that tablecloth is now pulled with an acceleation geate than µ s g, it will ovecome the static fiction between the object and itself, making it possible to pull the tablecloth out fom unde the object. Howeve, while the object emains in contact with the tablecloth, it will be acceleated by kinetic fiction at the ate µ k g. If we call the length of time the object ides the tablecloth t c, then the object attains a peak speed of v max = µ k gt c and is pulled along ove the table fo a distance x 1 = ½ µ k g t c befoe the tablecloth comes out fom unde it. With the object now sliding acoss the bae tabletop, kinetic fiction will act to bing it to a stop. With a coefficient of kinetic fiction µ k between the object and the
14 tabletop, the acceleation of the object is now µ k g. The object will come to est in an inteval Δt given by v f = 0 = v i + aδt = v max + ( µ k ' g ) Δt = µ k gt c µ k ' g Δt µ k gt c = µ k ' g Δt Δt = µ k µ k ' t c. In that inteval, the object will continue to slide, befoe finally stopping, by a distance x = v max Δt + 1 ( µ k' g ) (Δt ) µ = (µ k gt c ) k µ k ' t c 1 µ µ k' g k µ k ' t c = µ k g t c µ k '. We can also find this by applying the velocity-squaed fomula: v f = v i + a (Δx ) 0 = (µ k gt c ) + ( µ k ' g ) x x = µ k g t c µ k '. We want to ensue that no object is caused to move moe than 8 cm., so we equie that x 1 + x 0.08 m. 1 µ k g t c + 1 µ k µ k ' g t c = 1 µ k ' + µ k µ k ' µ k g t c 0.08 m. t c 0.08 m. µ k g µ k ' µ k ' + µ k 0.08 m m. sec sec t c 0.6 sec. So the tablecloth must be pulled out quickly enough that no object ides on it fo longe than 0.6 second; this will plainly equie the tablecloth to be acceleated well above the amount needed to ovecome static fiction with the objects esting on it. 14) We can examine the plumb bob in the fist pat of this Poblem using eithe the stationay efeence fame of, say, a pedestian on the sidewalk watching the ca ound the cone, o the acceleating efeence fame of a passenge inside the ca. This consideation of the two diffeing points of view will be of use in the second pat of the Poblem.
15 In the stationay efeence fame, the tuning vehicle is seen undegoing a centipetal acceleation of mi. ft. a C = v ( R = h. mi. 1 h sec. ) 7 ft. (16.1 ft. sec. ) 7 ft ft. sec. o 9.64 ft. sec. 1 m. 3.8 ft..94 m. sec.. This acceleation is povided by the centipetal foce, which is a esultant foce in this efeence fame. In the vetical diection, the foce equation is T cos θ Mg = Ma V = 0 T cos θ = Mg T = Mg cos θ In the adial diection, we simply have T sin θ = F C, that is, the hoizontal component of the tension in the cod of the plumb line is the centipetal foce that pulls the bob aound the cuve. Fom this, we find that T sin θ = F C Mg sin θ = Ma cos θ C. tan θ = a C g 9.64 ft. sec. 3. ft. sec. o.94 m. sec m. sec θ 16.7 o fom the vetical.
16 Fom the viewpoint of a passenge in the tuning ca, thee is no acceleation of the bob; athe it is in static equilibium unde thee foces: gavity (its own weight), the tension in the cod, and a centifugal foce F C which pulls the bob to the outside of the tun. The adial foce equation in this efeence fame becomes T sin θ F C = Ma = 0, which then leads to the same esult fo θ. The Pinciple of Equivalence, discussed in Poblem 5, can be bought in to descibe this physical situation moe simply: since thee is no acceleation of the bob obseved, it is hanging staight down in a gavitational field which happens to be tilted 16.7º with espect to a vetical diection defined by the body of the ca. Objects will fall towad the outside of the tun simply because this gavitational field pulls them that way. (We would also find that the acceleation due to gavity in this tilted field is g' = m. sec.. ) We can now apply some of this easoning to the second pat of this Poblem. Fom a stationay efeence fame, the Eath is seen to otate with a peiod of one sideeal day, which is 86,164 seconds (using the mean sola day of 86,400 seconds will only change ou esults in the thid decimal place), giving an angula speed fo the suface of the Eath of ω = π π ad. 5 ad T sec. sec.. So the plumb bob (and Minneapolis) ae obseved to tavel on a cicle about the Eath s otation axis of adius = R E cos 45º (whee R E is the Eath s equatoial adius) with a centipetal acceleation of a C = ω 5 ad. = ( sec. ) (6378 km m. km. ) m. sec..
17 Now we conside the view of an obseve in acceleating Minneapolis looking at the plumb bob. The plumb line hangs vetically in a diection which is the vecto sum of Eath s adial gavitational field (magnitude g = 9.81 m./sec. making an angle 45º to the Equato; this takes the Eath to be a pefect sphee) and a centifugal acceleation (magnitude a C in a diection paallel to the Equato). We can apply the Law of Cosines to the indicated vecto tiangle to find g' = g + a C g a C cos 45 o (9.81) (0.040) ( ) m. sec. 4 g' m. sec.. The Law of Sines then gives us sin 45 o g' = sin φ a C sin φ = a C g' sin 45 o m. sec m. sec. ( ) φ o, which is the deviation of the local vetical diection in Minneapolis fom the adial diection (the acceleation due to appaent gavity is also a little bit less than it would be fo a non-otating Eath). To put it anothe way, down does not point exactly towad the Eath s cente, but athe at a place about 10 km. close to the South Pole along the otation axis. It is these slight deviations in the effective gavitational field of the spinning Eath that gives ou planet a slightly flattened shape (called an oblate spheoid), with the pola adius being about 1 km. (0.34%) smalle than the equatoial adius. 15) a) Without the action of ai dag, the fall of a aindop would simply obey consevation of mechanical enegy. If we assume that it leaves the cloud at an altitude of 3000 metes stating nealy at est, then the aindop would each the gound at a speed given by K i + U i = K f + U f ½ m 0 + mgh i = ½ m v f + mg 0 v f = gh i ( 9.81 m./sec. ) ( 3000 m. ) 59,000 m. /sec. g is petty nealy constant ove that height (see Poblem 3) v f 40 m./sec. (o 540 mi./h.). When ai esistance is taken into account, the aindop will cease to acceleate when the upwad dag foce becomes equal to the downwad weight foce acting on the dop. If we assume that the dag is popotional to coss-sectional aea and to the squae of the dop s velocity, and that the doplet etains its spheical shape while falling (this last assumption is not actually ealistic), then the dag foce is modeled by the equation F D = kav. The falling doplet will then each a teminal velocity when the dag and weight foces balance, given by
18 F D mg = ma = 0 k Av t = mg k π v t = ρ 4π 3 3 g we have used the aea of a cicle fo the coss-section of the spheical dop, m = ρv fo the mass of the dop, and the fomula fo the volume of a sphee v t = 4gρ 3k, whee ρ is the density of wate and k is the (assumed constant) dag coefficient. velocity thus: We can solve this to expess the adius of the doplet in tems of its teminal faste one, we find = 3kv t 4gρ ' =. If we now compae the slowe-falling aindop with the 3 k 4gρ 3 k 4gρ v' t = v t ' v v t t = 1 m. sec. 5 m. sec. = 1 5. As fo the fomation of the dop of wate within the aincloud, if its suface aea gows at a constant ate, then the dop s suface aea will be popotional to the time spent in the cloud; thus, S = 4π = Ct. Upon compaing the amounts of time equied fo each dop to fom, accoding to this model, we have S' S = 4π ' 4π = C t' C t ' = 1 = = t' t. Thus, a aindop which lands at 5 m./sec. is pedicted to be 5 times lage in diamete than one which lands at 1 m./sec. and equies 65 times longe to gow to that size befoe falling out of its cloud. This suggests that conditions within ainclouds must diffe consideably to poduce diffeent sizes of aindops. b) Hailstones will emain aloft as long as the ai dag fom the violent winds within a stom cloud can ovecome thei weight. If we make the same assumptions egading them that we did fo aindops (spheical shapes, constant densities, and constant dag coefficients egadless of size), we can use the esult we obtained ealie, v t = 4gρ, and apply this to the wind speed elative to an individual hailstone. 3k If we compae the wind speed equied to suppot baseball-sized hail to that needed to pea-sized hail aloft, we can estimate that v' t v t = 4gρ 3k 4gρ 3k ' = ' = ' = 7.5 cm cm. = 10 v t ' v t = So the tubulent wind speeds in a stom cloud need to be about thee times faste to suppot the lagest sots of hailstones than ae equied fo the moe typical sizes of hail. (Winds within the most violent cumulonimbus clouds can exceed 10 mph.)
19 16) a) The thee hoizontal foces acting on the ca ae the applied foce fom the engine, F A, the kinetic fictional foce between the ties and the oad, f k, and the dag foce fom the ai, F D. Since the vehicle is on a level oad hee, the nomal foce on it fom the oad suface is N = Mg, so the kinetic fictional foce is f k = µ k Mg. With the ca taveling at constant speed, the net hoizontal foce on the ca is F netx = F A + f k + F D = 0 F netx = F A f k F D = 0. The net powe applied to the ca will then also be zeo: P net = F netx v = F A v + f k v + = F A v f k v F D v = 0. F D v = F A v cos 0 o + f k v cos 180 o + F D v cos 180 o We ae told that the powe being povided by the engine is P A = F A v = 95,000 W. At a speed of v = 65 mi m. 1 h m. h. mi sec. sec., the potion of this powe that is being used to ovecome ai dag is P D = F D v = F A v f k v = P A µ k Mgv = 95,000 W 0.05 (900 kg.) (9.81 m. m. ) (9.05 sec. sec. ) 95, W 88,600 W. To maintain the ca at this speed, used to ovecome ai dag. 88,600 W 95,000 W of the engine s powe is being b) If the dag foce is popotional to the squae of the vehicle s speed, then at F D ' 80 mph, the dag is = v' 80 mph F D v = times stonge than it is at 65 mph 65 mph. The powe necessay to ovecome ai dag at this highe speed, howeve, is P D ' = F D 'v' P D F D v = v' v v' 3 v = 80 mph times geate than at 65 mph. This 65 mph gives us P D P D ,600 W 165,00 W. The total powe the engine must now supply is P A = P D + P f 165, W 171,600 W ( 30 hp ).
20 c) With the ca now moving up an incline, the nomal foce fom the oad suface becomes N = Mg cos θ, making the kinetic fictional foce f k = µ k Mg cos θ. In addition to the foces aleady descibed, the weight foce on the vehicle now also has a component paallel to the incline, W = Mg sin θ. The net foce on the ca is now F net = F A + F D + f k ' + W = 0, so the net powe becomes P net = F A v F D v ( µ k Mg cos θ ) v ( Mg sin θ ) v = 0. Since the ca is on a 1% gade, tan θ = 0.1 sin θ , cos θ We have seen that the powe exeted by ai dag is P D 88,600 W at 9.05 m./sec. (65 mph) and that it is popotional to v 3, so we can wite P D = 88,600 v 3 W. The 9.05 powe which must be deliveed by the engine in ode fo the vehicle to maintain a speed v on the 1% upwad incline is then P A = F A v = P D + (µ k Mg cos θ ) v + (Mg sin θ ) v 88, 600 v 3 + (0.05) (900 kg.) (9.81 m. m. ) (0.999) v + (900 kg.) ( ) (0.1191) v sec. sec v v v v v. We ae going to keep the powe fom the engine at the same level it had in pat (b) above, so we need to solve fo v the equation v v = 171,600. It is not vey convenient to solve this last equation diectly (we can, of couse, use gaphing softwae, as will be discussed below), but we don t need to esot to tial-andeo completely eithe. We know that this amount of powe on level gound allows the ca to tavel at 80 mph ( 35.8 m./sec. ), so we might expect that on a 1% climb, the ca might move, say, 10% o so slowe. Let s make a fist guess that the solution is v = 3 m./sec. Since the cubic tem in the equation, v 3, changes much moe apidly than the linea tem, 171v, we ll simply set this tem to fo the pesent and solve the following fo v : v = v = 171, v 3 = 130,900 v 3 36,00 v 33.1 sec. m., which is petty close to ou initial guess. If we adjust the linea tem to accommodate this new value, we find
21 3.614 v = v = 171,600 v 33.0 m. sec.. Ou solution has stabilized to one decimal place, so we may stop hee. A moe pecise solution, using gaphing softwae to find the x-intecepts fo the function v v 171,600, gives us v 3.98 m./sec., so ou esult is acceptable. As a check, we can calculate the individual powe tems: powe exeted against dag: P A ,900 W powe exeted against fiction: P f W powe exeted against gavity: P W ,700 W total powe to be deliveed by engine: 171,800 W, ageeing with the intended total value to about 0.1%. d) Without the applied foce fom the engine, and assuming that thee is no intenal fiction in the dive tain of the ca (a bit unealistic), the net foce on the vehicle acting paallel to the downwad incline is F net ' = W + F D + f k ' = 0 F net ' = W F D f k ' = 0, making the net powe P net = ( Mg sin θ ) v F D v ( µ k Mg cos θ ) v = 0. Each of these tems has the same value as it did in pat (c) above, giving us 105 v v v 83.3v v 3 = 0, which is much easie to solve fo v than the powe equation we found in the pevious pat. We can facto this as v ( v ) = 0, so eithe v = 0 (the uninteesting solution, since the powe tems ae of couse all zeo when the ca is paked) o v = 0 v 30.3 v 15. m./sec. (34 mph). In a eal vehicle, intenal fiction would make the downhill coasting speed somewhat lowe than this. 17) With the puck moving on a fictionless suface, it is possible to aange a steadystate situation (not a static, but athe a dynamic equilibium) in which the puck tavels on a cicle at a constant tangential speed v. The centipetal foce keeping the puck moving on its cicle is povided by the physical foce of tension in the cod connecting it to the hanging weight. If we set things up so that the suspended mass emains at est, then the vetical foce equation fo this mass is T Mg = Ma V = 0 T = Mg. The centipetal foce on the puck is thus F C = mv R = T = Mg.
22 The tangential speed which the puck should be given in ode to maintain this equilibium is then given by v = M m gr 1.3 kg. 0.3 kg. m. m. (9.81 ) (0.45 m.) 19.1 v 4.37 m. sec. sec. sec.. If we now intoduce a fictional foce by consideing the motion of the puck on a eal ai table, the kinetic fiction poduces a toque on the evolving puck, causing a change in its angula momentum. (Note: the fiction will also dissipate the mechanical enegy of the system, causing the hanging mass to descend and the puck to move towad the cente by spialing inwad; we ae not in a position in this couse, howeve, to analyze the changes in a system whee the ciculating object changes distance fom the axis of otation this equies somewhat moe advanced mechanics.) If the ate of change is not too apid, the puck s velocity, and hence the diection of the kinetic fictional foce, will be vey nealy pependicula to the cod. The angula momentum at the moment descibed in the Poblem is then L 0 = 0 p 0 L 0 R 0 mv 0 sin 90 o M = R 0 m m gr 0 1/ = and the fictional toque is τ f = f k τ f R µ k N sin 90 o = µ k mg R. ( gmm ) 1/ 3 / R 0 Since this toque acts in the opposite diection to the angula momentum, we can wite µ k mg R τ f = dl dt d dt gmm ( ) 1/ R 3 / [ ] = 3 gmm ( ) 1/ R 1/ dr dt
23 dr dt µ k mg R 3 gmm ( ) 1/ = R 1/ 3 µ k g m M 1/ R 1/ A A is a constant which we shall efe to below At the moment descibed in the Poblem, we find dr dt 0 3 µ k ( g m M ) 1/ 1/ R m. sec. o m. 0.3 kg. (0.0008) 9.81 sec. 1.3 kg mm. sec. 1/ (0.45 m.) 1/ Since the puck is getting close to the hole in the table and is attached to the hanging mass by the same cod, that mass is descending at this ate as well. The spialing-in of the puck is quite slow, so we can easonably apply the tangential speed elationship we found ealie, v = M m gr. We see fom this that the tangential speed becomes smalle as the adius at which the puck tavels shinks. So the puck is slowing down as it spials in towad the hole. We can examine this in somewhat moe detail by using the diffeential equation dr we have detemined, = AR 1/, with A > 0. By eaanging this equation (in dt what is called a sepaation of vaiables ), we can then integate both sides to solve fo the adius of the (appoximate) cicle on which the puck is moving as a function of time: dr dt = AR 1/ A being the constant shown above dr = A dt 1 1 R1/ = At + C ; R 1/ since, at t = 0, R(0) = R 0, we have R 0 1/ = A 0 + C C = R 0 1/, and hence R 1/ = 1 3 µ k g m M 1/ t + 1 R 1/ 0 1/ R(t ) = R µ k g m M 1/ t and, using the tangential speed equation above, we obtain v(t ) = g M m R1/ = g M m 1 3 µ k g m M 1/ t + R 1/ 0 = g M m R 1/ µ k g m M 3 / t.
24 We can now estimate the time T at which the adius of the cicle the puck moves along dops to zeo 1/ R µ k g m M 1/ T = 0 T = R1/ µ k g m M ( ) 1/ = 3 M R 0 µ k mg 1/ kg m kg m. sec. 1/ 1700 sec. ; we find that the tangential speed also becomes zeo at this time ( v ( T ) = 0 ). Related physical quantities, such as the angula momentum of the puck, L ( t ) = ( gmm ) 1/ R 3/, the fictional toque acting on it, τ ( t ) = µ k mgr, and the otational kinetic enegy of the puck, K ot (t ) = 1 Iω = 1 (mr ) v = 1 R m v = 1 m M gr = 1 m MgR, all fall smoothly to zeo as the time appoaches T. (It should be said that the time T is only an estimate since the appoximation that the puck is taveling on diminishing cicles beaks down when the adius becomes less than seveal centimetes.) 18) Fo the object olling on the suface of the globe, thee ae two foces acting upon it (we can neglect olling fiction): the nomal foce, N, and the weight foce, mg. What will be of inteest to us ae the components of these foces in the adial diection, as thei esultant povides the centipetal foce holding the object on its cicula motion as it tavels down a meidian of the globe s suface. If we call the adial diection towad the cente of the globe positive and θ is the pola angle measued downwad fom the top of the sphee, the foce in the inwad adial diection is mg cos θ N = F C = mv, with R being the adius of the globe. R
25 By itself, the foce equation is not of enough help to us: we also need to know something about the speed of the object as it olls. It is not convenient to detemine this fom analyzing the foces, since the acceleation is not constant (o easily integated). Since we ae teating fiction as negligible, mechanical enegy is conseved; if we compae this total enegy at the top of the sphee ( θ = 0, which we take to be the Noth Pole of the globe) and at pola angle θ, we find K i + U i = 0 + mgr = K f + U f = ( K lin + K ot ) + mgh. The linea kinetic enegy of the olling object is K lin = ½ mv, while its otational kinetic enegy is K ot = ½ I ω = ½ ( Cm ) ( v/ ) = ½ Cmv, whee C is the object s coefficient of otational inetia, and olling equies that ω = v/, being the adius of the object. With h = R cos θ, ou consevation of enegy equation becomes mgr = 1 mv + 1 C mv + mgr cos θ v = gr (1 cos θ ) 1 + C. We now etun to the foce equation. The olling object will emain on the suface of the globe as long as N > 0 ; it beaks contact with the suface at the point whee the nomal foce dops to zeo. At that position, we can say that mg cos θ 0 = mv R g cos θ = gr (1 cos θ ) 1 + C R cos θ = (1 cos θ ) 1 + C (1 + C ) cos θ = cos θ cos θ = 3 + C. Fo a mable of unifom density, which is a solid sphee, C = /5, so the mable leaves the suface of the globe at cos θ = 3 + = ( 17 = ) 17 θ 54.0o ; since the latitude on the globe is λ = 90º θ, the mable flies off the globe at λ 36.0º. The finge ing behaves nealy like an idealized hoop with C 1, so it leaves the globe s suface whee cos θ = = 4 = 1 θ = 60.0o λ = 30.0 o. By contast, a sliding block will have no otational kinetic enegy, so C = 0, giving cos θ = = 3 θ = 48.o λ = 41.8 o. Since this mass does not otate, it builds up speed on the globe s suface faste and so ovecomes ealie the ability of gavity to hold it along the cuved suface. -- G.Ruffa oiginal notes developed duing evised: August 010