Chapter 9 Rotation. Conceptual Problems


 Shavonne Hicks
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1 Chapte 9 otation Conceptual Poblems 7 Duing a baseball game, the pitche has a blazing astball. You have not been able to swing the bat in time to hit the ball. You ae now just tying to make the bat contact the ball, hit the ball oul, and avoid a stikeout. So, you decide to take you coach s advice and gip the bat high athe than at the vey end. his change should incease bat speed; thus you will be able to swing the bat quicke and incease you chances o hitting the ball. Explain how this theoy woks in tems o the moment o inetia, angula acceleation, and toque o the bat. Detemine the Concept By choking up, you ae otating the bat about an axis close to the cente o mass, thus educing the bat s moment o inetia. he smalle the moment o inetia the lage the angula acceleation (a quicke bat) o the same toque. he moto o a meygoound exets a constant toque on it. As it speeds up om est, the powe output o the moto (a) is constant, (b) inceases linealy with the angula speed o the meygoound, (c) is zeo. (d) None o the above. Detemine the Concept he powe deliveed by the constant toque is the poduct o the toque and the angula speed o the meygoound. Because the constant toque causes the meygoound to acceleate, the powe output inceases linealy with the angula speed o the meygoound. (b) is coect. A spool is ee to otate about a ixed axis, and a sting wapped aound the axle causes the spool to otate in a counteclockwise diection (see Figue 94a). Howeve, i the spool is set on a hoizontal tabletop, the spool instead (given suicient ictional oce between the table and the spool) otates in a clockwise diection and olls to the ight (Figue 94b). By consideing toque about the appopiate axes, show that these conclusions ae consistent with Newton s second law o otations. Detemine the Concept Fist, visualize the situation. he sting pulling to the ight exets a toque on the spool with a moment am equal in length to the adius o the inne potion o the spool. When the spool is eely otating about that axis, then the toque due to the pulling sting causes a counte clockwise otation. Second, in the situation in which the spool is esting on the hoizontal tabletop, one should (o ease o undestanding) conside toques not about the cental axle o the spool, but about the point o contact with the tabletop. n this situation, thee is only one oce that can poduce a toque the applied oce. he motion o the spool can then be undestood in tems o the oce applied by the sting and 69
2 7 Chapte 9 the moment am equal to the dieence between the oute adius and the inne adius. his toque will cause a clockwise otation about the point o contact between spool and table and thus the spool olls to the ight (wheeas we might have thought the spool would otate in a counteclockwise sense, and thus move let). Angula Velocity, Angula Speed and Angula Acceleation 9 A wheel, eleased om est, is otating with constant angula acceleation o.6 ad/s. At 6. s ate its elease: (a) What is its angula speed? (b) hough what angle has the wheel tuned? (c) How many evolutions has it completed? (d) What is the linea speed, and what is the magnitude o the linea acceleation, o a point.3 m om the axis o otation? Pictue the Poblem Because the angula acceleation is constant, we can ind the vaious physical quantities called o in this poblem by using constantacceleation equations. (a) Using a constantacceleation equation, elate the angula speed o the wheel to its angula acceleation: ω ω αδt o, when ω, ω αδt Evaluate ω when Δt 6. s: ω.6 ad/s ( 6.s) 6 ad/s 5.6 ad/s (b) Using anothe constantacceleation equation, elate the angula displacement to the wheel s angula acceleation and the time it has been acceleating: Δθ ω ( Δ ) Δt α t o, when ω, Δθ α Δ t ( ) Evaluate Δ θ when Δt 6. s: Δθ ( 6s) (.6 ad/s )( 6.s) 47 ad 46.8ad (c) Convet Δθ ( 6.s) om adians to evolutions: ev Δ θ π ad ( 6.s) 46.8ad 7.4ev
3 otation 7 (d) elate the angula speed o the paticle to its tangential speed and evaluate the latte when Δt 6. s: v ω 4.7 m/s (.3m)( 5.6 ad/s) elate the esultant acceleation o the point to its tangential and centipetal acceleations when Δt 6. s: a a t a c 4 α ω ( α ) ( ω ) evaluate a: a 4 (.3 m) (.6 ad/s ) ( 5.6 ad/s) 73m/s 37 he tape in a standad VHS videotape cassette has a total length o 46 m, which is enough o the tape to play o. h (Figue 944). As the tape stats, the ull eel has a 45mm oute adius and a mm inne adius. At some point duing the play, both eels have the same angula speed. Calculate this angula speed in adians pe second and in evolutions pe minute. (Hint: Between the two eels the tape moves at constant speed.) Pictue the Poblem he two tapes will have the same tangential and angula velocities when the two eels ae the same size, i.e., have the same aea. We can calculate the tangential speed o the tape om its length and unning time and elate the angula speed to the constant tangential speed and the adii o the eels when they ae tuning with the same angula speed. All the tape is on the spool to the let. he spools ae otating with the same angula speed.
4 7 Chapte 9 elate the angula speed o a tape to its tangential speed when the tapes have the same angula speed: At the instant both eels have the same aea: v ω () π π ( π π ) Solving o yields: Expess the tangential speed o the tape in tems o its length L and unning time : L v Substituting o and v in equation () gives: ω L L evaluate : 46 m ω 36 s. h h ad ev.4 s π ad ( 45mm) ( mm) 6s min 9.9 ev/min.4 ad/s. ad/s Calculating the Moment o netia 4 Fou paticles, one at each o the ou cones o a squae with.m long edges, ae connected by massless ods (Figue 945). he masses o the paticles ae m m 3 3. kg and m m 4 4. kg. Find the moment o inetia o the system about the z axis. Pictue the Poblem he moment o inetia o a system o paticles with espect to a given axis is the sum o the poducts o the mass o each paticle and the squae o its distance om the given axis.
5 otation 73 Use the deinition o the moment o inetia o a system o ou paticles to obtain: m i m i i m m 3 3 m 4 4 evaluate z : z ( 3.kg)(.m) ( 4.kg)( ) ( 3.kg)(.m) ( 4.kg)( m) 56kg m 53 [SSM] Use integation to show that the moment o inetia o a thin spheical shell o adius and mass m about an axis though its cente is m /3. Pictue the Poblem We can deive the given expession o the moment o inetia o a spheical shell by ollowing the pocedue outlined in the poblem statement. Find the moment o inetia o a sphee, with espect to an axis though a diamete, in able 9: Expess the mass o the sphee as a unction o its density and adius: Substitute o m to obtain: Expess the dieential o this expession: Expess the incease in mass dm as the adius o the sphee inceases by d: Eliminate d between equations () and () to obtain: m m π ρ π ρ 8 4 d π ρ d () 3 dm 4π ρ d () d 3 dm ntegate ove the mass o the spheical shell to obtain: spheical shell 3 m
6 74 Chapte 9 oque, Moment o netia, and Newton s Second Law o otation 59 A.5kg cmadius cylinde, initially at est, is ee to otate about the axis o the cylinde. A ope o negligible mass is wapped aound it and pulled with a oce o 7 N. Assuming that the ope does not slip, ind (a) the toque exeted on the cylinde by the ope, (b) the angula acceleation o the cylinde, and (c) the angula speed o the cylinde ate.5 s. Pictue the Poblem We can ind the toque exeted by the 7N oce om the deinition o toque. he angula acceleation esulting om this toque is elated to the toque though Newton s second law in otational om. Once we know the angula acceleation, we can ind the angula speed o the cylinde as a unction o time. Comment [EPM]: DAVD: HE EDS N HS POBLEM MAY EFFEC HE SOLUON, HE ANSWE, O BOH. (a) he toque exeted by the ope is: τ Fl ( 7 N)(.m).9 N m.87 N m (b) Use Newton s second law in otational om to elate the acceleation esulting om this toque to the toque: Expess the moment o inetia o the cylinde with espect to its axis o otation: Substitute o and simpliy to obtain: evaluate α: τ α M τ α M α (.87 N m) (.5kg)(.m). ad/s 4 ad/s (c) Using a constantacceleation equation, expess the angula speed o the cylinde as a unction o time: ω ω α t o, because ω, ω α t evaluate ω (.5 s): ω (.5s) ( 4 ad/s )(.5s) 6 ad/s
7 otation 75 Enegy Methods ncluding otational Kinetic Enegy 65 A.4kg 5cmdiamete solid sphee is otating about its diamete at 7 ev/min. (a) What is its kinetic enegy? (b) an additional 5. mj o enegy ae added to the kinetic enegy, what is the new angula speed o the sphee? Comment [EPM]: DAVD: HE EDS N HS POBLEM MAY EFFEC HE SOLUON, HE ANSWE, O BOH. Pictue the Poblem We can ind the kinetic enegy o this otating ball om its angula speed and its moment o inetia. n Pat (b) we can use the wokkinetic enegy theoem to ind the angula speed o the sphee when additional kinetic enegy has been added to the sphee. (a) he initial otational kinetic enegy o the ball is: Expess the moment o inetia o the ball with espect to its diamete: Substitute o to obtain: K i 5 ω M K i 5 i M ω i evaluate K: K i 5 ev min π ad ev min 6s (.4kg)(.75m) mj 85mJ (b) Apply the wokkinetic enegy theoem to the sphee to obtain: W ΔK K K o W ω K ω i i ( W K ) i Substitute o and simpliy to obtain: ω ( W K ) 5( W K ) 5 M i M i evaluate ω : ω ( 5. mj) (.4 kg)( 7.5 cm) mj ad ev 6 s 7.54 s π ad min 7 ev/min
8 76 Chapte 9 67 A kg block is lited at a constant speed o 8. cm/s by a steel cable that passes ove a massless pulley to a motodiven winch (Figue 953). he adius o the winch dum is 3 cm. (a) What is the tension in the cable? (b) What toque does the cable exet on the winch dum? (c) What is the angula speed o the winch dum? (d) What powe must be developed by the moto to dive the winch dum? Comment [EPM3]: DAVD: Slight modiication. Pictue the Poblem Because the load is not being acceleated, the tension in the cable equals the weight o the load. he ole o the massless pulley is to change the diection the oce (tension) in the cable acts. (a) Because the block is lited at constant speed: mg 9.6 kn ( kg)( 9.8m/s ) (b) Apply the deinition o toque to the winch dum: τ ( 9.6 kn)(.3m) 5.9kN m (c) elate the angula speed o the winch dum to the ate at which the load is being lited (the tangential speed o the cable on the dum): v.8 m/s ω.3m.7 ad/s (d) he powe developed by the moto in tems is the poduct o the tension in the cable and the speed with which the load is being lited: P v.6kw ( 9.6kN)(.8m/s) Pulleys, YoYos, and Hanging hings 7 he system shown in Figue 955consists o a 4.kg block esting on a ictionless hoizontal ledge. his block is attached to a sting that passes ove a pulley, and the othe end o the sting is attached to a hanging.kg block. he pulley is a uniom disk o adius 8. cm and mass.6 kg. Find the acceleation o each block and the tension in the sting. Pictue the Poblem he diagams show the oces acting on each o the masses and the pulley. We can apply Newton s second law to the two blocks and the pulley to obtain thee equations in the unknowns,, and a.
9 otation 77 F n 4 x m p m g m 4 x x g Apply Newton s second law in tanslational and otational om to the two blocks and the pulley to obtain: Substitute o p and α in equation () to obtain: Eliminate and between equations (), (3) and (4) and solve o a: evaluate a: F x m 4 a, () τ p ( ) pα, () and F x m g m a (3) M a (4) mg a m m 4 3.m/s p M p (.kg)( 9.8m/s ) a.kg 4.kg 3.m/s (.6 kg) Using equation (), evaluate : ( 4.kg)( 3.m/s ) N Solve equation (3) o : m ( g a) evaluate : (.kg)( 9.8m/s 3.m/s ) 3N 79 wo objects ae attached to opes that ae attached to two wheels on a common axle, as shown in Figue 96. he two wheels ae attached togethe so that they om a single igid object. he moment o inetia o the igid object is 4 kg m. he adii o the wheels ae. m and.4 m. (a) m 4 kg, ind m such that thee is no angula acceleation o the wheels. (b) kg is placed on top o m, ind the angula acceleation o the wheels and the tensions in the opes.
10 78 Chapte 9 Pictue the Poblem he ollowing diagam shows the oces acting on both objects and the pulley o the conditions o Pat (b). By applying Newton s second law o motion, we can obtain a system o thee equations in the unknowns,, and α that we can solve simultaneously. x x g m x F mg (a) When the system does not acceleate, m g and m g. Unde these conditions: Solving o m yields: evaluate m : (b) Apply Newton s second law in tanslational and otational om to the objects and the pulley: Eliminate a in avo o α in equations () and (3) and solve o and : Substitute o and in equation () and solve o α to obtain: τ m g mg m m m.m.4m ( 4kg) 7kg F x m g m a, () τ α, () and F x m g m a (3) ( g ) m α (4) and m ( g α ) (5) α ( m m ) m m g evaluate α: α [( 36kg)(.m) ( 7kg)(.4m) ]( 9.8m/s ) ( 36kg)(.m) ( 7kg)(.4m) 4kg m.37 ad/s.4 ad/s
11 otation 79 Substitute numeical values in equation (4) to ind : ( 36kg)[ 9.8m/s (.m)(.37 ad/s )].9kN Substitute numeical values in equation (5) to ind : ( 7kg)[ 9.8m/s (.4m)(.37 ad/s )].75kN 8 A uniom cylinde o mass m and adius is pivoted on ictionless beaings. A massless sting wapped aound the cylinde is connected to a block o mass m that is on a ictionless incline o angle θ, as shown in Figue 96. he system is eleased om est with the block a vetical distance h above the bottom o the incline. (a) What is the acceleation o the block? (b) What is the tension in the sting? (c) What is the speed o the block as it eaches the bottom o the incline? (d) Evaluate you answes o the special case whee θ 9 and m. Ae you answes what you would expect o this special case? Explain. Comment [EPM4]: DAVD: HE EDS N HS POBLEM MAY EFFEC HE SOLUON, HE ANSWE, O BOH. Pictue the Poblem Let the zeo o gavitational potential enegy be at the bottom o the incline. By applying Newton s second law to the cylinde and the block we can obtain simultaneous equations in a,, and α om which we can expess a and. By applying the consevation o enegy, we can deive an expession o the speed o the block when it eaches the bottom o the incline. F m θ m g F n x y (a) Apply Newton s second law in otational om to the cylinde and in tanslational om to the block: τ α () and F x m g sinθ m a () Substitute o α and in equation (), solve o, and substitute in equation () and solve o a to obtain: (b) Substituting o a in equation () and solve o yields: a g sinθ m m mg sinθ m m
12 8 Chapte 9 (c) Expess the total enegy o the system when the block is at the bottom o the incline in tems o its kinetic enegies: E bottom K K m v tan ot ω Substitute o ω and to obtain: v ( m ) m gh mv Solving o v yields: v gh m m (d) Fo θ 9 and m : a g,, and v gh Objects otating and olling Without Slipping 85 n 993 a giant 4kg yoyo with a adius o.5 m was dopped om a cane at a height o 57 m. One end o the sting was tied to the top o the cane, so the yoyo unwound as it descended. Assuming that the axle o the yoyo had a adius o. m, estimate its linea speed at the end o the all. Pictue the Poblem he oces acting on the yoyo ae shown in the igue. We can use a constantacceleation equation to elate the velocity o descent at the end o the all to the yoyo s acceleation and Newton s second law in both tanslational and otational om to ind the yoyo s acceleation. m x mg Comment [EPM5]: DAVD: HE EDS N HS POBLEM MAY EFFEC HE SOLUON, HE ANSWE, O BOH. Using a constantacceleation equation, elate the yoyo s inal speed to its acceleation and all distance: v v aδh o, because v, v aδh ()
13 otation 8 Apply Newton s second law in tanslational om and in otational om the yoyo to obtain: Use a α to eliminate α in equation (3) Eliminate between equations () and (4) to obtain: Substitute m o in equation (5): F x mg ma () and τ α (3) a (4) mg a ma (5) m g mg a ma a a 9.8m/s evaluate a: (.5m) (.m).864 m/s Substitute in equation () and evaluate v: v ( )( 57m) 3.m/s.864m/s 93 A uniom thin cylindical shell and a solid cylinde oll hoizontally without slipping. he speed o the cylindical shell is v. he cylinde and the hollow cylinde encounte an incline that they climb without slipping. the maximum height they each is the same, ind the initial speed v o the solid cylinde. Pictue the Poblem Let the subscipts u and h ee to the uniom and thinwalled sphees, espectively. Because the cylindes climb to the same height, thei kinetic enegies at the bottom o the incline must be equal. Expess the total kinetic enegy o the thinwalled cylinde at the h K tans K ot mhv hω m v m bottom o the inclined plane: ( ) h h h Expess the total kinetic enegy o the solid cylinde at the bottom o K v m v u K tans K ot muv' uω' m v' m the inclined plane: ( ) 3 u u 4 u K v' m v'
14 8 Chapte 9 Because the cylindes climb to the same height: 3 4 m v' and m v h u m gh m gh h u Divide the ist o these equations by the second: 3 4 muv' m v h mu gh m gh h Simpliy to obtain: 3v' 4v v' 4 v 3 99 wo lage geas that ae being designed as pat o a lage machine and ae shown in Figue 966; each is ee to otate about a ixed axis though its cente. he adius and moment o inetia o the smalle gea ae.5 m and. kg m, espectively, and the adius and moment o inetia o the lage gea ae. m and 6 kg m, espectively. he leve attached to the smalle gea is. m long and has a negligible mass. (a) a woke will typically apply a oce o. N to the end o the leve, as shown, what will be the angula acceleations o geas the two geas? (b) Anothe pat o the machine (not shown) will apply a oce tangentially to the oute edge o the lage gea to tempoaily keep the gea system om otating. What should the magnitude and diection o this oce (clockwise o counteclockwise) be? Comment [EPM6]: DAVD: deleted the labels in this poblem Pictue the Poblem he oces esponsible o the otation o the geas ae shown in the diagam to the ight. he oces acting though the centes o mass o the two geas have been omitted because they poduce no toque. We can apply Newton s second law in otational om to obtain the equations o motion o the geas and the not slipping condition to elate thei angula acceleations. l N F F (a) Apply τ α to the geas to obtain thei equations o motion:. N m F α () and F α () whee F is the oce keeping the geas om slipping with espect to each othe.
15 otation 83 Because the geas do not slip elative to each othe, the tangential acceleations o the points whee they ae in contact must be the same: Divide equation () by to obtain: Divide equation () by to obtain: Adding these equations yields: α α o α α α. N m F α F α. N m α α (3) Use equation (3) to eliminate α : Solving o α yields: evaluate α :. N m α. N m α α.kg m.4 ad/s α. N m.5m.m ( ) ( 6kg m ).4 ad/s Use equation (3) to evaluate α : ( ) α.4 ad/s.ad/s (b) o countebalance the.n m toque, a counte toque o. N m must be applied to the ist gea:. N m F. N m F evaluate F:. N m F.5m 4. N, clockwise olling With Slipping 5 A.6kg billiad ball whose adius is 3. cm is given a shap blow by a cue stick. he applied oce is hoizontal and the line o action o the oce passes though the cente o the ball. he speed o the ball just ate the blow is
16 84 Chapte 9 4. m/s, and the coeicient o kinetic iction between the ball and the billiad table is.6. (a) How long does the ball slide beoe it begins to oll without slipping? (b) How a does it slide? (c) What is its speed once it begins olling without slipping? Pictue the Poblem Because the impulse is applied though the cente o mass, ω. We can use the esults o Example 96 to ind the olling time without slipping, the distance taveled to olling without slipping, and the speed o the ball once it begins to oll without slipping. (a) Fom Example 99 we have: t v 7 μ g k t 4. m/s evaluate t : 7 (.6)( 9.8m/s ).9s (b) Fom Example 99 we have: evaluate s : s v 49 μ g k ( 4. m/s) s 49 (.6)( 9.8m/s ).67m (c) Fom Example 96 we have: evaluate v : Geneal Poblems 5 v v 7 5 v 7 ( 4. m/s).9m/s 3 A uniom kg disk with a adius equal to.4 m initially otates with an angula speed o ev/min. A constant tangential oce is applied at a adial distance o.6 m om the axis. (a) How much wok must this oce do to stop the wheel? (b) the wheel is bought to est in.5 min, what toque does the oce poduce? What is the magnitude o the oce? (c) How many evolutions does the wheel make in these.5 min? Pictue the Poblem o stop the wheel, the tangential oce will have to do an amount o wok equal to the initial otational kinetic enegy o the wheel. We can ind the stopping toque and the oce om the aveage powe deliveed by the oce duing the slowing o the wheel. he numbe o evolutions made by the wheel as it stops can be ound om a constantacceleation equation.
17 otation 85 (a) elate the wok that must be done to stop the wheel to its kinetic enegy: ( m ) ω m W ω ω 4 evaluate W: W ev min π ad ev min 6s ( kg)(.4m) 78kJ 7.8 kj 4 (b) Expess the stopping toque in tems o the aveage powe equied: Pav τω av Pav τ ω av evaluate τ : τ 78kJ.5 min ( )( 6s/min) ( ev/min)( π ad/ev)( min/6s) 9.3N m 9 N m elate the stopping toque to the magnitude o the equied oce and evaluate F: (c) Using a constantacceleation equation, elate the angula displacement o the wheel to its aveage angula speed and the stopping time: 9.3N m F τ.6m Δθ ω Δt av.5 kn evaluate Δθ: ev/min Δθ.4 3 ev (.5min) 9 You ae paticipating in league bowling with you iends. ime ate time, you notice that you bowling ball olls back to you without slipping on the lat section o tack. When the ball encountes the slope that bings it up to the ball etun, it is moving at 3.7 m/s. he length o the sloped pat o the tack is.5 m. he adius o the bowling ball is.5 cm. (a) What is the angula speed o the ball beoe it encountes the slope? (b) the speed with which the ball emeges at the top o the incline is.4 m/s, what is the angle (assumed constant),
18 86 Chapte 9 that the sloped section o the tack makes with the hoizontal? (c) What is the magnitude o the angula acceleation o the ball while it is on the slope? Pictue the Poblem he pictoial epesentation shows the bowling ball slowing down as it olls up the slope. Let the system include the ball, the incline, and Eath. hen W ext and we can use consevation o mechanical enegy to ind the angle o the sloped section o the tack. m θ L U g (a) Because the bowling ball olls without slipping, its angula speed is diectly popotional to its linea speed: evaluate ω: (b) Apply consevation o mechanical enegy to the system as the bowling ball olls up the incline: v ω whee is the adius o the bowling ball. 3.7 m/s ω 3.7 ad/s.5 m 3. ad/s Wext ΔK ΔU o, because W ext, K K K K U U t, t,,, Substituting o the kinetic and potential enegies yields: mv mv ω mglsinθ ballω ball Solving o θ yields: ( ) ( ) Because m : θ sin v ball ω mgl m v ω ball 5 m( v v ) 5 m ( ω ω ) θ sin sin ( v v ) 7 gl mgl
19 evaluate θ : (c) he angula acceleation o the bowling ball is diectly popotional to its tanslational acceleation: Use a constantacceleation equation to elate the speeds o the ball at points and to its acceleation: Substitute in equation () to obtain: otation 87 ( 3.7 m/s) (.4 m/s) ) ( 9.8m/s )(.5 m) 7 θ sin 3 a α () v v v v al a L α v v L evaluate α : α (.4 m/s) ( 3.7 m/s) (.5 m)(.5 m) Comment [DN7]: 4 ad/s A popula classoom demonstation involves taking a metestick and holding it hoizontally at the.cm end with a numbe o pennies spaced evenly along its suace. the hand is suddenly elaxed so that the metestick pivots eely about the.cm mak unde the inluence o gavity, an inteesting thing is seen duing the ist pat o the stick s otation: the pennies neaest the.cm mak emain on the metestick, while those neaest the cm mak ae let behind by the alling metestick. (his demonstation is oten called the aste than gavity demonstation.) Suppose this demonstation is epeated without any pennies on the metestick. (a) What would the initial acceleation o the.cm mak then be? (he initial acceleation is the acceleation just ate the elease.) (b) What point on the metestick would then have an initial acceleation geate than g? Pictue the Poblem he diagam shows the oce the hand suppoting the metestick exets at the pivot point and the oce Eath exets on the metestick acting at the cente o mass. We can elate the angula acceleation to the acceleation o the end o the metestick using a Lα and use Newton s second law in otational om to elate α to the moment o inetia o the metestick. F hand cm P x L Mg L Comment [EPM8]: DAVD: HE EDS N HS POBLEM MAY EFFEC HE SOLUON, HE ANSWE, O BOH. Deleted: the eath
20 88 Chapte 9 (a) elate the acceleation o the a end o the metestick to the angula acceleation o the metestick: Apply τ α about an axis pependicula to the page and though the let end o the mete stick to obtain: Fom able 9, o a od pivoted at one end, we have: Substitute o P in the expession o α to obtain: a Lα () L MgL Mg P α α P P ML α 3 3MgL ML 3g L Substitute o α in equation () to 3g a obtain: evaluate a: (b) Expess the acceleation o a point on the metestick a distance x om the pivot point: Expess the condition that the metestick have an initial acceleation geate than g: ( ) 3 9.8m/s a 3g a α x x L 3g L x > g x > L m/s Substitute the numeical value o L and evaluate x: ( ).cm x > cm 5 Let s calculate the position y o the alling load attached to the winch in Example 98 as a unction o time by numeical integation. Let the y diection be staight downwad. hen, v(y) dy/dt, o N y t d y v ( y ) v ( y i ) Δ y i whee t is the time taken o the bucket to all a distance y, Δy is a small incement o y, and y NΔy. Hence, we can calculate t as a unction o d by numeical summation. Make a gaph o y vesus t between s and. s. Assume m w. kg,.5 m, m b 5. kg, L. m, and m c 3.5 kg. Use Δy. m. Compae this position to the position o the alling load i it wee in eeall.
21 otation 89 Pictue the Poblem As the load alls, mechanical enegy is conseved. As in Example 97, choose the initial potential enegy to be zeo and let the system include the winch, the bucket, and Eath. Apply consevation o mechanical enegy to obtain an expession o the speed o the bucket as a unction o its position and use the given expession o t to detemine the time equied o the bucket to tavel a distance y. Apply consevation o mechanical enegy to the system to obtain: U K U K () i i Expess the total potential enegy U U b Uc U w when the bucket has allen a distance y y: mgy m c'g whee m c ' is the mass o the hanging pat o the cable. Assume the cable is uniom and expess m c ' in tems o m c, y, and L: mc ' mc mc o mc ' y y L L Substitute o m c ' to obtain: U mc gy mgy L Noting that bucket, cable, and im o the winch have the same speed v, expess the total kinetic enegy when the bucket is alling with speed v: mv m v ( M ) Substituting in equation () yields: m gy mgy L K Kb K mv mv c K m v w c c m v c mv mcv 4 Mv c 4 ω Mv v Solving o v yields: v mcgy 4mgy L M m m c A speadsheet solution is shown below. he omulas used to calculate the quantities in the columns ae as ollows: Cell Fomula/Content Algebaic Fom D9 y
22 9 Chapte 9 D D9$B$8 y Δy E9 v E ((4*$B$3*$B$7*D*$B$7*D^/(*$B$5))/ mcgy ($B$*$B$3*$B$4))^.5 4mgy L M m mc F F9$B$8/((EE9)/) vn vn tn Δy J9.5*$B$7*H9^ gt A B C D E F G H J M kg.5 m 3 m 5 kg 4 m c 3.5 kg 5 L m 6 7 g 9.8 m/s 8 dy. m y v(y) t(y) t(y) y.5gt he solid line on the ollowing gaph shows the position o the bucket as a unction o time when it is in ee all and the dashed line shows its position as a unction o time unde the conditions modeled in this poblem.
23 otation 9 y (m) y' ee all n poblems dealing with a pulley with a nonzeo moment o inetia, the magnitude o the tensions in the opes hanging on eithe side o the pulley ae not equal. he dieence in the tension is due to the static ictional oce between the ope and the pulley; howeve, the static ictional oce cannot be made abitaily lage. Conside a massless ope wapped patly aound a cylinde though an angle Δθ (measued in adians). t can be shown that i the tension on one side o the pulley is, while the tension on the othe side is ( > ), the maximum value o that can be maintained without the ope slipping is max e μ s Δθ, whee μ s is the coeicient o static iction. Conside the Atwood s machine in Figue 977: the pulley has a adius.5 m, the moment o inetia is.35 kg m, and the coeicient o static iction between the wheel and the sting is μ s.3. (a) the tension on one side o the pulley is N, what is the maximum tension on the othe side that will pevent the ope om slipping on the pulley? (b) What is the acceleation o the blocks in this case? (c) the mass o one o the hanging blocks is. kg, what is the maximum mass o the othe block i, ate the blocks ae eleased, the pulley is to otate without slipping? t (s)
24 9 Chapte 9 Pictue the Poblem Feebody diagams o the pulley and the two blocks ae shown to the ight. Choose a coodinate system in which the diection o motion o the block whose mass is M (downwad) is the y diection. We can use the given μsδθ elationship ' max e to elate the tensions in the ope on eithe side o the pulley and apply Newton s second law in both otational om (to the pulley) and tanslational om (to the blocks) to obtain a system o equations that we can solve simultaneously o a,,, and M. μsδθ (a) Use ' max e to evaluate the maximum tension equied to pevent the ope om slipping on the pulley: (b) Given that the angle o wap is π adians, expess in tems o : Because the ope doesn t slip, we can elate the angula acceleation, α, o the pulley to the acceleation, a, o the hanging masses by: Apply τ α about an axis pependicula to the plane o the pulley and though its cente to obtain: Substitute o om equation () in equation () to obtain: Solving o yields: Apply F y may to the block whose mass is m to obtain: ' max m M mg ( ) (.3 N e ) 6 N (.3)π e π 5.66 N Mg () a α a () ( ) (.3) π ( e ) (.3) ( e π ) a mg ma and ma mg (3) a
25 otation 93 Equating these two expessions o and solving o a gives: a g ( ).3 π ( e ) m evaluate a: a (.3) π ( e )(. kg)(.5 m).98 m/s 9.8 m/s.35 kg m. m/s (c) Apply F ma to the block whose mass is M to obtain: y y Mg Ma M g a Substitute o (om equation ()) and (om equation (3)) yields: M ( ) (.3 g e ) m a g a π evaluate M: M ( )( ) (.3. kg.98 m/s 9.8 m/s ) π e 3. kg 9.8 m/s.98 m/s
26 94 Chapte 9
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