r 6.10 m 5.2 m / s 9.80 m / s

Transcription

1 Week 4 homewok IMPORAN NOE ABOU WEBASSIGN: In the WebAign veion of thee poblem, vaiou detail have been hanged, o that the anwe will ome out diffeently. he method to find the olution i the ame, but you will need to epeat pat of the alulation to find out what you anwe hould have been. WebAign Poblem 1: In a kating tunt known a ak-the-whip, a numbe of kate hold hand and fom a taight line. hey ty to kate o that the line otate about the kate at one end, who at a the pivot. he kate fathet out ha a ma of 80.0 kg and i 6.10 m fom the pivot. He i kating at a peed of 6.80 m/. Detemine the magnitude of the entipetal foe that at on him. REASONING he magnitude F of the entipetal foe that at on the kate i given by Equation 5.3 a F = mv /, whee m and v ae the ma and peed of the kate, and i the ditane of the kate fom the pivot. Sine all of thee vaiable ae known, we an find the magnitude of the entipetal foe. SOLUION he magnitude of the entipetal foe i ( 80.0 kg ) ( 6.80 m/ ) mv F = = = 6.10 m 606 N WebAign Poblem : A blok i hung by a ting fom the inide oof of a van. When the van goe taight ahead at a peed of 8 m/, the blok hang vetially down. But when the van maintain thi ame peed aound an unbanked uve (adiu = 150 m), the blok wing towad the outide of the uve. hen the ting make an angle θ with the vetial. Find θ. REASONING AND SOLUION he entipetal aeleation of the blok i a = v / = (8 m/) /(150 m) = 5. m/ he angle θ an be obtained fom F θ = tan H G a I = KJ tan 1 C 1 g F H G 5. m / 9.80 m / I = 8 KJ WebAign Poblem 3: he dawing how a baggage aouel at an aipot. You uitae ha not lid all the way down the lope and i going aound at a ontant peed

2 on a ile ( = 11.0 m) a the aouel tun. he oeffiient of tati fition between the uitae and the aouel i 0.760, and the angle θ in the dawing i How muh time i equied fo you uitae to go aound one? REASONING he entipetal foe F equied to keep an objet of ma m that move with peed v on a ile of adiu i F = mv / (Equation 5.3). Fom Equation 5.1, we know that v = π /, whee i the peiod o the time fo the uitae to go aound one. heefoe, the entipetal foe an be witten a m F = ( π / ) m = 4 π (1) hi expeion an be olved fo. Howeve, we mut fit find the entipetal foe that at on the uitae. SOLUION hee foe at on the uitae. hey ae the weight mg of the uitae, the foe of tati fition f MAX, and the nomal foe F N exeted on the uitae by the ufae of the aouel. he following figue how the fee body diagam fo the uitae. In thi diagam, the y axi i along the vetial dietion. + y he foe of gavity at, then, in the y dietion. F M A X N f he entipetal foe that aue the uitae to θ move on it iula path i povided by the net θ foe in the +x dietion in the diagam. Fom the diagam, we an ee that only the foe F N and f MAX have hoizontal omponent. hu, we have θ MAX F = f o θ F inθ N, whee the minu ign indiate that the x omponent of F N point to the left in the diagam. Uing Equation 4.7 fo the m g maximum tati fitional foe, we an wite thi eult a in equation (). F = µ F o θ F in θ = F ( µ o θ in θ ) () N N N If we apply Newton' eond law in the y dietion, we ee fom the diagam that + x

3 MAX F o θ + f in θ mg = ma = 0 o F o θ + µ F in θ mg = 0 N y N N whee we again have ued Equation 4.7 fo the maximum tati fitional foe. Solving fo the nomal foe, we find mg F = N oθ + µ inθ Uing thi eult in equation (), we obtain the magnitude of the entipetal foe that at on the uitae: mg( µ o θ in θ ) F = F ( µ o θ in θ ) = N oθ + µ inθ With thi expeion fo the entipetal foe, equation (1) beome Solving fo the peiod, we find mg( µ o θ in θ ) mπ = 4 oθ + µ inθ ( + ) ( + ) 4π oθ µ inθ 4 π (11.0 m) o in 36.0 = = = g( µ o θ in θ ) 9.80 m/ o 36.0 in 36.0 ( ) ( ) 45 WebAign Poblem 4: Conult Multiple-Conept Example 14 fo bakgound petinent to thi poblem. In deigning otating pae tation to povide fo atifiial-gavity envionment, one of the ontaint that mut be onideed i motion ikne. Studie have hown that the negative effet of motion ikne begin to appea when the otational motion i fate than two evolution pe minute. On the othe hand, the magnitude of the entipetal aeleation at the atonaut feet hould equal the magnitude of the aeleation due to gavity on eath. hu, to eliminate the diffiultie with motion ikne, deigne mut hooe the ditane between the atonaut feet and the axi about whih the pae tation otate to be geate than a etain minimum value. What i thi minimum value? REASONING wo piee of infomation ae povided. One i the fat that the magnitude of the entipetal aeleation a i 9.80 m/. he othe i that the pae tation hould not otate fate than two evolution pe minute. hi ate of twie pe minute oepond to thity eond pe evolution, whih i the minimum value fo the peiod of the motion. With thee data in mind, we will bae ou olution on Equation 5., whih give the entipetal aeleation a a = v /, and on Equation 5.1, whih peifie that the peed v on a iula path of adiu i v = π /. SOLUION Fom Equation 5., we have

4 v v a = o = a Subtituting v = π / into thi eult and olving fo the adiu give ( π / ) a ( 9.80 m/ ) ( 30.0 ) v = = o = = = a a 4π 4π 3 m WebAign Poblem 5: A olle oate at an amuement pak ha a dip that bottom out in a vetial ile of adiu. A paenge feel the eat of the a puhing upwad on he with a foe equal to twie he weight a he goe though the dip. If = 0.0 m, how fat i the olle Paenge oate taveling at the bottom of the dip? REASONING Aoding to Equation 5.3, the magnitude F of the entipetal foe that at on eah paenge i F = mv /, whee m and v ae the ma and peed of a paenge and i the adiu of the tun. Fom thi elation we ee that the peed i given by v = F / m. he entipetal foe i the net foe equied to keep eah paenge moving on the iula path and point towad the ente of the ile. With the aid of a fee-body diagam, we will evaluate the net foe and, hene, detemine the peed. SOLUION he fee-body diagam how a paenge at the bottom of the iula dip. hee ae two foe ating: he downwad-ating weight mg and the upwadating foe mg that the eat exet on he. he net foe i +mg mg = +mg, whee we have taken up a the poitive dietion. hu, F = mg. he peed of the paenge an be found by uing thi eult in the equation above. Subtituting F = mg into the elation v = F / m yield ( ) F mg v = g ( 9.80 m/ ) ( 0.0 m) 14.0 m/ m = m = = = mg mg WebAign Poblem 6:. In an automati lothe dye, a hollow ylinde move the lothe on a vetial ile (adiu = 0.3 m), a the dawing how. he appliane i deigned o that the lothe tumble gently a they dy. hi mean that when a

5 piee of lothing eahe an angle of θ above the hoizontal, it loe ontat with the wall of the ylinde and fall onto the lothe below. How many evolution pe eond hould the ylinde make in ode that the lothe loe ontat with the wall when θ = 70.0? REASONING he dawing at the ight how the two foe that at on a piee of lothing jut befoe it loe ontat with the wall of the ylinde. At that intant the entipetal foe i povided by the nomal foe F N and the adial omponent of the weight. Fom the dawing, the adial omponent of the weight i given by F N φ m g θ C l o t h e mg o φ = mg o (90 θ ) = mg inθ heefoe, with inwad taken a the poitive dietion, Equation 5.3 ( F = m v / ) give F + mg in θ = N mv At the intant that a piee of lothing loe ontat with the ufae of the dum, F N = 0, and the above expeion beome mg in θ = mv Aoding to Equation 5.1, v = π /, and with thi ubtitution we obtain g in θ = ( π / ) 4π = hi expeion an be olved fo the peiod. Sine the peiod i the equied time fo one evolution, the numbe of evolution pe eond an be found by alulating 1/.

6 SOLUION Solving fo the peiod, we obtain 4π 0.3 m = = π = π = g in θ g inθ 9.80 m / in 70.0 h heefoe, the numbe of evolution pe eond that the ylinde hould make i 1 1 = = 0.85 ev / WebAign Poblem 7: he eath otate one pe day about an axi paing though the noth and outh pole, an axi that i pependiula to the plane of the equato. Auming the eath i a phee with a adiu of, detemine the peed and entipetal aeleation of a peon ituated (a) at the equato and (b) at a latitude of 30.0 noth of the equato. REASONING AND SOLUION a. At the equato a peon tavel in a ile whoe adiu equal the adiu of the eath, = R e = m, and whoe peiod of otation i = 1 day = We have he entipetal aeleation i a v = πr e / = 464 m/ ( 464 m/ ) v = = = m/ m 6 b. At 30.0 latitude a peon tavel in a ile of adiu, hu, = R e o 30.0 = m v = π / = 40 m/ and a = v / =.9 10 m/ WebAign Poblem 8: At amuement pak, thee i a popula ide whee the floo of a otating ylindial oom fall away, leaving the bak of the ide plateed againt the wall. Suppoe the adiu of the oom i 3.30 m and the peed of the wall i 10.0 m/ when the floo fall away. (a) What i the oue of the entipetal foe

7 ating on the ide? (b) How muh entipetal foe at on a 55.0-kg ide? () What i the minimum oeffiient of tati fition that mut exit between a ide bak and the wall, if the ide i to emain in plae when the floo dop away? REASONING AND SOLUION a. he entipetal foe i povided by t h e n o m a l f o e e x e t e d o n t h e i d e b y t h e w a l l. b. Newton' eond law applied in the hoizontal dietion give F N = mv / = (55.0 kg)(10.0 m/) /(3.30 m) = N. Newton' eond law applied in the vetial dietion give µ F N mg = 0 o µ = (mg)/f N = Patie oneptual poblem: 3. he equation of kinemati (Equation (3.3a) (3.4a) (3.5a) (3.6b)) deibe the motion of an objet that ha a ontant aeleation. hee equation annot be applied to unifom iula motion. Why not? REASONING AND SOLUION he equation of kinemati (Equation ) annot be applied to unifom iula motion beaue an objet in unifom iula motion doe not have a ontant aeleation. While the aeleation veto i ontant in magnitude a = v / h, it dietion hange ontantly -- it alway point towad the ente of the ile. A the objet move aound the ile the dietion of the aeleation mut ontantly hange. Beaue of thi hanging dietion, the ondition of ontant aeleation that i equied by Equation i violated. 8. What i the hane of a light a afely ounding an unbanked uve on an iy oad a ompaed to that of a heavy a: woe, the ame, o bette? Aume that both a have the ame peed and ae equipped with idential tie. Aount fo you anwe. REASONING AND SOLUION Fom Example 7, the maximum afe peed with whih a a an ound an unbanked uve of adiu i given by v = µ g. hi expeion i independent of the ma (and theefoe the weight) of the a. hu, the hane of a light a afely ounding an unbanked uve on an iy oad i the ame a that fo a heavie a (auming that all othe fato ae the ame).

8 1. Explain why a eal aiplane mut bank a it flie in a ile, but a model aiplane on a guideline an fly in a ile without banking. REASONING AND SOLUION A model aiplane on a guideline an fly in a ile beaue the tenion in the guideline povide the hoizontal entipetal foe neeay to pull the plane into a hoizontal ile. A eal aiplane ha no uh hoizontal foe. he ai on the wing on a eal plane exet an upwad lifting foe that i pependiula to the wing. he plane mut bank o that a omponent of the lifting foe an be oiented hoizontally, theeby poviding the equied entipetal foe to aue the plane to fly in a ile. 14. A tone i tied to a ting and whiled aound in a ile at a ontant peed. I the ting moe likely to beak when the ile i hoizontal o when it i vetial? Aount fo you anwe, auming the ontant peed i the ame in eah ae. REASONING AND SOLUION When the ting i whiled in a hoizontal ile, the tenion in the ting, F, povide the entipetal foe whih aue the tone to move in a ile. Sine the peed of the tone i ontant, mv / = F and the tenion in the ting i ontant. When the ting i whiled in a vetial ile, the tenion in the ting and the weight of the tone both ontibute to the entipetal foe, depending on whee the tone i on the ile. Now, howeve, the tenion ineae and deeae a the tone tavee the vetial ile. When the tone i at the lowet point in it wing, the tenion in the ting pull the tone upwad, while the weight of the tone at downwad. heefoe, the entipetal foe i mv / = F mg. Solving fo the tenion how that F = mv / + mg. hi tenion i lage than in the hoizontal ae. heefoe, the ting ha a geate hane of beaking when the tone i whiled in a vetial ile.