O -x 0. 4 kg. 12 cm. 3 kg

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Eugenia Horton
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1 Anwer, Key { Homework 9 { Rubin H andau 1 Thi print-out houd have 18 quetion. Check that it i compete before eaving the printer. Ao, mutipe-choice quetion may continue on the net coumn or page: nd a choice before making your eection. Some oution may be found on the ca home page. AP B 1998 MC 7 1:01, trigonometry,mutipe choice, < 1 min. 001 Three force act on an object. IF the object i in tranationa equiibrium, which of the foowing mut be true? I. The vector um of the three force mut equa zero. II. The magnitude of the three force mut be equa. III. A three force mut be parae. 1. I ony correct. III ony 3. I and II ony 4. II and III ony 5. I, II and III If an object i in tranationa equiibrium, the vector um of a force acting on it mut equa zero. Thu, I ony i the correct anwer. Reting Beam 1:01, cacuu, numeric, > 1 min. 003 A uniform beam of weight w and ength ha weight W 1 and W at two poition, a in the gure. The beam i reting at two point. For what vaue of wi the beam be baanced at P uch that the norma force at O i zero? 1. (W 1 + W )d + W 1 W. (W + W )d + W 1 W correct 3. (W 1 + W )d + W W 4. (W 1 + W )d + W 1 W 5. (W 1 + W )d + W W 1 Baic Concept: In equiibrium, X ~ F 0 X ~ 0 Take the torque about the point P. X P 0:,N 0 + d +W 1 + d +Wd,W 0 We want to nd for which N 0 0 (W 1 + W )d + W 1 W Baanced Rod 1:01, cacuu, mutipe choice, > 1 min. 004 Two object are to be dragged aong a tabe by attaching them to the end of a(n) :3 m rod. One weigh 6 N and ha a coecient of friction of 0:. The other weigh N and ha a coecient of friction of 0:54. W 1 W How far from the 6 N attachment houd the tring be poitioned to drag them eveny (a ange 90 ) and uniformy? Correct anwer: 1:555 m. Baic Concept Rotationa and Tranationa Equiibrium cw, ccw 0

2 Anwer, Key { Homework 9 { Rubin H andau Soution et the fucrum be at the point of attachment of the puing tring. µ W 1 1 y µ W The weight W 1 eert a frictiona force of 1 W 1 on the bar at a ditance from the point of attachment. The weight W eert a frictiona force of W on the bar at a ditance `, from the point of attachment. Thu by rotationa equiibrium 1 W 1, W (`, ) 0 ( 1 W 1 + W ) W ` ) W ` 1 W 1 + W Agorithm ` :3 m 5 (1) W 1 6N 0 40 () 1 0: 0:18 0:4 W N 15 5 (4) 0:54 0:1 0:7 (5) W ` 1 W 1 + W (6) h0:54ihih:3i h0:i h6i + h0:54i hi 1:555 m hmi hi hnihmi hi hni + hi hni Find for a inge bock. Correct anwer: 9:95 cm. Baic Concept: The denition of the center of ma (n brick): cm nx i1 nx i1 i m i m i 1 n nx i1 i ; where i i the center of ma poition of the i th brick and m i i the ma of the i th brick. Soution: The center of ma of a inge brick i in it midde or 1 of a brick' ength from it maimum overhang. Since 1 1, a meaured from the maimum overhang, cm n Brick on the Brink 1:03, cacuu, numeric, > 1 min. 007 Three identica uniform brick of ength 19:9 cm are tacked over the edge of a horizonta urface with the maimum overhang poibe without faing. Brick on the Brink 1:03, trigonometry, numeric, > 1 min. 005 A uniform brick of ength 19:9 cm i paced over the edge of a horizonta urface with the maimum overhang poibe without faing. y Find for three bock. Correct anwer: 18:417 cm.

3 Anwer, Key { Homework 9 { Rubin H andau 3 The top two brick can etend 3 of a brick' 4 ength from the maimum overhang. When the top two brick etended 3 of their ength 4 pat the third brick, the center of ma of the top three brick i in their midde or 11 1 of a brick' ength from the maimum overhang. Since 3 cm + 1 n , a meaured from the maimum overhang, cm n : h1: ih9i 100:0 1: Pa hpai hnm ih%i hi dp atm dp pa h Pa atmi h1: i h101300i 1750:5 atm hatmi hpai hpaatmi (6) 009 When water freeze, it epand about 9%. The buk moduu of ice i 1: Nm. What woud be the preure increae inide your automobie engine bock if the water in it froze? Correct anwer: 1750:5 atm. Baic Concept: Soution: P B V V P B V V (1: Nm )(0:09) 1: Pa 1750:5 atm : Agorithm hatmi Pa Paatm (1) 9:0% () :0 h9i 100:0 0:09 hi h%i hi B 1: Nm dp pa B 100:0 3:10 9 (4) (5) 010 A 450 kg oad i hung on a wire of ength 4:5 m, cro ectiona area :8 10,5 m, and Young' moduu 6: Nm. What i it increae in ength? Correct anwer: 10:48 mm. Baic Concept: F A Y F A Y, and for we have mg A Y (450 kg)(9:8 m ) (4:5 m) (:8 10,5 m ) (6: Nm ) 0:01048 m 10:48 mm Agorithm h mm m i 1000 mmm (1) g 9:8 m () m 450 kg :5 m 3:5 5:5 (4) A :8 10,5 m 1:510,5 4:510,5 (5) Y 6: Nm 6: : (6) F mg (7) h450ih9:8i 4410 N

4 Anwer, Key { Homework 9 { Rubin H andau 4 hni hkgihm i d m F (8) AY h4410ih4:5i h:8 10,5 ih6: i 0:01048 m hmi hnihmi hm ihnm i d mm d m h mm m i (9) h0:01048ih1000i 10:48 mm hmmi hmihmmmi 011 The dipacement in impe harmonic motion i maimum when the 1. acceeration i zero. kinetic energy i a maimum 3. veocity i a maimum 4. potentia energy function ha a maimum 5. veocity i zero correct 6. inear momentum i a maimum The maimum dipacement occur at the turning point, which are the point where the veocity i zero. Piton in Harmonic Motion 1:04, cacuu, numeric, > 1 min. 01 A piton in an automobie engine i in impe harmonic motion. If it ampitude of ociation from centerine i 4:6 cm and it ma i :97 kg, nd the maimum veocity of the piton when the auto engine i running at the rate of 1850 revmin. Correct anwer: 8:95041 m. The impe harmonic motion i decribed by A co!t where! i the frequency in rad/ if t i in econd. To nd! from the given vaue in rev/min, note: 60 revmin 1 rev 1 Hz, o! f (30:8334 rev) 193:73 rad : Dierentiating above we nd the veocity: v d dt,a! in(!t) meaning the maimum veocity i v ma A! (0:046 m) (193:73 rad) 8:95041 m ; ince ine ha a maimum vaue of 1. Piton in Harmonic Motion 1:04, cacuu, numeric, > 1 min. 013 Find the maimum acceeration of the piton when the auto engine i running at thi rate. Correct anwer: 1:73398 km. Simiary, the acceeration i a d dt,a! co(!t) o the maimum acceeration i a ma A! (0:046 m) (193:73 rad) 1:73398 km ; ince coine ha a maimum vaue of 1. Agorithm h m cmi 0:01 mcm (1) h min i 0: min () h km m i 0:001 kmm A 4:6 cm 8 (4) m :97 kg 0:8 3: (5) f 1850 revmin 1440 A u A h m (6) 5760 cmi (7)

5 Anwer, Key { Homework 9 { Rubin H andau 5 h4:6ih0:01i 0:046 m hmi hcmihmcmi f u f h min i (8) h1850ih0: i 30:8334 rev hrevi hrevminihmini! :0 f u (9) :0 h3:141596ih30:8334i 193:73 rad hradi hi hi hrevi v ma A u! (10) h0:046ih193:73i 8:95041 m hmi hmihradi a ma A u! :0 (11) h0:046ih193:73i :0 1733:98 m hm i hmihradi :0 a u a ma h km m i (1) h1733:98ih0:001i 1:73398 km hkm i hm ihkmmi. 18 cm 3. 4 cm 4. 9cm cm When the 3.0 kg bock i repaced by 4.0 kg bock, the equiibrium point i 1 cm 3:0kg 4:0 kg 16 cm beow the untretched point. However, the point when the bock revere it direction i twice a far a the equiibrium point. Thu the bock wi go 3 cm before it direction i revered. AP B 1998 MC 8 13:01, cacuu, numeric, > 1 min. 018 The graph beow repreent the potentia energy a a function of dipacement for an object on the end of a pring ociating in impe harmonic motion with ampitude 0. AP B 1998 MC 38 13:01, noarithmetic, mutipe choice, < 1 min. 014 A bock of ma 3.0 kg i hung from a tring, cauing it to tretch 1 cm at equiibrium, a hown. 1 cm 4 kg - 0 O - 0 Which of the foowing graph repreent the kinetic energy K of the object a a function of dipacement? 3 kg The 3.0 kg bock i then repaced by a 4.0 kg bock, and the new bock i reeaed from the poition hown above, at which the pring i untretched. How far wi the 4.0 kg bock fa before it direction i revered? 1. 3 cm correct

6 Anwer, Key { Homework 9 { Rubin H andau O O - 0. correct O O - 0 In impe harmonic motion of an object on the end of a pring, the tota energy i conerved. At the maimum dipacement 0, the kinetic energy i 0, thu, E ( 0 ). + K E ) K(), () : - 0 O - 0 Thu, K() ook ike an upidedown (). 019 Aume a car' upenion behave a a arge pring with a pring contant k. The ma of the car without any paenger i 1400 kg. When three peope with a combined ma of 74 kg it in the car, the car drop 1:1 cm ower on it pring. When they eave the car, the car bounce up and down. What i the frequency of the car' ociation? Correct anwer: :10159 Hz.

7 Anwer, Key { Homework 9 { Rubin H andau 7 We nd the pring contant from o that Then k mg mg k; (74 kg) (9:8 m ) (1:1 cm) Nm : r f 1 k M 1 :10159 Hz : Nm 1400 kg Agorithm h m cmi 0:01 mcm (1) g 9:8 m () m 74 kg :1 cm 0: (4) 1: cmi (5) u h m h1:1ih0:01i 0:011 m hmi hcmihmcmi M 1400 kg (6) k mg u (7) h74ih9:8i h0:011i Nm hnmi hkgihm i hmi q k M f 1:0 :0 q 1:0 h44109i h1400i :0 h3:141596i :10159 q Hz hhzi hi hnmi hkgi hi hi (8) AP M 1993 MC 33 13:0, trigonometry,mutipe choice, < 1 min. 00 A impe penduum conit of a 1.0-kiogram bra bob on a tring about 1.0 meter ong. It ha a period of.0 econd. The penduum woud have a period of 1.0 econd if the 1. tring were repaced by one about.0 meter ong. tring were repaced by one about 0.5 meter ong correct 3. bob were repaced by a 0.5-kg bra phere 4. bob were repaced by a 4.0-kg bra phere 5. ampitude of the motion were increaed The period of the motion i T g : Ony the change of wi reut in the change of the period. So if we want to reduce the period to haf of it origina vaue, we houd repace the tring with one about 0.5 meter ong. AP M 1998 MC 10 13:0, cacuu, numeric, > 1 min. 01 A penduum with a period of 1 on Earth, where the acceeration due to gravity i g, i taken to another panet, where it period i S. The acceeration due to gravity on the other panet i mot neary 1. 4g. g 3. g 4. g

8 Anwer, Key { Homework 9 { Rubin H andau 8 5. g 4 correct For a penduum, the reationhip between ` the period and the acceeration i T g. Now T 1 : T 1:,og 1 : g 4:1,which give the acceeration on the other panet i g 4. 0 A uniform dik of radiu :7 m and ma 9:1kg i upended from a pivot 0:43 m above it center of ma. ai Find the anguar frequency! for ma ociation. Correct anwer: 1:05115 rad. Baic Concept The phyica penduum: I,mgd in d dt o that the anguar frequency for ma ociation (in ) i r mgd! : I Parae ai theorem I I 0 + ma Soution: We need the moment of inertia of the dik about the pivot point, which we ca P. The moment of inertia of a uniform dik about it center i I dik 1 mr ; but here the dik i rotating about P, a ditance d from the center of ma. The parae ai theorem et u move the ai of rotation a ditance d: I P 1 mr + md m R + d : Then uing the formua for the ma ange ociation frequency of a phyica penduum (ee Baic Concept above), we obtain or!! mgd I P vu u t vu u gd vu u t R + d m mgd R + d u t (9:8 m )(0:43 m) (:7 m) +(0:43 m) 1:05115 rad : Agorithm R :7 m 1 5 (1) m 9:1 kg 1 10 () R 1 0:16 0:1 0:7 d R 1 R (4) h0:16ih:7i 0:43 m hmi hi hmi g 9:8 m (5)! hradi gd R :0 + (6) :0 d:0 h9:8ih0:43i h:7i :0 + h0:43i :0 :0 1:05115 rad vu u t hm ihmi hmi :0 hi + hmi :0 05 The period of a impe penduum winging through a ma ange i independent of

9 Anwer, Key { Homework 9 { Rubin H andau 9 1. the acceeration due to gravity. the height above ea eve 3. the ocation on the earth' urface 4. the ength of the penduum 5. the ma of the penduum bob correct The period of a penduum winging in a ma arc i T g Ociation 13:03, trigonometry,mutipe choice, < 1 min. 06 An object with a potentia energy () can ociate around 1. Any point provided that the retoring force eerted on the object i given byhooke' aw. Any tabe equiibrium point correct 3. Any point 4. Any untabe equiibrium point 5. Certain tabe equiibrium point 6. Any equiibrium point Ociation can ony be done around point of tabe equiibrium.

HO 25 Solutions. = s. = 296 kg s 2. = ( kg) s. = 2π m k and T = 2π ω. kg m = m kg. = 2π. = ω 2 A = 2πf

HO 25 Solutions. = s. = 296 kg s 2. = ( kg) s. = 2π m k and T = 2π ω. kg m = m kg. = 2π. = ω 2 A = 2πf HO 5 Soution 1.) haronic ociator = 0.300 g with an idea pring T = 0.00 T = π T π π o = = ( 0.300 g) 0.00 = 96 g = 96 N.) haronic ociator = 0.00 g and idea pring = 140 N F = x = a = d x dt o the dipaceent