Physics 100A, Homework 12-Chapter 11 (part 2)

Transcription

1 Phic A, Homework -Chapter (part ) Torque on a Seeaw A) Marcel i helping hi two children, Jacque and Gille, to balance on a eeaw o that the will be able to make it tilt back and orth without the heavier child, Jacque, impl inking to the ground. Given that Jacque, whoe weight i W, i itting at ditance L to the let o the pivot, at what ditance L hould Marcel place Gille, whoe weight i w, to the right o the pivot to balance the eeaw? B)ind the torque τ about the pivot due to the weight C)Determine the um o the torque on the eeaw. The torque produced b Gille weight τ G wl The torque produced b Jacque weight τ J WL w o Gille on the eeaw. The total torque about the pivot point mut equal zero in equilibrium. WL wl L WL/ w D) Gille ha an identical twin, Jean, alo o weight w. The two twin now it on the ame ide o the eeaw, with Gille at ditance rom the pivot and Jean at ditance L. L 3 Where hould Marcel poition Jacque to balance the eeaw? WL wl wl3 L ( w/ W)( L + L3) E) When Marcel ind the ditance L rom the previou part, it turn out to be greater than Lend, the ditance rom the pivot to the end o the eeaw. Hence, even with Jacque at the ver end o the eeaw, the twin Gille and Jean eert more torque than Jacque doe. Marcel now elect to balance the eeaw b puhing idewa on an ornament (hown in red) that i at height h above the pivot. WL h w( L + L ) end 3 ( WL w( L + L ))/h end 3.4) A hand-held hopping baket 6. cm long ha a.8 kg carton o milk at one end, and a.7 kg bo o cereal at the other end. Where hould a.8 kg container o orange juice be placed o that the baket balance at it center? Picture the Problem The bo o cereal i at the let end o the baket and the milk carton i at the right end. 3. cm r Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. m cereal Cereal g mjuiceg Milk mmilk g

2 Chapter Rotational Dnamic and Static Equilibrium Strateg Place the origin at the center o the L.6 m baket. Write Newton Second Law or torque with the pivot ai at the center o the baket. Set the net torque equal to zero and olve or the ditance r o the orange juice rom the center o the baket. The orange juice will be placed on the cereal ide o the baket becaue the cereal ha le ma and eert le torque than doe the milk. Solution Set olve or r τ and τ + ( L) mcereal g+ rm juice g ( L ) mmilk g L( mcereal + mmilk ) (.6 m)( kg) r.87 m 8.7 cm m.8 kg juice Inight Another wa to olve thi quetion i enure that the center o ma o the baket i at it geometric center, in a manner imilar to problem 46 in Chapter 9. However, the balancing o the torque i actuall a bit impler in thi cae..44) Maimum Overhang Three identical, uniorm book o length L are tacked one on top the other.ind the maimum overhang ditance d in the igure uch that the book do not all over. Picture the Problem The book are arranged in a tack a depicted at right, with book on the bottom and book 3 at the top o the tack. Strateg It i helpul to approach thi problem rom the top down. The center o ma o each et o book mut be above or to the let o the point o upport. ind the poition o the center o ma or ucceive tack o book to determine d. Meaure the poition o the book rom the right edge o book 3 (right hand dahed line in the igure). I the center o ma o the book above an edge i to the right o that edge, there will be an unbalanced torque on the book and the ll topple over. Thereore we can olve the problem b orcing the center o ma to be above the point o upport. Solution. The center o ma o book 3 need to be above the right end o book d. The reult o tep mean that the center o ma o book i located at d L + L Lrom the right edge o book The center o ma o book 3 and need to be above the right end o book X 3 cm,3 L + m( L) m L 3 L m 4 4. The reult o tep 3 mean that the center o ma o book i located at d 3L 4+ L 5L The center o ma o book 3,, and need to be above the right end o the table + + m L m L m 5L 4 d X L cm,3 3m Inight A we learned in problem 87 o Chapter 9, i ou add a ourth book the maimum overhang i ( 5 4 ) L. I L L L L 5 ou eamine the overhang o each book ou ind an intereting erie d L. The erie give ou a hint about how to predict the overhang o even larger tack o book..49) You pull downward with a orce o 35 N on a rope that pae over a dik-haped pulle o ma.5 kg and radiu.75m. The other end o the rope i attached to a.87 kg ma. Picture the Problem You pull traight downward on a rope that pae over a dik-haped pulle and then upport a weight on the other ide. The orce o our pull rotate the pulle and accelerate the ma upward. Strateg Write Newton Second Law or the hanging ma and Newton Second Law or torque about the ai o the pulle, and olve the two epreion or the tenion at the other end o the rope. We are given in the problem that T Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher.

3 Chapter Rotational Dnamic and Static Equilibrium T 35 N. Let m be the ma o the pulle, r be the radiu o the pulle, and M be the hanging ma. or the dikhaped pulle the moment o inertia i I mr.. Solution. (a) The tenion in the rope i not the ame in both ide o the pulle. The tenion in the rope on the other end o the rope accelerate the hanging ma, but the tenion on our ide both impart angular acceleration to the pulle and accelerate the hanging ma. Thereore, the rope on our ide o the pulle ha the greater tenion.. (b) A tated in the problem, T 35 N or the rope on our ide o the pulle. τ α or the pulle τ α ( ) ( ) 3. Set m a or the hanging ma T Mg Ma 4. Set I 5. Subtitute the epreion or a rom tep 4 into the one rom tep 3, and olve or T (the tenion on the other ide o the pulle rom ou) rt rt I mr a r a T T m T Mg M T T m mt mmg MT MT M ( T + mg) T M + m (.87 kg) ( 35 N) + (.5 kg)( 9.8 m/ ).87 kg +.5 kg Inight The net orce on the hanging ma i thu T Mg 3 (.87)(9.8) 4. N, enough to accelerate it upward at a 4. / m/. The angular acceleration o the pulle i thu ar ( 6.3 m/ ) (.75 m) 7 rad/. 3 N.5) You pull downward with a orce o 35 N on a rope that pae over a dik-haped pulle o ma.5 kg and radiu.75 m. The other end o the rope i attached to a.87 kg ma. Thi i the ame problem a.49. The anwer or the acceleration i above..54) A.5 kg record with a radiu o 5 cm rotate with an angular peed o 33 rpm. 3 ind the angular momentum o the record. Picture the Problem The dik-haped record rotate about it ai with a contant angular peed. Strateg Ue equation - and the moment o inertia o a uniorm dik rotating about it ai, I MR, to ind the angular momentum o the record. Solution Appl equation - directl L Iω rev π rad min MR ω.5 kg.5 m 33 3 min rev 6 4 L 5.9 kg m / Inight The angular momentum o a compact dik rotating at 3 rev/min i about kg m /. The compact dik (m 3 g, r 6. cm) i maller than a record, but it pin ater, o the angular momenta are imilar. Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 3

4 Chapter Rotational Dnamic and Static Equilibrium Spinning Situation. Suppoe ou are tanding on the center o a merr-go-round that i at ret. You are holding a pinning biccle wheel over our head o that it rotation ai i pointing upward. The wheel i rotating counterclockwie when oberved rom above. or thi problem, neglect an air reitance or riction between the merr-go-round and it oundation. Suppoe ou now grab the edge o the wheel with our hand, topping it rom pinning. What happen? Conider ourel, the merr-go-round, and the biccle wheel to be a ingle tem. When ou top the wheel rom pinning, the angular momentum o the tem about the vertical ai remain unchanged. Then to conerve angular momentum the merr-go-round begin to rotate counterclockwie (a een rom above). Change in Angular Velocit Ranking Tak A merr-go-round o radiu R, hown in the igure, i rotating at contant angular peed. The riction in it bearing i o mall that it can be ignored. A andbag o ma m i dropped onto the merr-go-round, at a poition deignated b r. The andbag doe not lip or roll upon contact with the merr-go-round. Rank the ollowing dierent combination o m and r on the bai o the angular peed o the merr-go-round ater the andbag "tick" to the merr-go-round. The guiding principle i that angular momentum i conerved. L I ω i mgr i ( mgr + ) L I mr ω L Li Imgrωi ω The value o ω depend o the value o the moment o inertia o the andbag mr. ( I + mr ) mgr cae m (kg) r (R) mr In decreaing order o omega (increaing order o the moment o inertia) 6, 3, (,), 5, 4 Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 4

5 Chapter Rotational Dnamic and Static Equilibrium.65) A an ice kater begin a pin, hi angular peed i 3.7 rad/. Ater pulling in hi arm, hi angular peed increae to 5.46 rad/. ind the ratio o the kater' inal moment o inertia to hi initial moment o inertia. Picture the Problem The kater pull hi arm in, decreaing hi moment o inertia and increaing hi angular peed. Strateg The angular momentum o the kater remain the ame throughout the pin becaue there i aumed to be no torque o an kind acting on hi bod. Ue the conervation o angular momentum (equation -5) together with equation -, to ind the ratio I I i. Solution Set Li I ωi 3.7 rad/ L and olve or I I i Iiωi Iω.58 I ω 5.46 rad/ Inight B rearranging hi ma, epeciall b bringing hi arm and leg in cloe to hi ai o rotation, the kater ha reduced hi moment o inertia b an impreive 4% and increaed hi angular peed b 7%..8) To prepare homemade ice cream, a crank mut be turned with a torque o 3.95 N m. How much work i required or each complete turn o the crank? i. Picture the Problem The torque acting through an angular diplacement doe work on the ice cream crank. Strateg Ue equation -7 to ind the work done b the torque acting through the given angular diplacement. One complete turn correpond to an angular diplacement o π radian. Solution Appl equation -7 directl W τ θ ( π ) Δ 3.95 N m rad 4.8 J Inight The work done on the ice cream crank i diipated a heat via riction in the vicou ice cream miture. Introduction to Rotational Work and Power. Conider a motor that eert a contant torque o 5. N m to a horizontal platorm whoe moment o inertia i 5. kg m. Aume that the platorm i initiall at ret and the torque i applied or. rotation. A) How much work doe the motor do on the platorm during thi proce? W τδ θ (5)(rev)(πrad/rev),885 J B)What i the rotational kinetic energ o the platorm K rot, j at the end o the proce decribed above? rom the work energ theorem the total work i equal to the change in kinetic energ. So i the platorm i initiall at ret, the inal kinetic energ i equal to the work or,885 J. Now the low approach τ Iα α τ / I 5 / 5.5 rad/ ω ω + αδθ i ω ( + (.5rad/)(rev)(πrad/rev)) 8.68 rad/ Krot, Iω (5)(868),885 J C) What i the angular velocit ω o the platorm at the end o thi proce? A ound above ω 8.68 rad/ Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 5

6 Chapter Rotational Dnamic and Static Equilibrium D) How long doe it take or the motor to do the work done on the platorm calculated in Part A? ω ω + α i t t ω α With ω i / 8.68 / E) What i the average power delivered b the motor in the ituation above? /,885 / Watt P W t avg P ω, o P increae a ) Note that the intantaneou power delivered b the motor i directl proportional to the platorm pin ater and ater. How doe the intantaneou power P being delivered b the motor at the time t compare to the average power P avg calculated in Part E? P τω (5)(868) 7 Watt P P avg Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 6

7 Chapter Rotational Dnamic and Static Equilibrium 8. Picture the Problem The drill pin the bit at a rapid rate while eerting a torque on the bit to keep it pinning. Strateg The power produced b the drill equal the torque it produce time it angular peed (equation -9). Solution. Convert τ into unit o N m. Convert ω into unit o rad/ lb 4.45 N m τ 3.68 oz in.6 N m 6 oz lb 39.4 in rev π rad min ω 4, rad/ min rev 6 3. Appl equation -9 directl P τω Inight The ame torque applied at 45 rev/min require onl.6 W o power..6 N m 445 rad/ 6 W 8. Picture the Problem The object gain rotational kinetic energ rom an applied torque acting through an angular diplacement. Strateg ind the kinetic energ that the L-haped object ha when it i rotated at.35 rad/ about the,, and z ae. The work that mut be done on the object to accelerate it rom ret equal it inal kinetic energ (equation -8 and -7). rom problem 5 we note that I 9. kg m, I kg m, and I 9 kg m. z Solution. (a) ind K or rotation about the ai W K I ω 9. kg m.35 rad/ 5 J. (b) ind 3. (c) ind K or rotation about the ai W K I ω kg m.35 rad/ 8 J K or rotation about the z ai W K I ω 9 kg m.35 rad/ 5 J z z Inight The larger the moment o inertia, the more work i required to obtain the ame rotation rate. 83. Picture the Problem The object gain rotational kinetic energ rom an applied torque acting through an angular diplacement. Strateg ind the kinetic energ that the rectangular object ha when it i rotated at.5 rad/ about the,, and z ae. The work that mut be done on the object to accelerate it rom ret equal it inal kinetic energ (equation -8 and -7). The power required to accomplih thi in 6.4 i the work divided b the time (equation -9). rom problem 8 we note that I.8 kg m, I.5 kg m, and I 4.3 kg m. z Solution. (a) ind P or rotation about the ai (.8 kg m )(.5 rad/) W I ω P t t W. (b) ind P or rotation about the ai ( ) I ω.5 kg m.5 rad/ W P. W t t 6.4 W I 4.3 kg m.5 rad/ 3. (c) ind P or rotation about the z ai z ω z P t t 6.4 Inight The larger the moment o inertia, the more work i required to obtain the ame rotation rate.. W Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 7

8 Chapter Rotational Dnamic and Static Equilibrium 84. Picture the Problem The aw blade rotate on it ai and gain rotational kinetic energ due to the torque applied b the electric motor. Strateg The torque applied through an angular diplacement give the blade it rotational kinetic energ. Ue equation -7 and -7 to relate the kinetic energ to the torque applied b the motor. Then ue equation -7 again to ind the kinetic energ and angular peed ater the blade ha completed hal a man revolution. Solution. (a) ind. Set W ΔK and olve orτ ω in unit o rad/ec rev π rad min ω rad/ min rev 6 W τ Δ θ Iω and I mr mr ω τ 5.8 N m Δ θ 6.3 rev rad/rev (.755 kg)(.5 m) ( 379 rad/) ( π ) 3. (b) The time to rotate the irt 3.5 revolution i greater than the time to rotate the lat 3.5 revolution becaue the blade i peeding up. So more than hal the time i pent in the irt 3.5 revolution. Thereore, the angular peed ha increaed to more than hal o it inal value. Ater 3.5 revolution, the angular peed i greater than 8 rpm. 4. (d) Set W ΔK and olve orω τ Δ θ Iω mr ω 4 4τ Δθ 4( 5.8 N m)( 3.5 rev π rad/rev) ω mr (.755 kg)(.5 m) 6 rev ( 68 rad/) 56 rev/min min π rad Inight The angular peed increae linearl upon time ( ω ω + αt αt) but depend upon the quare root o the angular diplacement ω ω + αδ θ αδθ. 85. Picture the Problem A uniorm dik tand upright on it edge, and ret on a heet o paper placed on a tabletop. The paper i pulled horizontall to the right. Strateg Ue Newton Second Law or linear motion and or torque to predict the behavior o the dik. Solution. (a) There are three orce that act upon the clinder, the orce o riction rom the paper, the orce o gravit on the center o ma, and the normal orce rom the tabletop. The paper orce i the onl one that eert a torque about the clinder center o ma, and it act in the counterclockwie direction to rotate the dik.. (b) The normal orce and the orce o gravit balance each other and do not produce an acceleration. The paper orce i unbalanced and produce an acceleration that will caue the center o the dik to move to the right. Inight When the paper i removed the dik i tranlating toward the right but i rolling toward the let. What happen net depend upon the rotation and tranlation peed a well a the magnitude o the riction orce on the dik. 86. Picture the Problem The two rotating tem hown at right each conit o a ma m attached to a rod o negligible ma pivoted at one end. On the let, the ma i attached at the midpoint o the rod; to the right, it i attached to the ree end o the rod. The rod are releaed rom ret in the horizontal poition at the ame time. Strateg Ue Newton Second Law or torque τ I α to predict the behavior o the two rotating tem. Solution The angular acceleration o each tem i given b α τ I. We can ee that the right hand tem eperience a larger torque due to it larger moment arm, but it alo ha a larger moment o inertia. Quantiing the two tem, we ind that τ let ( L )( mg) and ( ) I m L ml α mg L m L g L, and τ mgl and I right let, o 4 ml o right ( mgl) ( ml ) g L. right, let 4 α We can ee that the let hand tem ha the larger angular acceleration, and we conclude that when the rod to the let reache the vertical poition, the rod to the right i not et vertical (location A). Inight The greater eect i the moment o inertia, becaue it depend on the quare o the ditance rom the ai o rotation, wherea the torque depend onl on the irt power o the ditance. Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 8

9 Chapter Rotational Dnamic and Static Equilibrium Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 9

10 Chapter Rotational Dnamic and Static Equilibrium 87. Picture the Problem A dik and a biccle wheel o equal radiu and ma each have a tring wrapped around their circumerence. Hanging rom the tring, halwa between the dik and the hoop, i a block o ma m, a hown at right. The dik and the hoop are ree to rotate about their center. Strateg Ue Newton Second Law or torque τ I α to predict the behavior o the two rotating tem. Solution. (a) Upon it releae the ma eert equal torque on the dik and the wheel. However, the dik ha a maller moment o inertia than the wheel and eperience the larger angular acceleration α τ I. The tring on the dik will unravel ater than the tring on the biccle wheel, and we conclude that when the block i allowed to all, it will move toward the let.. (b) The bet eplanation i II. The wheel ha the greater moment o inertia and unwind more lowl than the dik. Statement I i ale, and tatement III i true, but irrelevant. Inight Statement III i onl true in term o ma and radiu. In term o moment o inertia, the tem i not mmetric, and that act i what lead to the oberved behavior. 88. Picture the Problem A beetle it at the rim o a turntable that i at ret but i ree to rotate about a vertical ai. Strateg Ue the conervation o angular momentum to anwer the conceptual quetion. Solution. (a) A the beetle begin to walk, it eert a orce and a torque on the turntable. The turntable eert an equal but oppoite orce and torque on the beetle. There are no torque on the beetle-turntable tem, o there i no net change in it linear or angular momentum. I the turntable i much more maive than the beetle, it will barel rotate backward a the beetle move orward. The beetle, then, will begin to circle around the perimeter o the turntable almot the ame a i it were on olid ground.. (b) I the turntable i virtuall male, it will rotate backward with a linear peed at the rim that i almot equal to the orward linear peed o the beetle. The beetle will progre ver lowl relative to the ground in thi cae though a ar a it i concerned, it i running with it uual peed. In the limit o a male turntable, the beetle will remain in the ame location relative to the ground. Inight In either cae, maive turntable or nearl male turntable, the angular momentum o the beetle in the laborator rame o reerence i balanced b the angular momentum o the turntable. The angular momentum o the beetle-turntable tem mut remain zero becaue there are no eternal torque on the tem. 89. Picture the Problem A beetle it at the rim o a turntable that i at ret but i ree to rotate about a vertical ai. Strateg Ue the conervation o angular momentum to anwer the conceptual quetion. Solution The angular momentum L Iω o the tem mut remain contant becaue there are no eternal torque acting on it. Thu, a the beetle walk toward the ai o rotation, which reduce the moment o inertia o the tem, the angular peed o the turntable will increae. Inight The beetle mut do work againt the centriugal orce, or rom another perpective the orce o riction (that upplie the centripetal orce to keep the beetle moving in a circle) doe work on the beetle a it move toward the center. The kinetic energ o the beetle thereore increae. A imilar eect occur when an ice kater doe work to move her arm inward toward her bod, and gain kinetic energ a he pin ater. 9. Picture the Problem The Earth i imagined to magicall epand, doubling it radiu while keeping it ma the ame. Strateg Ue the conervation o angular momentum to anwer the conceptual quetion. Solution The angular momentum L Iω o the Earth mut remain contant becaue there are no eternal torque acting on it. The moment o inertia I MR would increae ater the epanion, o the angular peed ω would 5 decreae and the length o a da would increae. Inight The moment o inertia o the Earth in thi cae would increae b a actor o our, producing a da that i our time longer, or 96 hour! Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher.

11 Chapter Rotational Dnamic and Static Equilibrium 9. Picture the Problem The work the hamter doe on the eercie wheel give the wheel rotational kinetic energ. Strateg ind the rotational kinetic energ o the wheel to determine the work done b the hamter (equation -8). Ue Table - to ind the moment o inertia o a hoop, I mr. The hamter run without lipping relative to the circumerence o the eercie wheel, o that ω vr(equation -5) relate it linear peed with the angular peed o the wheel. Solution Set W ΔK and ubtitute or I and ω (.65 kg)(.3 m/) ω W Δ K I mr v r mv J 5.5 mj Inight Note that in thi pecial cae the rotational kinetic energ o the wheel in the laborator rame o reerence equal the linear kinetic energ the hamter ha in the rotating rame o reerence o the wheel. 9. Picture the Problem The peron weight i upported b the hinge and the wire in the manner hown in the igure at right. Strateg Set the um o the torque about the hinge equal to zero and olve or the moment arm o the peron relative to the hinge. Let L length o the rod, m ma o the rod, m ma o the peron, and r p r T Tma 4 N p ditance rom the hinge to the peron. Let equation -6 to olve or r p. and ue r p mrg T θ mpg Solution Set τ olve or r p and ( θ) ( ) τ LTin Lmg r mg LT inθ Lmr g rp mg p r p p ( 4.5 m)( 45 N) in ( 3. ) ( 4.5 m)( 47. kg)( 9.8 m/ ) ( 68. kg)( 9.8 m/ ) 3.5 m Inight Note that when the peron i 3.5 m rom the hinge the tenion in the cable (45 N) i more than twice the weight o the peron (667 N). Thi i becaue about hal the tenion i pulling horizontall toward the hinge and not upporting the downward weight o the peron and the rod. 93. Picture the Problem The puck travel in a circular path about the hole in the table, but the radiu o the path can be adjuted b pulling on the tring rom underneath the table, a hown in the igure at right. Strateg Let the angular momentum o the puck remain contant, and ue equation - to ind the inal peed o the puck. Solution. (a) The angular momentum o the puck doe not change becaue the tring eert no torque on the puck, but it moment o inertia decreae a the radiu o it path decreae. Becaue L mvrwe conclude the linear peed o the puck mut increae in order or L to remain the ame while r decreae.. (b) Set L L and olve or i v mvr mv r r r v v v v r r Inight The puck gain kinetic energ in thi proce becaue pulling on the tring eert a orce in the ame direction a the radial diplacement and thereore doe work on the puck. Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher.

12 Chapter Rotational Dnamic and Static Equilibrium 94. Picture the Problem The maeter mucle and the biting orce each produce a torque about the joint in a manner depicted b the igure at right. Strateg ind the torque produced b the two orce b inding the portion o each orce that i perpendicular to the horizontal moment arm hown in the igure (equation -3). The torque rom the biting orce mut be the ame magnitude a the torque rom the maeter mucle in order or the torque to be in equilibrium. Ue the torque produced b the biting orce together with the moment arm to ind the magnitude o that orce. inall, appl Newton Second Law in the horizontal and vertical direction to ind the component o the orce J that the mandible eert on the joint. Solution. (a) The vertical component o M i the portion o the orce that produce a torque about the moment arm r D d. M r ( co D d M θ) τ m 455 N co N m. (b) Ue equation -3 again to ind B 3. (c) Set to ind 4. (d) Set to ind J, J, τ r D B B τ 3.3 N m 3 N D.85 m + M, J, J, M, M inθ 455 N in N + + B M, J, J, B M, B M coθ 3 N 455 N co N Inight While the biting orce i large (3 N i equal to 7.6 lb) the 348-N total orce on the joint i the ame a 78.3 lb, and i an indicator o how trong the joint and mucle mut be in order or the jaw to work correctl! 95. Picture the Problem The orce rom the elatic cord produce a torque about the elbow joint in the manner indicated b the igure at right. Strateg Ue the geometr in the igure to determine the component o the moment arm that i perpendicular to the orce, and then ue equation -3 to determine the that will produce the deired torque. inall, ue Hooke Law (equation 6-4) to ind the pring contant rom the orce and the tretch ditance. Let a be the 38-cm length o the peron arm. The perpendicular component o the moment arm i r ain θ. A careul anali o the geometr reveal that θ The tretch ditance i the dierence between the 44-cm tretched length and the 3-cm untretched length o the elatic cord. Solution. Solve equation -3 or. Solve equation 6-4 or k τ 8 N m 57 N r.38 m in 57 N k 44 N/m 4.4 kn/m.44.3 m Inight The 57 N o orce the elatic cord eert on the hand i equivalent to 3 lb. A good workout! Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher.

13 Chapter Rotational Dnamic and Static Equilibrium 96. Picture the Problem Thi i a unit converion problem. Strateg The ormula i a verion o equation -9 but with non-metric unit. The contant C impl convert the unit rom rev/min to rad/ and rom t lb/ to horepower. Ue equation -9 to ind the value o C, then ue the given ormula and the known value o C to ind the engine torque in t lb. Solution. (a) Ue equation -9 to ind C P( hp) τ( t lb) ω( rev/min ). (b) Ue the given ormula to ind τ π rad min hp rev 6 55 t lb/ Torque RPM Torque RPM HP hp 55 t lb rpm/hp C C 55 t lb rev/min/hp 55 t lb rev/min/hp ( 55 t lb rpm/hp)( 3 hp) ( 65 rpm) C HP Torque 59 t lb RPM Inight The contant C can be conidered unitle becaue it i baicall power divided b power, but we retained the unit to indicate how to accomplih the converion. We bent the rule or igniicant igure or C a bit in tep to avoid rounding error. 97. Picture the Problem The torque about the hip joint rom the weight o the tail balance the torque rom the weight o the upper toro o the dinoaur. Strateg Write Newton Second Law or torque about the hip joint and olve or the ma o the tail. Let m U be the ma o the upper toro, let mt be the ma o the tail, and let M mu + mt be the total ma o the T. re. Solution. Set τ and rmg T T rm U Ug ubtitute or rm r M m m U T T U T. Now olve or m T m rm (.4 m)( 54 kg) 3 U T rt + r U m kg. kg Inight Such a maive tail would not be necear i the creature tood upright like human do, placing it ma over the point o upport o it eet. Other creature like monke have large tail or better balance when doing acrobatic in the tree top. 98. Picture the Problem The weight o the pen, the thumb orce, and the inde inger orce act on the pen in the manner indicated b the igure. Strateg Ue Newton Second Law or torque and Newton Second Law or orce in the vertical direction to determine the magnitude o the orce. The orce and torque are each in equilibrium. The weight o the pen will act at the center o ma, 7. cm rom the end o the pen. Solution. (a) The orce rom the inde inger will be greater in magnitude than the orce rom the thumb, becaue the inger orce ha to counteract both the thumb orce and the pen weight.. (b) Set τ and olve or 3. Set and olve or t τ r r mg cm r mg ( 7. cm)(.8 kg)( 9.8 m/ ) cm r 3.5 cm t.55 N t mg mg.55 N.8 kg 9.8 m/.7 N Inight The larget orce,.55 N, amount to onl. oz. The 8 g pen weigh about. oz. mg 7. cm 3.5 cm t Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 3

14 Chapter Rotational Dnamic and Static Equilibrium 99. Picture the Problem The peron tand on the 6.-N ladder in the manner depicted b the igure at right. Strateg The problem can be olved b etting the vector um o the orce and the torque equal to zero. The onl dierence between thi problem and Active Eample - 3 are the addition o a vector m g at the center o ma o the ladder, and the modiication o the ditance b. The horizontal ditance between the bae o the ladder and the vector m g i c ( 4. m) ( 8 m).6 m. 3. τ a3 bmg cmg bmg + cm g c( mg+ m g) Solution. (a) Set τ and olve or 3. Let b c becaue the peron i halwa up the ladder. Determine the numerical value o a 3. Set and olve or 3.5 kn 4. Set and olve or.6 m 85 kg 9.8 m/ + 6. N 3.8m 46 N.5 kn 3 mg m g mg m g + 85 kg 9.8 m/ + 6. N 894 N.89 kn a b 5. (b) Set τ and olve or 3. Let 3 3 ( 4. m ) (.94 m)( 85 kg)( 9.8 m/ ) + (.6 m)( 6. N) b 3.8 m.94 m bmg + cm g a 6 N. kn 6. Let 3 a in tep 3 3. kn 7. The orce i unchanged 894 N.89 kn Inight A the peron climb higher on the ladder both 3 and increae. The ladder lean with more orce 3 againt the wall and relie more heavil on the tatic riction orce to keep the bae o the ladder rom liding out. 3.8 m Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 4

15 Chapter Rotational Dnamic and Static Equilibrium. Picture the Problem The ign i upported b a rope a indicated in the igure at right. Strateg Set the net torque about the bolt equal to zero and olve or the tenion in the rope. The torque due to the rope i poitive and the torque due to the weight i negative. Then write Newton Second Law in the vertical and horizontal direction to ind the vertical and horizontal component o the orce eerted b the bolt on the ign. τ ( L)( T θ ) ( L)( mg) Solution. (a) Set and olve or T. (b) Let horizontal orce um to zero and olve or 3. (c) Let vertical orce um to zero and olve or τ in mg ( 6. kg)( 9.8 m/ ) T 9 N inθ in Tcoθ mg mg Tcoθ coθ inθ tanθ mg+ Tinθ ( 6. kg)( 9.8 m/ ) tan. 6 N mg mg T inθ mg inθ mg inθ 6. kg 9.8 m/ 78.5 N Inight Note that the 9-N tenion in the rope i almot.5 time larger than the 57-N weight o the ign becaue the rope i alo pulling horizontall, and onl the vertical portion i upporting the weight o the ign. It would take an ininite orce to upport the ign with a rope that i horizontal (θ. )!. Picture the Problem The diver o ma m tand at the end o the diving board o negligible ma a hown at right. The pillar are d. m apart, the ma o the diver i 67. kg, and the magnitude o 88 N. Strateg Write Newton Second Law or rotation with the pivot point at the econd pillar and olve or L. Then write Newton Second Law in the vertical direction and olve or. Solution. (a) Set about pillar and olve or L τ. (b) Set and olve or τ d + L d mg d mg+. m 67. kg 9.8 m/ + 88 N L.49 m ( ) mg ( 67. kg)( 9.8 m/ ) + mg mg+ 67. kg 9.8 m/ + 88 N 49 N.49 kn Inight Pillar mut eert a downward orce in order to balance the torque produced b the diver weight. Pillar mut thereore eert a large orce upward to balance the two downward orce and mg. Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 5

16 Chapter Rotational Dnamic and Static Equilibrium. Picture the Problem The diver o ma m 9. kg tand at a ditance rom the let end o the diving board o ma M 85 kg and length L 5. m a hown at right. The pillar are d.5 m apart. Strateg Write Newton Second Law in the vertical direction and Newton Second Law or rotation with the pivot point at the let end o the board. The two equation can then be combined to ind the two unknown and a unction o. Solution. Set and olve or. Set τ and olve or 3. Ue the value o in the equation rom tep to ind + mdiver g Wboard m g+ W diver board τ + ( d) ( ) mdiver g ( L ) mboard g g ( m g+ m gl) ( m + m L) diver board diver board d d 9.8 m/ ( 9. kg) ( 85 kg )( 5. m).5 m ( 589 N/m) 39 N (.589 kn/m).4 kn ˆ 9. kg 9.8 m/ + 85 kg 9.8 m/ 589 N/m + 39 N 589 N/m + 33 N (.589 kn/m) +.33 kn ˆ Inight A the diver move toward the end o the board, increae, become larger in the negative (downward) direction, and become larger in the upward direction, with maimum value o.6 kn and 4.3 kn. 3. Picture the Problem The weight o the peron i ditributed between the heel and the toe in dierent wa becaue o the hape o the hoe a hown in the igure at right. Strateg Write Newton Second Law or torque about point A and olve or B. Then write Newton Second Law in the vertical direction to ind the orce A. Note that the orce and A B are upward orce on the oot eerted b the loor. Solution. (a) Set τ and olve or B rw r w B B r w B rb 4. cm 79 N 3.5 cm B 83. N. Now et and olve or A A + B w A w B 79 N 83. N 96 N 3. (b) Repeat tep or the high heel rw w r B B rw 3.53 cm B w ( 79 N ) 95.6 N rb.3 cm 4. Repeat tep or the high heel A + B w A w B 79 N 95.6 N 83 N 5. (c) The high heel ha hited more o the woman weight to her toe. Inight Note that even a lat hoe eert more orce on the heel than the toe becaue w i located cloer to the heel. Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 6 w

17 Chapter Rotational Dnamic and Static Equilibrium 4. Picture the Problem The quadricep mucle eert a orce jut below the knee that upport the lower leg in the manner indicated in the igure. Strateg Write Newton Second Law or torque about the knee joint and olve or Q. Note that the moment arm or the quadricep orce i r,q ( cm) in cm and or the weight o the leg it i ( 35 cm)co39 7 cm. r,w Solution Set τ and olve or Q r r mg,q Q,W r 7 cm 3.4 kg 9.8 m/ 55 N.6 kn,w Q mg r,q 5.8 cm Inight Note that in order to produce the ame torque a the leg weight, but with a much maller moment arm, the mucle mut eert a orce that i 4.7 time greater than the weight o the leg. 5. Picture the Problem The deltoid mucle eert a orce jut below the houlder that upport the weight o the upper and lower arm, hand, and top ign in the manner indicated b the diagram at right. Strateg Write Newton Second Law or torque about the houlder joint and olve or d. Note that the moment arm or the deltoid orce i r,d ( 4 cm)in8 4.3 cm, and the moment arm or the weight are jut thoe component that are labeled in the diagram. Then write Newton Second Law in the horizontal and vertical direction to ind the orce and. Solution. (a) The magnitude o d i greater than the magnitude o becaue although mut equal the magnitude o the horizontal component o d (becaue the are the onl two horizontal orce and the arm i in equilibrium), d alo ha a vertical component.. (b) Set τ and olve or d r,d d rw u u rw l l rh ( Wh + W) rw u u + rw l l + rh ( Wh + W) d r 3. (c) Set and olve or 4. (d) Set and olve or d,d ( 8 cm)( 8 N) + ( 4 cm)( N) + ( 65 cm)( cm) 38 N.38 kn co8 d u l h d d 4.3 cm co8.38 kn co8.36 kn + din8 Wu Wl Wh W W + W + W + W in N 38 N in8 8 N.8 kn Inight The negative value o indicate it actuall act in the downward direction on the houlder joint, not upward a indicated in the igure. The rule o ubtraction leave u with jut one igniicant igure or the anwer to part (d). Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 7

18 Chapter Rotational Dnamic and Static Equilibrium 6. Picture the Problem The tricep mucle eert an upward orce on the ulna at a point jut behind the elbow joint a indicated in the igure at right. Strateg Write Newton Second Law or torque about the elbow joint and olve or T. Solution. Set τ and olve or T. Inert the numerical value r r Mg+ r T T T cm r rcmmg r T ( cm)( 89. N) ( cm)( 5.6 N) T 96 N.78 cm Inight The 96-N (6-lb!) orce eerted b the tricep mucle i much greater than the 89.-N (.-lb) orce eerted b the hand becaue the moment arm o the tricep orce i much maller than that o the hand. 7. Picture the Problem The book are arranged in a tack a depicted at right, with book on the bottom and book 4 at the top o the tack. Strateg It i helpul to approach thi problem rom the top down. The center o ma o each et o book mut be above or to the let o the point o upport, otherwie there will be a net torque on the tem and it will tip. ind the poition o the center o ma or ucceive tack o book to determine d. Meaure the poition o the book rom the right edge o book (right hand dahed line in the igure). Solution. (a) The center o ma o book 4 L need to be above the right end o book 3. d 3. The reult o tep mean that the center o ma o book 3 i located at L + L Lrom the right edge o book. 3. The center o ma o book 4 and 3 need to be above the right end o book + m( L) m L 3 d Xcm,43 L m 4 4. The reult o tep 3 mean that the center o ma o book i located at 3L 4+ L 5L The center o ma o book 4, 3, and need to be above the right end o book + + m L m L m 5L 4 d X cm,43 L 3m 6. The reult o tep 3 mean that the center o ma o book i located at L + L 7L. 7. The center o ma o all our book need to be above the right edge o the table m L m L m 5L 4 m 7L 5 d Xcm,43 L 4m 4 8. (b) I the ma o each book i increaed b the ame amount, the anwer to part (a) will ta the ame becaue it onl depend upon the aumption that each book ha the ame ma, irregardle o the value o that ma. L L L L 5 Inight I ou eamine the overhang o each book ou ind an intereting erie d L. The erie give ou a hint about how to predict the overhang o even larger tack o book! Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 8

19 Chapter Rotational Dnamic and Static Equilibrium 8. Picture the Problem The Earth pin on it ai with a nearl contant angular peed. Strateg Becaue the melting o the polar ice cap reditribute the Earth ma a little bit but doe not eert an eternal torque on the planet, the angular momentum o the Earth would remain contant. Combine equation -5, -, and -5 to ind the new rotation period or the Earth. Solution. (a) With conervation o angular momentum, an increae in the moment o inertia lead to a decreae in the peed o rotation. The length o a da would thereore increae.. (b) Set Li L and ubtitute L Iω Iiωi Iω 3. Now let ω π T and olve or T π π I I Ii I T Ti Ti Ti T Ii Ii 4. ind T T T E E Δ T T T T i I.33M R Δ i i 86,4 6 Ii.33MERE Inight The longer da would be noticeable over time, a 6 i equivalent to 4.35 min. The longer da would caue grie or time-enitive atronomical obervation and would mean that geonchronou atellite would be in the wrong orbit and would drit lowl acro the k (ee Active Eample -). 9. Picture the Problem The orce i applied to the ai o the wheel in order to lit it over the tep a hown in the igure at right. Strateg In order to ind the minimum orce that will lit the wheel over the tep, we mut balance the torque. The torque about the corner o the tep that i produced b mut balance the torque produced b the downward orce o gravit acting at the ale. The moment arm or the orce i r, R and the moment arm or 4 ( R) R 4 5 the weight i r,w Rcoθ, where coθ. R 6 Solution Set τ and olve or min τ r,w Mg r, min r Mg ( R 5 6 ) Mg,W min r, R 4 5 Mg Inight Le orce i required i the tep i maller. or intance, a tep height o R would onl require a orce o min Mg.. Picture the Problem The o-o hang in equilibrium under the inluence o the two orce T and T a indicated in the diagram at right. Strateg Write Newton Second Law or torque about the ai o the o-o, and then Newton Second Law in the vertical direction or the o-o and or the hanging ma to obtain epreion or, T, and m. The problem tate that R 5.6 r. T Solution. Set τ and olve or T r T r T r R T T T r r 5.6r T r T 5.6 T Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 9

20 Chapter Rotational Dnamic and Static Equilibrium. Set or the o-o and olve or T 3. Ue the epreion rom tep to ind T T T Mg ( 5.6) T T Mg (. kg)( 9.8 m/ ) Mg T. N 5.6 / 5.6 T ( ) (. kg)( 9.8 m/ ) T Mg Mg N 4. Set or the hanging ma and olve or m T mg T Mg M. kg. kg. g m g g Inight I the hanging ma were not there, the weight o the o-o would create a torque with moment arm r relative to the point where T contact the ai, and the o-o would rotate counterclockwie and decend the tring.. Picture the Problem The variou orce are applied to the rod, which i in equilibrium, a hown in the igure at right. Strateg Let L the rod length and write Newton Second Law or torque about the bottom o the rod in order to determine the wire tenion T. Then write Newton Second Law in the horizontal and vertical direction to determine the normal orce N and the tatic riction orce. Then determine the maimum orce that can be applied to the rod without cauing it to lip. Solution. (a) Set olve or T. Set, ubtitute the epreion or T rom tep, and olve or N 3. Set and ubtitute or μ N μ Mg+ τ and τ LT ( co 45 ) ( L ) T co45 4. Now olve or ( μ ) N Mg Tin 45 N Mg + T in 45 Mg + Mg + T co 45 μn + Tco 45 μ μ + ( Mg ) μ Mg + μ Mg μ Mg μ Mg μ M g μ + μ μ ( ) ( ) Inight The maimum orce increae with μ until it become ininite when μ. I the coeicient o tatic riction i one or larger, it i impoible to pull the bottom o the rod out while appling the orce at the midpoint; ou would have to pull on a point below the midpoint. Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher.

21 Chapter Rotational Dnamic and Static Equilibrium. Picture the Problem The variou orce are applied to the rod, which i in equilibrium, a hown in the igure at right. Strateg Let L the rod length and write Newton Second Law or torque about the bottom o the rod in order to determine the wire tenion T. Then write Newton Second Law in the horizontal and vertical direction to determine the normal orce N and the tatic riction orce. Then how the maimum orce can be ininitel large and the rod will till not lip. Solution. (a) Ue the epreion rom problem 96 to ind the maimum. (b) Set τ and olve or T μ Mg ( 7 ).3 kg 9.8 m/ μ l 7 7 LT ( 8 ) 3. Set, ubtitute the epreion or T rom tep, and olve or N 4. Set and ubtitute or μ N μ Mg+ 5. Now olve or 7 ( μ ) τ co 45 L T co 45 8 N Mg Tin N 7 7 N Mg + T in 45 Mg + Mg T co 45 μn + Tco 45 7 μ ( Mg ) μmg ( μ ) 8 + μmg μ Mg μ Mg 8μMg μ + 7μ 7μ 7 8( ) 8( ) 6. Now i we inert μ 7into the above epreion, the denominator become and the orce become ininite. Thu the bottom o the rod will not lip under thee condition, no matter how hard ou pull! Inight On the other hand, i the urace were rictionle ( μ ) the rod would lip with the mallet orce applied anwhere along the length o the rod. Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher.

22 Chapter Rotational Dnamic and Static Equilibrium 3. Picture the Problem The clinder rotate and all downward along the length o the tring. Strateg Write Newton Second Law or torque about the center o the clinder, then Newton Second Law in the vertical direction or the clinder in order to ind it linear acceleration. rom Table - the moment o inertia or a clinder rotated about it ai i I mr. Let upward be the poitive direction. Solution. Set τ Iα and olve or T rt Iα ( mr )( a r) T ma mg T. Let ma and olve or a T mg ma ma mg ma a+ a g a g 3 Inight Two idea can help eplain the lowing o the clinder acceleration () the tring eert an upward orce on the clinder, reducing the net orce that i accelerating it downward; and () the rotation o the clinder tore ome o the gravitational potential energ in the orm o rotational a oppoed to tranlational kinetic energ. 4. Picture the Problem The phere rotate and all downward along the length o the tring. Strateg Write Newton Second Law or torque about the center o the phere, then Newton Second Law in the vertical direction or the phere in order to ind it linear acceleration. rom Table - the moment o inertia or a phere rotated about it ai i I mr. Let upward be the poitive direction. 5 Solution. Set τ Iα and olve or T rt Iα ( mr 5 )( a r) T ma 5 mg T. Let ma and olve or a 5 T mg ma ma mg ma a+ a g a g Inight Two idea can help eplain the lowing o the phere acceleration () the tring eert an upward orce on the phere, reducing the net orce that i accelerating it downward; and () the rotation o the phere tore ome o the gravitational potential energ in the orm o rotational a oppoed to tranlational kinetic energ. Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher.

23 Chapter Rotational Dnamic and Static Equilibrium 5. Picture the Problem You pull traight downward on a rope that pae over a dik-haped pulle and then upport a weight on the other ide. The orce o our pull rotate the pulle and accelerate the ma upward. Strateg Write Newton Second Law or the hanging ma and Newton Second Law or torque about the ai o the pulle. Let be the tenion on the right ide o the pulle and T be the tenion on the let ide. Let m be the ma o T the pulle, r be the radiu o the pulle, and M be the hanging ma. The tenion pulling orce. or the dik-haped pulle the moment o inertia Solution. (a) Set ma or the hanging ma. Set τ Iα I mr T Mg Ma T (Table -). T M g+ a or the pulle τ rt rt Iα ( mr )( a r) a ( T T ) m 3. Subtitute the epreion or T rom tep into the one rom tep, and olve or a 4. (b) The tenion on the right ide o the pulle i T on the right ide mut equal the ( ) M ( g+ a) ( ) T T Mg M a a m m m m a M m Mg m Mg ( Mg) Mg m + M m M + m M + m ( + ) ( ) a becaue there can onl be one tenion along the rope. 5. (c) Subtitute or a in the epreion rom tep Mg T Mg+ Ma Mg+ M M + m Mg ( M + m) M M g + M + m M + m M g+ mmg+ M M g M + mmg T M + m M + m 6. (d) A m, a M g and T. Thee are the epected reult or a male, rictionle pulle. A M Mg m, a and T + + Mg Mg. Thee are the epected reult or a pulle that i too M + M + maive to rotate, o that the hanging ma i in equilibrium at ret. Inight The tenion in the rope on the let ide accelerate the hanging ma, but the tenion on the right ide both impart angular acceleration to the pulle and accelerate the hanging ma. Thereore, the right hand rope ha the greater tenion T. 6. Picture the Problem The brick are tacked in the manner indicated b the igure at right. Strateg Concentrate on the brick arthet to the right. The um o the torque about the pivot point at the right edge o the bottom brick mut be zero. There are two torque to conider, one caued b hal the weight o the top brick acting on the upper-let corner, and one caued b the weight o the brick itel acting on the center o ma. An eamination o the diagram reveal that a L and b L. Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 3

24 Chapter Rotational Dnamic and Static Equilibrium a( mg) b( mg ) ( ) ( ) Solution Set τ and olve or L mg bmg L mg L L L 3 L Inight The anwer i independent o the ma o the brick. It onl aume that the brick all have the ame ma and are placed mmetricall o that the weight o the top brick i evenl ditributed between the two middle brick. 7. Picture the Problem A tooth i both moved and rotated b the application o two orce. The graph at right how the value o the two orce necear to produce a given torque, where the torque i meaured about the center o the tooth. Strateg A counterclockwie torque i deired to correct the clockwie rotation o the tooth. Thi mean that the orce mut be larger than. Solution Requiring that > mean that graph I correpond to and graph II correpond to. Inight The orce a drawn do not have an component, but i the did, the magnitude o the two component would need to cancel in order to avoid hiting the tooth in the direction. 8. Picture the Problem A tooth i both moved and rotated b the application o two orce. The graph at right how the value o the two orce necear to produce a given torque, where the torque i meaured about the center o the tooth. Strateg Inpect the graph o line II to determine the value o the torque that correpond to one o the orce being equal to zero. Solution Line II, correponding to orce, croe the zero orce mark at a torque o.3 N m. Inight Although thi arrangement put le tre on the tooth, the torque i inuicient to rotate the tooth properl. We could alo ue equation to ind the torque. Let +.8 N.8 N i. Then the torque τ + D d mm.8 N.3 N m. on the tooth i 9. Picture the Problem A tooth i both moved and rotated b the application o the two orce indicated in the igure at right. Strateg Set the torque about the center o the tooth equal to zero and the um o the orce equal to.8 N in order to determine the magnitude o the orce. Solution. Set τ and τ d + ( D d) ubtitute D d d total total. Now rearrange and olve or ( D d) total d + ( D d) ( D d) ( ) D mm.8 N.5 N 4.5 mm total 3. Solve or total.8 N.5 N.3 N Inight Although thi arrangement put the correct orce on the tooth, there i no torque to rotate the tooth properl. 3 Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 4

25 Chapter Rotational Dnamic and Static Equilibrium. Picture the Problem A tooth i both moved and rotated b the application o the two orce indicated in the igure at right. Strateg Set the torque about the center o the tooth equal to.99 N m and the um o the orce equal to.8 N in order to determine the magnitude o the orce. Solution. Set τ and τ d + ( D d) τ total ubtitute D d τ d total total total. Now rearrange and olve or ( D d) total τ total d + ( D d) ( D d) τ 3. Solve or total total total.45.3 m.8 N.99 N m D.45 m.7 N.8 N.7 N 3.5 N Inight Although thi arrangement put the correct orce on the tooth, there i no torque to rotate the tooth properl.. Picture the Problem The cart lide along a rictionle track becaue o a contant orce eerted b a tring that i paed over a pulle. A in Eample -7, the cart ha a ma o.3 kg, the pulling orce i. N, and the pulle radiu i. m. However, the pulle ma i doubled to.6 kg. Strateg Appl Newton Second Law independentl to the pulle and to the cart and olve or T. The pulle i a dik with moment o inertia I mr (Table -). Solution. (a) The value o T will decreae when the ma o the pulle i doubled becaue a larger net torque will be required to rotate the pulle, orcing T to decreae i T remain the ame.. (b) Set ma T T Ma a or the cart and olve or a M 3. Set τ Iα and olve or T rt rt Iα ( mr )( a r) T T ma 4. Now ubtitute or a uing the epreion rom tep ( ) T T m T M T + m M T T. N T.87 N +.6 kg.3 kg + mm Inight A predicted, the tenion T decreaed rom.97 N to.87 N when the ma o the pulle wa doubled. I the ma o the pulle were ininitel large the tenion would be zero and o would the acceleration o the tem. T Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 5

26 Chapter Rotational Dnamic and Static Equilibrium. Picture the Problem The cart lide along a rictionle track becaue o a contant orce eerted b a tring that i paed over a pulle. A in Eample -7, the pulle ha a ma o.8 kg, the pulling orce i. N, and the pulle radiu i. m. However, the cart ma i doubled to.6 kg. Strateg Appl Newton Second Law independentl to the pulle and to the cart and olve or T. The pulle i a dik with moment o inertia I mr (Table -). Solution. (a) The value o T will increae when the ma o the cart i doubled becaue a larger net orce will be required to accelerate the cart, orcing to increae i T remain the ame. T T. (b) Set ma or the cart and olve or a T Ma a M 3. Set τ Iα and olve or rt rt Iα mr a r T T ma 4. Now ubtitute or a uing the epreion rom tep T ( ) T T m T M T + m M T T. N T. N +.8 kg.6 kg + mm Inight A predicted, the tenion T increaed rom.97 N to. N when the ma o the cart wa doubled. I the ma were ininitel large, the tenion T would be. N, and the acceleration would be zero becaue there would be no net torque on the pulle (and the cart i jut too maive to accelerate). 3. Picture the Problem The child run tangentiall to the merr-go-round and hop on. A in Active Eample -5, the child ha a ma o 34. kg, the merr-go-round ha a moment o inertia o 5 kg m and a radiu o.3 m, but the child initial peed i dierent than.8 m/. Strateg Ue equation -5 together with equation - to conerve angular momentum beore and ater the child jump on the merr-goround. Solve the reulting epreion or the initial peed v o the child. Solution Set Li L and olve or the initial peed v + rmv Iω ( I + mr ) ω ( I mr ) ω 5 + ( 34.)(.3 ) kg m (.45 rad/) + v 3.75 m/ rm (.3 m)( 34. kg) Inight A we would epect, the child need to run ater in order to get the merr-go-round pinning ater. The 34% increae in linear peed o the child reult in a 34% increae in the angular peed o the merr-go-round becaue the initial and inal angular momentum o the tem depend linearl upon the peed o the child. Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 6

27 Chapter Rotational Dnamic and Static Equilibrium 4. Picture the Problem The child run at an angle to the merr-go-round and hop on. A in Active Eample -5 the child ha a ma o 34. kg, the merr-go-round ha a moment o inertia o 5 kg m and a radiu o.3 m, and the child initial peed i.8 m/. Strateg Ue equation -5 together with equation - to conerve angular momentum beore and ater the child jump on the merr-goround. The moment arm o the child angular momentum i r rinθ. Solve the reulting epreion or the approach angle θ o the child. Solution. Set Li L and olve or inθ + rmvinθ Iω ( I + mr ) ω inθ ( I + mr ) rmv ω. Solve or θ, keeping in mind that the calculator will return an angle equal to 8 θ θ 8 in (.3 m)( 34. kg)(.8 m/) kg m.7 rad/ Inight I the child approache at an angle θ that i greater than 9, hi initial angular momentum i maller and the merr-go-round end up pinning at a lower rate. I θ 8, the initial angular momentum would be zero and the merr-go-round would not rotate at all; in thi cae the child approache the merr-go-round along the radial direction. Copright Pearon Education, Inc. All right reerved. Thi material i protected under all copright law a the currentl eit. No portion o thi material ma be reproduced, in an orm or b an mean, without permiion in writing rom the publiher. 7