# Discussion Examples Chapter 10: Rotational Kinematics and Energy

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3 Chapers -: Roaional Dynamics and Saic Equilibrium 5. Use equaion 7- o ind he energy loss rae (power): W P. J 7 y 3.6 s/y 3.5 W 3.5 TW Insigh: Your irs insinc migh be o ind he angular speed a hundred years ago assuming a period o 4. hrs ( rad/s) and igure ou he angular speed in ( rad/s), bu as you can see, aemping o subrac hese numbers requires us o ignore he rules or signiican igures. Using he approximaion oulined above allows us o avoid he subracion problem and keep wo signiican igures. The huge energy loss is due primarily o idal ricion, as he ocean ides dissipae he kineic energy o he Earh s roaion ino hea. 7. IP Awood s Machine The wo masses ( m 5. kg and m 3. kg ) in he Awood s machine shown below are released rom res, wih m a a heigh o.75 m above he loor. When m his he ground is speed is.8 m/s. Assuming ha he pulley is a uniorm disk wih a radius o cm, (a) ouline a sraegy ha allows you o ind he mass o he pulley. (b) Implemen he sraegy given in par (a) and deermine he pulley s mass. Picure he Problem: The larger mass alls and he smaller mass rises unil he larger mass his he loor. Sraegy: Use conservaion o mechanical energy, including he roaional energy o he pulley, o deermine he mass o he pulley. Because he rope does no slip on he pulley, here is a direc relaionship v r beween he roaion o he pulley and he linear speed o he rope and masses. Soluion:. (a) Equae he iniial and inal mechanical energies, hen solve or he mass o he pulley.. (b) Se Ei E and le vr p : p U K U K m gh m gh m v m v I i i p m m gh m m v m prp v rp 3. Rearrange he equaion and solve or m : p 4 m v m m gh m m v 4 p m m gh m m v mp v 4. kg 9.8 m/s.75 m 8. kg.8 m/s.8 m/s m. kg p Insigh: By he ime he masses reach.8 m/s,.8 J or % o he 5 J o oal kineic energy is sored in he kineic energy o he pulley, so he pulley plays a minor role in he energy balance o he sysem. Copyrigh Pearson Educaion, Inc. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion o his maerial may be reproduced, in any orm or by any means, wihou permission in wriing rom he publisher. 3

4 Chapers -: Roaional Dynamics and Saic Equilibrium Chaper : Roaional Dynamics and Saic Equilibrium 3. A.6 kg bowling rophy is held a arm s lengh, a disance o.65 m rom he shoulder join. Wha orque does he rophy exer abou he shoulder i he arm is (a) horizonal, or (b) a an angle o.5 below he horizonal? Picure he Problem: The arm exends ou eiher horizonally or a some angle below horizonal, and he weigh o he rophy is exered sraigh downward on he hand. Sraegy: The orque equals he momen arm imes he orce according o equaion -3. In his case he momen arm is he horizonal disance beween he shoulder and he hand, and he orce is he downward weigh o he rophy. Find he horizonal disance in each case and muliply i by he weigh o he rophy o ind he orque. In par (b) he horizonal disance is r r cos.65 m cos m. Soluion:. (a) Muliply he momen arm by he weigh: r mg.65 m.6 kg 9.8 m/s 9.56 N m. (b) Muliply he momen arm by he weigh: r mg.559 m.6 kg 9.8 m/s 8.83 N m Insigh: The orque on he arm is reduced as he arm is lowered. The orque is exacly zero when he arm is verical. 4. IP A wheel on a game show is given an iniial angular speed o. rad/s. I comes o res aer roaing hrough.75 o a urn. (a) Find he average orque exered on he wheel given ha i is a disk o radius.7 m and mass 6.4 kg. (b) I he mass o he wheel is doubled and is radius is halved, will he angle hrough which i roaes beore coming o res increase, decrease, or say he same? Explain. (Assume ha he average orque exered on he wheel is unchanged.) Picure he Problem: The wheel roaes abou is axis, decreasing is angular speed a a consan rae, and comes o res. Sraegy: Use Table - o ind he momen o ineria o a uniorm disk and calculae I. Then use equaion - o ind he angular acceleraion rom he iniial angular speed and he angle hrough which he wheel roaed. Use I and ogeher in equaion -4 o ind he orque exered on he wheel. Soluion:. (a) Use Table - o ind I MR : I MR 6.4 kg.7 m.6 kg m. Solve equaion - or :. rad/s.58 rad/s.75 rev rad rev 3. Apply equaion -4 direcly: I.6 kg m.58 rad/s.5 N m 4. (b) I he mass o he wheel is doubled and is radius is halved, he momen o inerial will be cu in hal (doubled because o he mass, cu o a ourh because o he radius). Thereore he magniude o he angular acceleraion will increase i he ricional orque remains he same, and he angle hrough which he wheel roaes beore coming o res will decrease. Insigh: I he momen o ineria is cu in hal, he angular acceleraion will double o.3 rad/s and he angle hrough which he wheel roaes will be cu in hal o.38 rev. This is because he wheel has less roaional ineria bu he ricional orque remains he same. We ben he rules or signiican igures in sep o avoid rounding error in sep 3. Copyrigh Pearson Educaion, Inc. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion o his maerial may be reproduced, in any orm or by any means, wihou permission in wriing rom he publisher. 4

5 Chapers -: Roaional Dynamics and Saic Equilibrium 6. IP BIO A Person s Cener o Mass To deermine he locaion o her cener o mass, a physics suden lies on a lighweigh plank suppored by wo scales.5 m apar, as indicaed in Figure 6xx3. I he le scale reads 9 N, and he righ scale reads N, ind (a) he suden s mass and (b) he disance rom he suden s head o her cener o mass. Picure he Problem: The person lies on a lighweigh plank ha ress on wo scales as shown in he diagram a righ. Sraegy: Wrie Newon s Second Law in he verical direcion and Newon s Second Law or roaion o obain wo equaions wih wo unknowns, m and x cm. Solve each o ind m and x cm. Using he le side o he plank as he origin, here are wo orques o consider: he posiive orque due o he righ hand scale and he negaive orque due o he person s mass. Soluion:. (a) Wrie Newon s Second Law in he verical direcion o ind m: Fy F F mg F F 9 N m 4 kg g 9.8 m/s. (b) Wrie Newon s Second Law or r F xcmmg roaion and solve or x cm : xf.5 m N x.74 m mg cm 4 kg9.8 m/s Insigh: The equaion in sep does no depend on he axis o roaion ha we choose, bu he equaion in sep does. Neverheless, we ind exacly he same x cm i we choose he oher scale, near her ee, o be he axis o roaion. 49. IP You pull downward wih a orce o 8 N on a rope ha passes over a disk-shaped pulley o mass. kg and radius.75 m. The oher end o he rope is aached o a.67-kg mass. (a) Is he ension in he rope he same on boh sides o he pulley? I no, which side has he larges ension? (b) Find he ension in he rope on boh sides o he pulley. Picure he Problem: You pull sraigh downward on a rope ha passes over a disk-shaped pulley and hen suppors a weigh on he oher side. The orce o your pull roaes he pulley and acceleraes he mass upward. Sraegy: Wrie Newon s Second Law or he hanging mass and Newon s Second Law or orque abou he axis o he pulley, and solve he wo expressions or he ension T a he oher end o he rope. We are given in he problem ha T 5 N. Le m be he mass o he pulley, r be he radius o he pulley, and M be he hanging mass. For he diskshaped pulley he momen o ineria is I mr. Soluion:. (a) No, he ension in he rope on he oher end o he rope acceleraes he hanging mass, bu he ension on your side boh impars angular acceleraion o he pulley and acceleraes he hanging mass. Thereore, he rope on your side o he pulley has he greaer ension.. (b) As saed in he problem, T 8 N or he rope on your side o he pulley. or he pulley: 3. Se F ma or he hanging mass: Fy T Mg Ma 4. Se I 5. Subsiue he expression or a rom sep 4 ino he one rom sep 3, and solve or T (he ension on he oher side o he pulley rom you): 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.67 kg 8 N. kg9.8 m/s.67 kg. kg 8 N Insigh: The ne orce on he hanging mass is hus T Mg N.4 N, enough o accelerae i upward a ar 7 m/s.75 m 3 rad/s. 7 m/s. The angular acceleraion o he pulley is hus Copyrigh Pearson Educaion, Inc. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion o his maerial may be reproduced, in any orm or by any means, wihou permission in wriing rom he publisher. 5

6 Chapers -: Roaional Dynamics and Saic Equilibrium 69. A disk-shaped merry-go-round o radius.63 m and mass 55 kg roaes reely wih an angular speed o.64 rev/s. A 59.4-kg person running angenial o he rim o he merry-go-round a 3.4 m/s jumps ono is rim and holds on. Beore jumping on he merry-go-round, he person was moving in he same direcion as he merry-go-round s rim. Wha is he inal angular speed o he merry-go-round? Picure he Problem: A child runs angenially o a roaing merry-go-round and hops on. Sraegy: Use conservaion o angular momenum because here is no ne orque on he sysem as long as he sysem includes boh he person and he merry-go-round. Find he momens o ineria o he disk-shaped merry-go-round, I M r, and he sysem aer he person hops on I M r mr mgr, where M is he mass o he merry-go-round, m is he mass o he person, and r is he radius o he merry-go-round. Se Li L and solve or he inal angular speed, where he iniial angular speed is: i.64 rev/s rad rev 4.3 rad/s. Soluion:. Se Li L L L L disk person inal and rearrange M r i mv r M r mr. Now solve or : M r mv r Mr mv +mr i i M r + mr M r 55 kg.63 m4.3 rad/s 59.4 kg3.4 m/s 55 kg.63 m 59.4 kg.63 m.84 rad/s Insigh: The merry-go-round has slowed down because he iniial linear speed o he person (3.4 m/s) is less han he iniial linear speed o he rim o he merry-go-round (.6 m/s). 79. A person exers a angenial orce o 36. N on he rim o a disk-shaped merry-go-round o radius.74 m and mass 67 kg. I he merry-go-round sars a res, wha is is angular speed aer he person has roaed i hrough an angle o 3.5 Picure he Problem: The merry-go-round is a uniorm disk ha is given an angular acceleraion abou is cener o mass by he applicaion o an unbalanced orque. Sraegy: The work done by he applied orque impars kineic energy o he merry-go-round. Se he orque imes he angular displacemen equal o he inal kineic energy o he merry-go-round (equaions -7 and -8) and solve or. The momen o ineria o he merry-go-round is aken o be I MR, as indicaed in Table - or a uniorm disk roaing abou is axis. Soluion:. Se W K, applying equaions -7, -3, and -7:. Solve or : rf K K I rf RF I MR i 67 kg.74 m.74 m 36. N 3.5 rev 8.43 rad/s Insigh: This roaion rae corresponds o a linear speed o only.6 m/s or he rim o he merry-go-round. The applied orce did 56. J o work o give he merry-go-round 56. J o roaional kineic energy. Copyrigh Pearson Educaion, Inc. All righs reserved. This maerial is proeced under all copyrigh laws as hey currenly exis. No porion o his maerial may be reproduced, in any orm or by any means, wihou permission in wriing rom he publisher. 6

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