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Physics 218 Exam 3 Fall  2010,  Sections  521-­524  

 

Do  not  fill  out  the  information  below  until  instructed  to  do  so!  

 

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Rules  of  the  exam:  

1. You  have  the  full  class  period  to  complete  the  exam.  2. When  calculating  numerical  values,  be  sure  to  keep  track  of  units.    3. You  may  use  this  exam  or  come  up  front  for  scratch  paper.    4. Be  sure  to  put  a  box  around  your  final  answers  and  clearly  indicate  your  work  to  your  grader.  5. Clearly  erase  any  unwanted  marks.  No  credit  will  be  given  if  we  can’t  figure  out  which  answer  you  are  

choosing,  or  which  answer  you  want  us  to  consider.  6. Partial  credit  can  be  given  only  if  your  work  is  clearly  explained  and  labeled.  7.  All  work  must  be  shown  to  get  credit  for  the  answer  marked.  If  the  answer  marked  does  not  

obviously  follow  from  the  shown  work,  even  if  the  answer  is  correct,  you  will  not  get  credit  for  the  answer.  

Put  your  initials  here  after  reading  the  above  instructions:__________  

 

 

 

 

 

 

 

 

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Table  to  be  filled  by  the  graders  

Part   Score  Part  1  (25)    Part  2  (25)    Part  3  (25)    Part  4  (25)    Bonus  (5)    Exam  Total    

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Part  1: (25p)  Energy    Problem  1.1: Two  cylinders,  each  with  mass  m  and  radius  R,  are  nailed  to  the  wall  and  allowed  to  rotate  freely  but  initially  at  rest.    A  slab  of  homogenous  material  with  mass  M  and  length  L  is  initially  at  rest  and  placed  between  the  two  cylinders  as  shown  in  the  picture  below.  The  slab  then  falls  between  the  cylinders  making  them  rotate  without  slipping.  

Question  1.1.1: (4p)   Identify   the  external   forces   to   the  cylinders+slab  system.  

The   gravity   on   the   slab   and   the  force   that   each   nail   exerts   on   the  cylinders  to  keep  them  in  place.  

 

 

Question  1.1.2: (4p)   Is   the  mechanical   energy   of   the  

cylinders+slab  system  conserved  ?  Explain  your  reasoning.  

The  internal   friction  force  does  not  produce  work  as  the  disks  are  rotating  without  slipping.  Yes,  the  mechanical  energy  is  conserved  as  the  work  of  the  non-­conservative  forces  is  zero.  

Question  1.1.3:  (15p)  At  the  moment  in  which  the  slab  losses  contact  with  the  cylinders,  find  the  magnitude  of  the  angular  velocity  of  the  cylinders.  

                                           

€

Because mechanical energy is conserved,

Ei = E f ⇒ Mg L2

= −Mg L2

+ 2(12Idω f

2 ) +12Mv 2 ⇒ MgL = Idω f

2 +12Mv 2

Since the slab went without slipping we know v =ω f R

⇒ MgL = Idω f2 +

12Mω f

2R2 =ω f2 (Id +

12MR2)⇒ω f =

MgL

Id +12MR2

ω f =MgL

12mR2 +

12MR2

=2MgL

R2(m + M)

 

 

Question  1.1.4: (2p)     Is   the  direction  of   the  angular  velocity  of  each  cylinder    opposite  or   the  same  to  each  other  ?  

Obviously  the  cylinders  rotate  in  opposite  ways.  

 

 

g  

   R    m  

   R    m  

L  

M  

   R    m  

   R    m  

L  

M  

Initially   Sometime  after  

g  

   R    m  

   R    m  

L  

M  

L  

Y  

X  

  4  

 

Part  2: (25p)  Energy  and  Kinematics  Question  2.1.1: A  disk  of  mass  m  and  radius  R  is  spinning  with  angular  velocity  ω .  The  disk  is  spinning   and   slipping   with   respect   to   the   surface   of   the   ground.   Initially   the   disk   is   just  spinning  but  not  moving  in  the  horizontal  direction.  There  is  friction  between  the  disk  and  the  table  with  kinetic  coefficient  µK.    

Question  2.1.2: (5p)   Draw   a  coordinate   system,   the  orientation   in   which   the  disk   rotates,   and   a   free  body  diagram  of  the  disk.  

 

 

Question  2.1.3: (10p)  Find  the  angular  acceleration  of   the  disk.   Is   it  constant  ?,   is   it  parallel  or  antiparallel  to  the  angular  velocity  ?  If  you   write   a   torque   specify   the   point   from   where   the   torque   is  computed.  

€

From the center of the disk the torque is

τ = +Rf = Idα ⇒ α =RfId

=RµKNId

=RµKmgId

=2RµKmgmR2 =

2µKgR

which is clearly constant as the all the components are constant.In the coordinate system depicted above the angular velocity is negative, while the angularacceleration found is positive, so the angular acceleration is antiparallel to the angular velocity. As expected as the object is deaccelerating.

 

Question  2.1.4: (10p)   After   an   N   number   of   revolutions   the   disk   is   found   rotating   without  slipping.  Using  the  work-­‐energy  theorem  find  the  angular  velocity  at  which  the  disk  is  now  rotating.  Assume  the  distance  travelled  in  the  horizontal  direction  is  zero.  

€

The work - energy theorem states that :Ki + Ri +Wnon−cons = K f + Rf

In our case we have Ki = 0,Ri =12Iω 2and Rf =

12Iω f

2 ,K f =12mv f

2 =12mR2ω f

2

After N revolutions the work of the torque is Wnon−cons = −τ2πN = −RµKmg2πNwhere the minus sign comes from the fact that the torque was going against the angle displacement.12Iω 2 − RµKmg2πN =

12Iω f

2 +12mR2ω f

2 ⇒ Iω 2 − 2RµKmg2πN =ω f2 (I +mR2)

⇒ω f =Iω 2 − 4RµKmgπN

(I +mR2)=

mR2ω 2

2 − 4RµKmgπN

(mR2

2 +mR2)=

Rω 2 − 8µKgπN3R

=

 

g  

   R    m  

Ground  

   R  Y  

X  Z  f  

N  w  

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Part  3: (25p)  Angular  Momentum  Problem  3.1: A  tandem  rotor  helicopter  has  two  sets  of  blades  powered  by  the  same  motor  that  rotate   in  opposite  direction  with  angular  velocity  of  magnitude  ωb.  The  moment  of   inertia  of  each  set  of  helicopter  blades  with  respect  to  each  set’s  center  is  Ib  and  the  moment  of  inertia  of  the  helicopter  around  its  center  of  mass  is  Ih,.  The  helicopter  is  initially  hovering  at  high  altitude  without  moving   in   the  horizontal  or  vertical  position  and  without  rotation  of   the  body  of   the  helicopter.  Ignore  any  air  resistance.  

Question  3.1.1: (5p)   Find   the  total   angular  momentum  of  the   helicopter,   and   show  how   you   reached   that  number.  

The     total   angular   momentum   is  

€

L = Ibω b − Ibω b = 0   as   the   body  is  not  rotating  and  the  blades  are  rotating  in  opposite  direction.  

 

Question  3.1.2: (10p)   If   a  malfunction   suddenly   stops   one   set   of   blades   to   a   screeching   halt,  find  the  angular  velocity  the  body  of  the  helicopter  develops  around  its  center  of  mass.  

 

€

The angular momentum has to be conserved as there are no external forces.

Li = 0 = Lf = Ibω b + Ihω h ⇒ω h =−Ibω b

Ih  

 

Question  3.1.3: (10p)  To  stop  the  spinning,  the  helicopter  posses  a  jet  engine  that  can  provide  a  constant  torque  τjet  on  the  body  of  the  helicopter  and  with  respect  to  the  center  of  mass  in  the  direction  of  the  angular  velocity  of  the  remaining  set.  Find  the  time  it  takes  to  bring  the  broken  helicopter  to  the  status  of  no-­‐body  rotation.  

 

€

Basically find the time the torque τ jet reduces the angular velocity from ω h to zero.

The torque τ jet produces a angular acceleration of α =τ jet

Ih

ω f = 0 =ω h +αt⇒ t =−ω h

α=Ibω bIh

Ihτ jet=Ibω b

τ jet

 

 

Ib  

Ih  Ib  

CM  

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Part  4: (25p)  Collisions  Problem  4.1: A   square   paddle   is   set   through   axis  A   to   a   table   and   allowed   to   rotate   free.   The  moment  of   inertia  of  the  paddle  around  axis  A   is  IA.   Initially  the  paddle  is  at  rest.  A  point-­‐like  particle  of  mass  m  is  moving  with  velocity  v0  towards  the  paddle  at  a  distance  R  with  respect  to  the  axis  of  the  paddle  as  shown  in  the  figure.    The  particle  collisions  with  the  paddle  and  sticks  to  the  paddle’s  sticky  surface.    

Question  4.1.1: (5p)   Is   any  component   of   the   angular  momentum   of   the   ball+paddle  system   as   computed   from   the  point  where  the  axis  intercept  the  table,   the   same   before   and   after  the  collision?,  why  ?  

Yes.  Since  no  external  force  can  exert  torque  on  the  z  component  the  angular  

momentum  on  the  z  component  must  be  conserved.  In  addition  the  axis  can  provide  force  in  the  x  and  y  direction,  but  the  torque  of  these  forces  as  computed  from  the  axis  is  also  zero,  so  the  angular  momentum  is  conserved  in  x,y  and  z.  

Question  1.1.1: (10p)   Find   the   angular   velocity   of   the     paddle+particle   system   after   the  collision.    

From  conservation  of  angular  momentum:  

     

€

mv0R = (IA +mR2)ω f ⇒ω f =mv0R

(IA +mR2)                

 

 

Question  1.1.2: (10p)  What  is  the  ratio  of  the  energy  before  the  collision  to  the  energy  after  the  collision?  Is  this  an  elastic  or  an  inelastic  collision  ?  

€

Ei =12mv0

2,

E f =12(IA +mR2)ω f

2 =12(IA +mR2) m2v0

2R2

(IA +mR2)2=12

m2v02R2

(IA +mR2)=12

mv02R2

(IA m + R2)

E f

Ei

=1

( IAmR2

+1)≤1

 

Because  the  final  kinetic  energy  is  less  than  the  initial  kinetic  energy  the  collision  was  inelastic.  

 

A  

R  

V0  

y  

x  

m