# Linear Momentum - Classical Mechanics - Solved Exam, Exams for Classical Mechanics. Shree Ram Swarup College of Engineering & Management

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Physics 218 Final Exam Spring  2010,  Sections  521-­525

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

Name   :  ________Solutions_________

Signature   :______________________________

Student  ID   :______________________________

E-­mail   :______________________________

Section  #   :______________________________

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.

2

Table  to  be  filled  by  the  graders

Part   Score   Part  1  (40)     Part  2  (30)     Part  3  (35)     Part  4  (30)     Part  5  (30)     Part  6  (35)     Bonus  (5)     Exam  Total

3

Part  1:  (30p)  General  questions   Problem  1.1: Linear  Momentum:  An  object  is  moving  with  constant  velocity.

Question  1.1.1: (3p)  Consider  two  inertial  coordinate  systems  that  are  at  rest  with  respect   to   each   other.   Would   both   coordinate   systems   measure   the   same   linear   momentum   of   the   object  ?

Yes,  the  linear  momentum  is  the  same  for  all  coordinate  systems  that  do  not  move  wrt  each  other.

Question  1.1.2: (3p)  In  what  situations  a  system  of  particles  conserves  its  linear  momentum?

When  the  sum  of  external  forces  to  the  system  is  zero.

Problem  1.2: Moment   of   Inertia:   A   weird-­‐shaped   solid   object   has   mass  M   and   moment   of   inertia  ICM  around  its  center  of  mass.   Question  1.2.1: (3p)  The  end  of  a  massless  rod  of  length  L  is  welded  to  the  center  of  mass  of  the   object.  What   is   the  moment  of   inertia  of   the  rod+object  system  around   the   free  end  of   the   rod?

From  the  parallel  axis  theorem,  Ifree-­end-­of-­rod=ICM+ML2

Question  1.2.2: (3p)  Objects  S   and  H   shown  below  have   the  same  shape,   the  same  mass,  and   the   same   position   of   their   center   of   masses.   Object   S   is   solid   and   its   mass   is   distributed   homogenously  throughout  its  volume.  Object  H  is  hollow  and  its  mass  is  distributed  only  on   the   surface   shell.  Which   object  will   have   the   larger  moment   of   inertia  with   respect   to   the   center  of  mass?,  and  why  ?

Since  the  definition  of  moment  of  inertia  involves  summing  of  the   squares  of  distances,  the  object  that  has  the  masses  farther  away   will  be  have  the  larger  moment  of  inertia.  In  this  case  H.

Problem  1.3: Angular   Moment:   an   object   is   spinning   around  its  center  of  mass.   Question  1.3.1: (3p)   In  general,  does  the  angular  momentum  of  a  body  or  system  of  particles   depend  on  the  inertial  coordinate  system  in  which  it  is  evaluated?

Yes,  as  the  definition  of  angular  momentum  depends  on  distances  from  the  coordinate  system.

Question  1.3.2: (3p)   In   general,   when   is   the   angular   momentum   of   a   system   of   particles   conserved?

When  the  sum  of  all  external  torques  is  zero.

Question  1.3.3: (3p)  Consider  two  inertial  coordinate  systems  that  are  at  rest  with  respect   to   each  other.  Does  the  angular  momentum  of  an  object  spinning  around  its  center  of  mass  with   no   translational   movement   depend   on   the   inertial   coordinate   system   in   which   it   is   evaluated?

No.  In  this  case  the  angular  momentum  can  be  considered  intrinsic.

S   cm

H   cm

4

Problem  1.4: Torques:  A  motor  is  attached  to  end  of  an  axle  and  provides  a  constant  torque  of   τ  along  the  axle.    The  other  end  of  the  axle  is  attached  to  the  center  of  a  solid  disk  of  mass  M   and  radius  R  initially  at  rest.   Question  1.4.1: (3p)  Find  the  magnitude  of  the  angular  acceleration  of  the  disk  around  the  axis.   Does  it  depend  on  the  time?

Simply

α = τ I

= 2τ MR2

Question  1.4.2: (3p)   Is   the   vector   of   the   angular   acceleration   parallel   or   transverse   to   the   torque  vector?

The  vector  of  angular  acceleration  is  parallel  to  the  torque  vector.

Question  1.4.3: (3p)   Is   the   velocity   vector   of   a   particle   in   the   rim   of   the   disk   parallel   or   transverse  to  the  angular  velocity?

The  angular  velocity  is  parallel  to  the  angular  acceleration  which  is  parallel  to  the  torque.  The  velocity   vector  of  a  particle  in  the  rim    is  then  transverse  to  the  angular  velocity.

Problem  1.5: Gravitation:    Objects  A  and  B  attract  each  other  under  the  force  of  gravity.   Question  1.5.1: (3p)  If   the  objects  A  and  B  are  spherically  symmetrical,  what   is   the  force  that   object  A  exerts  on  B  and  what  is  the  force  that  object  B  exerts  on  A?

FA  on  B=-­FB  on  A  with    |FBonA|=Gmamb/R2

Question  1.5.2: (3p)  What  is  the  gravitational  potential  energy  contained  in  the  system  of  A+B?

Simply  U=-­Gmamb/R+arbitrary  constant

Question  1.5.3: (4p)  If  objects  A  and  B  orbit  around  each  other  following  each  an  elliptical  path.   What  geometrical  point  always  lies  at  the  focus  of  both  ellipses?

The  center  of  mass  of  the  system  A+B.

5

Part  2: (30p)  Collisions   Problem  2.1: 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   spinning  with   an   angular  velocity  ω 0  around  axis  A.  A  ball  of  mass  m  moving  with  velocity  v0  towards  the  paddle   at   a  distance  R  with   respect   to   the  axis  of   the  paddle.    The  ball   collisions  elastically  with   the   paddle     when   the   paddle   is   exactly   in   the   y   direction   as   shown   in   the   figure,   and   after   the   collision  the  ball  moves  along  the  x  axis  (positive  or  negative).

Question  2.1.1: (6p)   Identify   all   the   external   forces   acting   on   the   ball+paddle   system.   Be   careful,   if   you   get   this   one   wrong   you   will   likely  get  the  rest  wrong.

Besides  gravity  and  normal  reactions  forces,  the   force  the  axis  exerts  on  the  spinwheel  at  the   moment  of  the  collision.

Question  2.1.2: (5p)   Is   the   linear  momentum  of   the  ball+paddle   system   the   same  before   and   after  the  collision?,  why  ?

No.  The  sum  of  external  forces  in  the  plane  is  not  zero  as  the  axis  exerts  force.

Question  2.1.3: (5p)  Is  any  component  of  the  angular  momentum  of  the  ball+paddle  system  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.

Question  2.1.4: (5p)   Is   the   energy   of   the   ball+paddle   system   the   same   before   and   after   the   collision?,  why  ?

Yes.  The  collision  is  elastic,  and  no  other  external  forces  perform  work  on  the  system.

Question  2.1.5: (9p)  Write  the  system  of  equations  that  would  in  principle  allow  you  to  solve   both  the  final  velocity  of  the  ball  and  the  final  angular  velocity  of  the  paddle.  Do  not  solve  it.

From  conservation  of  angular  momentum:

IAω 0 +mv0R = −IAω f +mv f R

From  conservation  of  energy:

1 2 mv0

2 + 1 2 IAω 0

2 = 1 2 mv f

2 + 1 2 IAω f

2

A

R

V0

y

x

m

ω0

6

Part  3: (35p)  Equilibrium   Problem  3.1: A  disk  of  radius  R  and  mass  m   lies  in  a  plane  with  friction  and  inclination  α ,  and   has  a  nail  attached  to  a  string  as  shown  in  the  figure  below.  The  other  end  of  the  string  connects   (via   a   pulley)   to   one   end   of   a   massless   bridge   of   length   L.   The   other   end   of   the   bridge   is   supported  by  a  column  and  a  mass  2m  is  located  at  a  distance  x  from  the  column.

Question  3.1.1: (10p)  Forgetting  about   the  bridge   setup,   find   the   tension  T   in   the   string   such   that   the   disk   does   not  move.

From  the  sum  of  forces  and  torques:

τz,cm = TR fR = 0∑ ⇒ T = f Fy = N mgcos(α)∑ = 0⇒ N = mgcos(α) Fx = mgsin(α)∑ −T f = mgsin(α) − 2T = 0

T = mgsin(α) 2

Question  3.1.2: (8p)   Forgetting   about   the   bridge   setup,   find   the   minimum   coefficient   µS   necessary  for  the  disk  to  be  in  equilibrium.

We  know  that  f  ≤µsN.  On  the  other  hand  from  3.1.1  we  got  f=T=mgsin(α)/2,  so

mgsin(α)/2≤µsN=µsmgcos(α)    tan(α)/2≤µs,  so  the  coefficient  must  be  greater  than  tan(α)/2

Question  3.1.3: (9p)  Considering   the  bridge  setup,   find   the  distance  x   necessary   to  achieve   equilibrium.

From  sum  of  forces  and  torques  over  the  column:

N1 +T − 2mg = 0⇒ N1 = mg 2 − sin(α) 2

⎛

⎝ ⎜

⎞

⎠ ⎟

−2mgx +TL = 0⇒ x = TL 2mg

= sin(α) 4

L

Question  3.1.4: (8p)  Is  that  equilibrium  stable?  What  would  happen  if  the  mass  in  the  bridge  is   moved  a  little  bit  off  the  position  of  equilibrium?

If  x  >  sin(α)*L/4  then  the  T  would  be  higher  than  needed  and  the  ball  rolls  up.

If  x  <  sin(α)*L/4  then  the  T  would  be  smaller  than  needed  and  the  ball  rolls  down.

Therefore  the  equilibrium  is  not  stable.

R  m

α

2m

L

x

X

Y

2m

L

x

ω

N1   T

7

Part  4:  (30p)  Linear  Kinematics   Problem  4.1: A   piston   follows   a   simple   harmonic   motion   with   angular   frequency   ω   and   amplitude  A.    At  t=0,  when  the  piston  is  at  its  minimum,  a  mass  m  is  quickly  put  on  top  of  it.  The   piston  does  not  changes  its  way  of  moving  after  the  mass  is  added.

Question  4.1.1: (5p)  Write  the  equation  of  position  of  the  piston  as   a  function  of  time  for  the  given  axis.  Make  sure  you  find  all  the   constants  from  the  initial  conditions.

y(t)=Acos(ωt  +  φ).  Because  at  t=0  the  piston  is  at  –A  we  get

y(t=0)=-­A=Acos(φ)  -­1=  cos(φ)φ=π

Question  4.1.2: (5p)   Write   the   velocity   and   acceleration   of   the   piston  as  a  function  of  time.

v(t)=dy(t)/dt=-­ωAsin(ωt  + π)

a(t)=dv(t)/dt=-­ω2Acos(ωt  + π)

Question  4.1.3: (3p)  Do  the  free-­‐body  diagram  of  the  mass  m.

Question  4.1.4: (5p)  Find  the  normal  force  as  a  function  of  time.

N-­mg=-­m ω2Acos(ωt  + π) N(t)=m(g-­ ω2Acos(ωt  + π))

Question  4.1.5: (6p)  Find  the  time  at  which  the  mass  leaves  the  piston.

The  mass  leaves  the  piston  when  the  normal  is  zero.  Let’s  call  t*  the  time  in  which  N(t*)=0,  then

N(t*) = m g −ω 2Acos(ωt *+π )( ) = 0⇒ g ω 2A

= cos(ωt *+π )⇒ t* = Arc cos( g

ω 2A ) −π

ω

Question  4.1.6: (6p)  Find  the  velocity  at  the  moment  the  mass  leaves  the  piston.

Need  to  find  the  velocity  at  time  t*,

v(t*) = −ωAsin(ωt *+π ) = −ωAsin(Arc cos( g ω 2A

))

g  A

Y

A

0

t=0   m

N

mg

8

Part  5: (30p)  Angular  Kinematics   Problem  5.1: A  disk  is  attached  to  a  roof  and  free  to  rotate  around  it  center  axis  as  shown  below.     The   disk   has   radius  R=0.4m   and   non-­‐homogenous  mass   distribution,   but   its   center   of   mass   coincides  with  the  axis  position.    A  rope  is  winded  around  the  disk.  The  other  end  of  the  rope  is   attached  to  a  mass  m2=12.5  Kg  (figure  A)  and  is  seen  that  the  mass  m2  accelerates  downwards   with  acceleration  am2=4.9m/s2  .  A  number  with  proper  units  is  expected  for  all  these  questions.

Question  5.1.1: (7p)   Find   the   moment   of   inertia   of   the   disk   around  its  axis.

Fy = T m2g = −m2∑ am2 ⇒ T = m2(g am2 )

τ = TR = Iα = I am2 R I = TR

2

am2ext ∑ =

m2(g am2 )R 2

am2 = 2 Kg m2

Problem  5.2: (20p)   Two   identical   smaller   objects   are   now   glued   to   the   side   of   the   disk   in   Problem  5.1:,  with   their   center   of  masses   at   a   distance  L=0.25m   from   the   axis   of   the  disk   as   shown  in  figure  below.  The  two  smaller  objects  are  exactly  on  opposite  sides  of  the  axis,  have  a   mass  of  mo=4Kg  and  a  moment  of  inertia  around  their  center  of  masses  of  Io=1  Kg  m2

Question  5.2.1: (10p)   Find   the   moment   of   inertia   of   the   disk+glued  objects  system  about  the  disk’s  center.

The  moment  of  inertia  is  the  sum  of  the  moment  of  inertia  of  the  disk   (calculated  in  5.1.1)  and  the  moment  of  inertia  of  each  smaller  object.   The  moment  of  each  object  is  obtain  just  by  using  the  parallel  axis   theorem:

I = 2 Kg m2 + 2 I0 +m0L 2( ) = 4.5 Kg m2

Question  5.2.2: (8p)  Find  the  acceleration  of  the  mass  m2   in  this  case.

From  5.1.1  we  get:

I = m2(g am2 )R

2

am2 ⇒ am2 I +m2R

2( ) = m2gR2 ⇒ am2 = m2gR

2

I +m2R 2( )

= 3.01 m/s2

Question  5.2.3: (5p)  Find   the  angular  velocity  of   the  disk+glued  objects  around   its  axis  when   the  mass  m2  dropped  a  distance  h=1m.

When  the  mass  m2  descended  one  meter  its  velocity  is

v f 2 = v0

2 + 2aΔyv f = 2aΔy ≅ 2.45 m s

For  that  velocity  the  angular  velocity  of  the  disk+glued  objects  is:

ω f = v f R

= 6.12s−1

Figure  B:

g

m2

R

c.m.

c.m.  L   L

g

m2

R

Figure  A:

Y

9

Part  6: (35p)  Gravitation   Problem  6.1: A  small  planetoid  of  mass  m  is  orbiting  around  a  much  bigger  sun  (of  mass  M)  on   an  elliptical  orbit.    In  its  orbit  the  distance  of  closest  approach  to  the  sun  is  Rp  (perihelion),  and   that  of  farthest  excursion  is  Ra  called  aphelion.  Answer  all  the  following  questions  in  terms  of   m,  M,  Rp,  Ra,  and  the  gravitational  constant  G.   Question  6.1.1: (8p)   Using   conservation   of   angular   momentum,   find   the   ratio   of   speeds   at   apogee  and  perigee.

L = mRava = mRpvp va vP

= RP Ra

Question  6.1.2: (9p)  Using  conservation  of  energy  obtain  the  speed  of  the  planetoid  at  aphelion   and  perihelion.

Let’s  write  the  equation  of  conservation  of  energy  and  then  include  the  result  from  6.1.1

E = 1 2 mvp

2 − GmM RP

= 1 2 mva

2 − GmM Ra

Using va = RP Ra

vP from 6.1.1 we get :

1 2 mvp

2 − GmM RP

= 1 2 m RP

2

Ra 2 vP

2 − GmM Ra

vp 2 −

2GM RP

= RP

2

Ra 2 vP

2 − 2GM Ra

vp 2 1− RP

2

Ra 2

⎛

⎝ ⎜

⎞

⎠ ⎟ = −2GM

1 Ra

− 1 RP

⎛

⎝ ⎜

⎞

⎠ ⎟ = vp

2 Ra 2 − RP

2

Ra 2

⎛

⎝ ⎜

⎞

⎠ ⎟ = −2GM

RP Ra RaRP

⎛

⎝ ⎜

⎞

⎠ ⎟

vp 2 = 2GM Ra RP

RaRP

⎛

⎝ ⎜

⎞

⎠ ⎟

Ra 2

Ra 2 − RP

2 = 2GM Ra + RP( )

Ra RP

vP = 2GM Ra + RP( )

Ra RP

The  result  for  the  aphelion  velocity  is  then  va=RpvP/Ra

Question  6.1.3: (9p)   A   big   interstellar   spaceship   carrier   (to   be   invented)  wants   to   “tow”   the   planetoid  far  away  from  the  sun.    What  is  the  total  work  the  carrier  performs  in  doing  this?

The  work  necessary  to  carry  the  planetoid  out  of  the  sun  equals  the  change  of  energy  of  the  system.  At  the   final  destination  the  potential  energy  is  zero  and  so  is  the  final  kinetic  energy.

Wcarrier = ΔE = E f Ei = −Ei

E = 1 2 mvp

2 − GmM RP

= 1 2 m 2GM

Ra + RP( ) Ra RP

GmM RP

= GmM RP

Ra Ra + RP( )

−1 ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ =

GmM RP

RP

Ra + RP( ) ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = −

GmM Ra + RP( )

W = −Ei = GmM Ra + RP( )

Question  6.1.4:  (9p)   If   instead   of   towing   the   planetoid,   the   spaceship   just   slows   the   orbital   velocity   of   the   planetoid   to   zero  when   it   is   at   the   perihelion.    What   is   the   velocity   of   the   planetoid  when  it  reaches  the  surface  of  the  sun  ?  (the  radius  of  the  sun  is  Rs)

Ei = Energy after the spaceship took all kinetic energy, Es = Energy at the surface of sun

Ei = Es = − GmM RP

= 1 2 mvs

2 − GmM Rs

vs 2 = 2GM 1

Rs

1 RP

⎛

⎝ ⎜

⎞

⎠ ⎟ ⇒ vs =

2GM RP Rs( ) RsRP

RP   Ra