Physics 111 Lecture 11 Today’s Agenda

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					                Physics 111: Lecture 11

                     Today’s Agenda

   Review
   Work done by variable force in 3-D
      Newton’s gravitational force
   Conservative forces & potential energy
   Conservation of “total mechanical energy”
      Example: pendulum
   Non-conservative forces
      friction
   General work/energy theorem
   Example problem



                                                Physics 111: Lecture 11, Pg 1
              Work by variable force in 3-D:

   Work dWF of a force F acting                      F
    through an infinitesimal
    displacement r is:
                                                              r
                 .
          dW = F r
   The work of a big displacement through a variable force will
    be the integral of a set of infinitesimal displacements:




                                                      
                                             WTOT = F r  .


                                                Physics 111: Lecture 11, Pg 2
              Work by variable force in 3-D:
              Newton’s Gravitational Force
   Work dWg done on an object by gravity in a displacement dr is
    given by:

                  .                 ^ .
          dWg = Fg dr = (-GMm / R2 r) (dR r ^+ Rd)
                                                          ^

          dWg = (-GMm / R2) dR            .   ^       .
                                  (since r ^ = 0, r ^ = 1)
                                                    r ^

                                              dR              ^
                                                              
                                                                      ^
                                                                      r
                                          Rd      dr
                                    Fg             m

                        d
                             R
                         M

                                                  Physics 111: Lecture 11, Pg 3
               Work by variable force in 3-D:
               Newton’s Gravitational Force
   Integrate dWg to find the total work done by gravity in a “big”
    displacement:


                
                    R2

                         
                             R2
          Wg = dWg = (-GMm / R2) dR = GMm (1/R2 - 1/R1)
                 R1      R1
                                            Fg(R2)        m



                                       R2
                                       Fg(R1)


                                  R1
                             M

                                                     Physics 111: Lecture 11, Pg 4
             Work by variable force in 3-D:
             Newton’s Gravitational Force
   Work done depends only on R1 and R2, not on the path taken.



                    1   1 
                   
           Wg  GMm      
                                                 m
                    R2 R1 


                                   R2




                            R1
                        M

                                            Physics 111: Lecture 11, Pg 5
                     Lecture 11, Act 1
                      Work & Energy
    A rock is dropped from a distance RE above the surface of the
     earth, and is observed to have kinetic energy K1 when it hits
     the ground. An identical rock is dropped from twice the height
     (2RE) above the earth’s surface and has kinetic energy K2
     when it hits. RE is the radius of the earth.

      What is K2 / K1?

    (a)   2
                                  2RE
          3
    (b)
          2
                                   RE
                             RE
    (c)   4
          3
                                                Physics 111: Lecture 11, Pg 6
                   Lecture 11, Act 1
                       Solution
   Since energy is conserved, K = WG.

                   1  1                         1  1 
          WG = GMm    
                  R R                   ΔK = c 
                                                 R   
                   2   1                        2 R1 
                                          Where c = GMm is the
                                          same for both rocks


                               2RE


                                 RE
                          RE


                                           Physics 111: Lecture 11, Pg 7
                                                              1  1 
                                                      ΔK = c 
                                                             R   
                    Lecture 11, Act 1                         2 R1 
                        Solution
                                1   1     1 1
   For the first rock: K1 = c 
                                      c 
                                        
                                RE 2RE    2 RE

                                    1   1     2 1
   For the second rock: K 2 = c 
                                          c 
                                            
                                    RE 3RE    3 RE

                    2
                 K2 3 4
        So:        = 
                 K1 1 3
                    2                                        2RE


                                                               RE
                                                      RE


                                               Physics 111: Lecture 11, Pg 8
              Newton’s Gravitational Force
               Near the Earth’s Surface:
   Suppose R1 = RE and R2 = RE + y

          R2  R1         R  y   RE           
         
Wg  GMm           GMm E                   m GM y
            R1R2           R  y R            R2 
                             E         E          E 

                              GM 
    but we have learned that  2   g
                              RE 

    So:   Wg = -mgy
                               m
                                   RE+ y
                          RE
                                            M

                                            Physics 111: Lecture 11, Pg 9
                   Conservative Forces:

   We have seen that the work done by gravity does not
    depend on the path taken.


                         m
                                              1   1 
                   R2                        
                                     Wg  GMm      
                                              R2 R1 
                                                     
              R1

         M
                                 m
                                     h   Wg = -mgh




                                             Physics 111: Lecture 11, Pg 10
                   Conservative Forces:

   In general, if the work done does not depend on the path
    taken, the force involved is said to be conservative.

                                                1   1 
   Gravity is a conservative force:           
                                       Wg  GMm      
                                                R2 R1 
                                                       

   Gravity near the Earth’s surface: Wg  mgy



                                                          k x2  x12 
                                                        1
   A spring produces a conservative force:   Ws            2

                                                        2




                                              Physics 111: Lecture 11, Pg 11
                   Conservative Forces:
   We have seen that the work done by a conservative force
    does not depend on the path taken.
                                         W2

         W1 = W2

                                                W1
   Therefore the work done in a
    closed path is 0.
                                         W2

         WNET = W1 - W2 = 0
                                                W1
   The work done can be “reclaimed.”


                                              Physics 111: Lecture 11, Pg 12
                  Lecture 11, Act 2
                 Conservative Forces
   The pictures below show force vectors at different points in
    space for two forces. Which one is conservative ?

         (a) 1              (b) 2             (c) both




y                               y

          x      (1)                    x      (2)
                                                Physics 111: Lecture 11, Pg 13
                      Lecture 11, Act 2
                          Solution
   Consider the work done by force when moving along different
    paths in each case:


           WA = WB                      WA > WB




                (1)                         (2)
                                             Physics 111: Lecture 11, Pg 14
                     Lecture 11, Act 2
   In fact, you could make money on type (2) if it ever existed:
      Work done by this force in a “round trip” is > 0!
      Free kinetic energy!!

             WNET = 10 J = K           W=0




                      W = 15 J                         W = -5 J




                                       W=0

                                                 Physics 111: Lecture 11, Pg 15
                      Potential Energy

   For any conservative force F we can define a
    potential energy function U in the following way:


                  W =    F.dr = -U
      The work done by a conservative force is equal and
       opposite to the change in the potential energy function.
                                                          r2      U2
   This can be written as:


                                F.dr
                                  r2
        U = U2 - U1 = -W = -
                                 r1           r1       U1


                                                   Physics 111: Lecture 11, Pg 16
               Gravitational Potential Energy

   We have seen that the work done by gravity near the Earth’s
    surface when an object of mass m is lifted a distance y is
    Wg = -mg y

   The change in potential energy of this object is therefore:

    U = -Wg = mg y




           j                         m
                                         y     Wg = -mg y




                                                 Physics 111: Lecture 11, Pg 17
               Gravitational Potential Energy

   So we see that the change in U near the Earth’s surface is:
    U = -Wg = mg y = mg(y2 -y1).

   So U = mg y + U0 where U0 is an arbitrary constant.

   Having an arbitrary constant U0 is equivalent to saying that
    we can choose the y location where U = 0 to be anywhere
    we want to.


           j                        m    y2
                                         y1     Wg = -mg y




                                                Physics 111: Lecture 11, Pg 18
                Potential Energy Recap:

   For any conservative force we can define a potential
    energy function U such that:

                                          F.dr
                                           S2
                  U = U2 - U1 = -W = -
                                           S1


   The potential energy function U is always defined only
    up to an additive constant.

      You can choose the location where U = 0 to be
       anywhere convenient.




                                                Physics 111: Lecture 11, Pg 19
          Conservative Forces & Potential Energies
                 (stuff you should know):
  Force                  Work           Change in P.E         P.E. function
    F                    W(1-2)          U = U2 - U1               U

              ^
Fg = -mg j            -mg(y2-y1)            mg(y2-y1)             mgy + C

         GMm ^        1   1                 1   1    GMm
Fg =         r      
                  GMm                     
                                         GMm            C
                       R2 R1                        
            2
          R                                               R
                                            R2 R1 

                        k x2  x12        k x2  x12 
                      1                   1                      1 2
Fs = -kx                   2                   2
                                                                   kx  C
                      2                   2                      2




                                                      Physics 111: Lecture 11, Pg 20
                    Lecture 11, Act 3
                    Potential Energy
   All springs and masses are identical. (Gravity acts down).
     Which of the systems below has the most potential energy
        stored in its spring(s), relative to the relaxed position?



                                            (a) 1
                                            (b) 2
                                            (c) same



        (1)         (2)



                                               Physics 111: Lecture 11, Pg 21
                       Lecture 11, Act 3
                           Solution
   The displacement of (1) from equilibrium will be half of that of (2)
    (each spring exerts half of the force needed to balance mg)




                                     0
                                     d
                                     2d


          (1)          (2)



                                                    Physics 111: Lecture 11, Pg 22
              Lecture 11, Act 3
                  Solution
                                         1
The potential energy stored in (1) is 2  kd2  kd2
                                         2
                                        1
                                          k2d  2kd2
                                               2
The potential energy stored in (2) is
                                        2




The spring P.E. is
twice as big in (2) !
                                                                  0
                                                                  d
                                                                  2d


                                 (1)            (2)
                                          Physics 111: Lecture 11, Pg 23
                  Conservation of Energy
   If only conservative forces are present, the total kinetic
    plus potential energy of a system is conserved.


       E=K+U

     E = K + U
       = W + U                 using K = W
       = W + (-W) = 0           using U = -W
    E = K + U is constant!!!

   Both K and U can change, but E = K + U remains constant.




                                                 Physics 111: Lecture 11, Pg 24
           Example: The simple pendulum

   Suppose we release a mass m from rest a distance h1
    above its lowest possible point.
      What is the maximum speed of the mass and where
       does this happen?
      To what height h2 does it rise on the other side?




                m

           h1                          h2

                           v
                                            Physics 111: Lecture 11, Pg 25
            Example: The simple pendulum

   Kinetic+potential energy is conserved since gravity is a
    conservative force (E = K + U is constant)
   Choose y = 0 at the bottom of the swing,
    and U = 0 at y = 0 (arbitrary choice)

    E = 1/2mv2 + mgy


      y




           h1                             h2
    y=0
                             v
                                               Physics 111: Lecture 11, Pg 26
              Example: The simple pendulum

   E = 1/2mv2 + mgy.
      Initially, y = h1 and v = 0, so E = mgh1.
      Since E = mgh1 initially, E = mgh1 always since energy is
       conserved.




        y




     y=0

                                                Physics 111: Lecture 11, Pg 27
             Example: The simple pendulum

 1/2mv2    will be maximum at the bottom of the swing.
    So at y = 0       1/ mv2 = mgh        v2 = 2gh1
                         2          1


                        v    2 gh1




       y


    y = h1
             h1
    y=0
                              v
                                               Physics 111: Lecture 11, Pg 28
                Example: The simple pendulum

       Since E = mgh1 = 1/2mv2 + mgy it is clear that the maximum
        height on the other side will be at y = h1 = h2 and v = 0.
       The ball returns to its original height.




          y


y = h1 = h 2

     y=0

                                                  Physics 111: Lecture 11, Pg 29
            Example: The simple pendulum                       Bowling

   The ball will oscillate back and forth. The limits on its
    height and speed are a consequence of the sharing of
    energy between K and U.

    E = 1/2mv2 + mgy = K + U = constant.




      y




                                                 Physics 111: Lecture 11, Pg 30
            Example: The simple pendulum

   We can also solve this by choosing y = 0 to be at the
    original position of the mass, and U = 0 at y = 0.

    E = 1/2mv2 + mgy.




      y


    y=0
           h1                             h2

                             v
                                               Physics 111: Lecture 11, Pg 31
              Example: The simple pendulum

   E = 1/2mv2 + mgy.
      Initially, y = 0 and v = 0, so E = 0.
      Since E = 0 initially, E = 0 always since energy is conserved.




        y


    y=0




                                                 Physics 111: Lecture 11, Pg 32
           Example: The simple pendulum

 1/2mv2   will be maximum at the bottom of the swing.
   So at y = -h1      1/ mv2 = mgh        v2 = 2gh1
                         2          1




                                        v     2 gh1
     y
                                       Same as before!

 y=0
           h1
y = -h1
                            v
                                              Physics 111: Lecture 11, Pg 33
            Example: The simple pendulum                Galileo’s
                                                       Pendulum
   Since 1/2mv2 - mgh = 0 it is clear that the maximum height
    on the other side will be at y = 0 and v = 0.
   The ball returns to its original height.




      y


y=0

                                          Same as before!


                                              Physics 111: Lecture 11, Pg 34
               Example: Airtrack & Glider

   A glider of mass M is initially at rest on a horizontal
    frictionless track. A mass m is attached to it with a
    massless string hung over a massless pulley as shown.
    What is the speed v of M after m has fallen a distance d ?


                                            v
    M


                      m

                                        d                     v



                                                Physics 111: Lecture 11, Pg 35
                Example: Airtrack & Glider                        Glider

   Kinetic+potential energy is conserved since all forces are
    conservative.
   Choose initial configuration to have U=0.

    K = -U         1
                        m  M v 2  mgd
                     2
                                                v
    M


                        m

                                            d



                                                    Physics 111: Lecture 11, Pg 36
                    Problem: Hotwheel

   A toy car slides on the frictionless track shown below. It
    starts at rest, drops a distance d, moves horizontally at
    speed v1, rises a distance h, and ends up moving
    horizontally with speed v2.
       Find v1 and v2.




                                               v2
d
                         v1                                      h



                                                Physics 111: Lecture 11, Pg 37
                 Problem: Hotwheel...

   K+U energy is conserved, so E = 0   K = - U
   Moving down a distance d, U = -mgd, K = 1/2mv12
   Solving for the speed:


                v1    2 gd




    d
                       v1                                   h



                                           Physics 111: Lecture 11, Pg 38
                   Problem: Hotwheel...

   At the end, we are a distance d - h below our starting point.
   U = -mg(d - h), K = 1/2mv22
   Solving for the speed:

                  v2    2 g d  h 




                              d-h             v2
    d
                                                                 h



                                                Physics 111: Lecture 11, Pg 39
               Non-conservative Forces:

   If the work done does not depend on the path taken, the
    force is said to be conservative.

   If the work done does depend on the path taken, the force
    is said to be non-conservative.

   An example of a non-conservative force is friction.
     When pushing a box across the floor, the amount of
       work that is done by friction depends on the path taken.
         » Work done is proportional to the length of the path!




                                              Physics 111: Lecture 11, Pg 40
          Non-conservative Forces: Friction

   Suppose you are pushing a box across a flat floor. The mass
    of the box is m and the coefficient of kinetic friction is k.
   The work done in pushing it a distance D is given by:
    Wf = Ff • D = -kmgD.




                            Ff = -kmg



                      D

                                               Physics 111: Lecture 11, Pg 41
             Non-conservative Forces: Friction

   Since the force is constant in magnitude and opposite in
    direction to the displacement, the work done in pushing the
    box through an arbitrary path of length L is just
    Wf = -mgL.
   Clearly, the work done depends on the path taken.

   Wpath 2 > Wpath 1
                                B


                path 1

                                     path 2
         A


                                              Physics 111: Lecture 11, Pg 42
         Generalized Work/Energy Theorem:

   Suppose FNET = FC + FNC (sum of conservative and non-
    conservative forces).

   The total work done is: WNET = WC + WNC

   The Work/Kinetic Energy theorem says that: WNET = K.
      WNET = WC + WNC = K
                   WNC = K - WC

   But WC = -U

    So              WNC = K + U = E


                                              Physics 111: Lecture 11, Pg 43
         Generalized Work/Energy Theorem:

                  WNC = K + U = E

   The change in kinetic+potential energy of a system is equal
    to the work done on it by non-conservative forces. E=K+U
    of system not conserved!

     If all the forces are conservative, we know that K+U
      energy is conserved: K + U = E = 0 which says that
      WNC = 0, which makes sense.

     If some non-conservative force (like friction) does work,
      K+U energy will not be conserved and WNC = E, which
      also makes sense.


                                              Physics 111: Lecture 11, Pg 44
         Problem: Block Sliding with Friction

   A block slides down a frictionless ramp. Suppose the
    horizontal (bottom) portion of the track is rough, such that
    the coefficient of kinetic friction between the block and the
    track is k.
       How far, x, does the block go along the bottom portion
        of the track before stopping?




    d                   k


                              x
                                                 Physics 111: Lecture 11, Pg 45
        Problem: Block Sliding with Friction...

   Using WNC = K + U
   As before, U = -mgd
   WNC = work done by friction = -kmgx.
   K = 0 since the block starts out and ends up at rest.
   WNC = U              -kmgx = -mgd



                                 x = d / k


    d                   k


                             x
                                               Physics 111: Lecture 11, Pg 46
                Recap of today’s lecture

   Work done by variable force in 3-D         (Text: 6-1)
      Newton’s gravitational force            (Text: 11-2)
   Conservative Forces & Potential energy     (Text: 6-4)
   Conservation of “Total Mechanical Energy”  (Text: 7-1)
      Examples: pendulum, airtrack, Hotwheel car
   Non-conservative forces                    (Text: 6-4)
      friction
   General work/energy theorem                 (Text: 7-2)
   Example problem

   Look at Textbook problems   Chapter 7: # 9, 21, 27, 51, 77



                                             Physics 111: Lecture 11, Pg 47

				
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