Physics 207: Lecture 2 Notes by HC120212201041

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									            Physics 207, Lecture 13, Oct. 15
Goals:
• Chapter 10
      Understand the relationship between motion and energy
      Define Potential Energy in a Hooke’s Law spring
      Develop and exploit conservation of energy principle
     in problem solving
•   Chapter 11
      Understand the relationship between force,
     displacement and work

Assignment:
 HW6 due Wednesday, Oct. 22
 For Monday: Read all of Chapter 11

                                            Physics 207: Lecture 13, Pg 1
                             Energy
   If only “conservative” forces are present, the total energy
    (sum of potential, U, and kinetic energies, K) of a system is
    conserved

For an object in a gravitational “field”

         ½ m vyi2 + mgyi = ½ m vyf2 + mgyf
              K ≡ ½ mv2         U ≡ mgy
                        Emech = K + U

                   Emech = K + U = constant

   K and U may change, but Emech = K + U remains a fixed value.

            Emech is called “mechanical energy”
                                                Physics 207: Lecture 13, Pg 2
        Example of a conservative system:
             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 207: Lecture 13, Pg 3
          Example: The simple pendulum.

 What is the maximum speed of the mass and where does
 this happen ?
   E = K + U = constant and so K is maximum when U is a
 minimum.




     y


   y=h1
   y=
     0
                                       Physics 207: Lecture 13, Pg 4
          Example: The simple pendulum.

 What is the maximum speed of the mass and where does
 this happen ?
   E = K + U = constant and so K is maximum when U is a
 minimum
   E = mgh1 at top
   E = mgh1 = ½ mv2 at bottom of the swing


      y

   y=h1
          h1
    y=0
                       v
                                       Physics 207: Lecture 13, Pg 5
           Example: The simple pendulum.

  To what height h2 does it rise on the other side?

   E = K + U = constant and so when U is maximum
    again (when K = 0) it will be at its highest point.

   E = mgh1 = mgh2 or h1 = h2


     y


y=h1=h2
   y=0

                                           Physics 207: Lecture 13, Pg 6
                         Example
                 The Loop-the-Loop … again
     To complete the loop the loop, how high do we have to let
      the release the car?
     Condition for completing the loop the loop: Circular motion
      at the top of the loop (ac = v2 / R)
      Use fact that E = U + K = constant !


  Ub=mgh
                                 Recall that “g” is the source of
                   Car has mass m the centripetal acceleration
  U=mg2R
                                   and N just goes to zero is
                                   the limiting case.
           h?                    Also recall the minimum speed
                                                              v
                           R       at the top is
                                                                            gR
y=0
                                                 Physics 207: Lecture 13, Pg 7
              Example
      The Loop-the-Loop … again
    Use E = K + U = constant
                                        v

   mgh + 0 = mg 2R + ½ mv2                                gR
    mgh = mg 2R + ½ mgR = 5/2 mgR


             h = 5/2 R


h?
               R




                                    Physics 207: Lecture 13, Pg 8
                            Example
                           Skateboard
   What speed will the skateboarder reach halfway down the hill
    if there is no friction and the skateboarder starts at rest?
   Assume we can treat the skateboarder as a “point”
   Assume zero of gravitational U is at bottom of the hill
       m = 25 kg
           ..
R=10 m

                 30°
                   R=10 m


                                               y=0

                                               Physics 207: Lecture 13, Pg 9
                            Example
                           Skateboard
   What speed will the skateboarder reach halfway down the hill
    if there is no friction and the skateboarder starts at rest?
   Assume we can treat the skateboarder as “point”
   Assume zero of gravitational U is at bottom of the hill
       m = 25 kg
           ..              Use E = K + U = constant
                             
R=10 m                      Ebefore = Eafter
                         0 + m g R = ½ mv2 + mgR (1-sin 30°)
                 30°
                            mgR/2 = ½ mv2
                  R=10 m        gR = v2  v= (gR)½
                           v = (10 x 10)½ = 10 m/s


                                               Physics 207: Lecture 13, Pg 10
             Potential Energy, Energy Transfer and Path
  A ball of mass m, initially at rest, is released and follows three
   difference paths. All surfaces are frictionless
1. Ball is dropped
2. Ball slides down a straight incline
3. Ball slides down a curved incline
After traveling a vertical distance h, how do the three speeds compare?

         1            2                           3


     h




    (A) 1 > 2 > 3    (B) 3 > 2 > 1   (C) 3 = 2 = 1 (D) Can’t tell
                                                   Physics 207: Lecture 13, Pg 11
            Potential Energy, Energy Transfer and Path
  A ball of mass m, initially at rest, is released and follows
   three difference paths. All surfaces are frictionless
1. The ball is dropped
2. The ball slides down a straight incline
3. The ball slides down a curved incline
After traveling a vertical distance h, how do the three speeds
   compare?
        1            2                         3


    h



    (A) 1 > 2 > 3   (B) 3 > 2 > 1   (C) 3 = 2 = 1 (D) Can’t tell
                                                Physics 207: Lecture 13, Pg 12
                           Example
                          Skateboard
   Now what is the normal force on the skate boarder?

                               N

       m = 25 kg             60°
           ..                      mg
R=10 m
                              S Fr = mar = m v2 / R
                   30°
                                          = N – mg cos 60°
                   R=10 m
                             N = m v2 /R + mg cos 60°
                             N = 25 100 / 10 + 25 10 (0.87)
                             N = 250 + 220 =470 Newtons
                                               Physics 207: Lecture 13, Pg 13
                Elastic vs. Inelastic Collisions

   A collision is said to be elastic when energy as well as
    momentum is conserved before and after the collision.
                         Kbefore = Kafter
      Carts colliding with a perfect spring, billiard balls, etc.


                                               vi




                                                    Physics 207: Lecture 13, Pg 14
                 Elastic vs. Inelastic Collisions



   A collision is said to be inelastic when energy is not
    conserved before and after the collision, but momentum is
    conserved.
                        Kbefore  Kafter
      Car crashes, collisions where objects stick together, etc.




                                                 Physics 207: Lecture 13, Pg 15
           Inelastic collision in 1-D: Example 1
   A block of mass M is initially at rest on a frictionless
    horizontal surface. A bullet of mass m is fired at the block
    with a muzzle velocity (speed) v. The bullet lodges in the
    block, and the block ends up with a speed V.

      What is the initial energy of the system ?
      What is the final energy of the system ?
      Is energy conserved?
                                                                        x
          v
                                                             V


                 before                       after

                                                Physics 207: Lecture 13, Pg 16
             Inelastic collision in 1-D: Example 1

What is the momentum of the bullet with speed v ?              
                                                              mv
  What is the initial energy of the system ?    1   1 2
                                                   mv  v  mv
                                                 2         2
  What is the final energy of the system ?     1
                                                  (m  M )V 2
                                                2
  Is momentum conserved (yes)?          mv  M 0  (m  M )V
  Is energy conserved?     Examine Ebefore-Eafter
  1       1               1      1      m     1          m
  2
    mv 2  [( m  M )V]V  mv 2  (mv)
          2               2      2     mM
                                           v  mv 2 1 
                                              2
                                                          (
                                                        mM
                                                                            )
         v                       No!                           V

                                                                              x
                before                          after
                                                  Physics 207: Lecture 13, Pg 17
 Variable force devices: Hooke’s Law Springs
   Springs are everywhere, (probe microscopes, DNA, an
    effective interaction between atoms)

                                 Rest or equilibrium position
                 F
                          Ds

   In this spring, the magnitude of the force increases as the
    spring is further compressed (a displacement).
   Hooke’s Law,
        Fs = - k Ds
Ds is the amount the spring is stretched or compressed
  from it resting position.
                                              Physics 207: Lecture 13, Pg 19
                       Home Exercise
                        Hooke’s Law




                         8m
  9m


                         What is the spring constant “k” ?
             Fspring
                         SF =  0 = Fs – mg = k Ds - mg
        50 kg            Use k = mg/Ds = 5 N / 0.01 m
(A) 50 N/m   (B) 100 N/m (C) 400 N/m (D) 500 N/m
         mg                                  Physics 207: Lecture 13, Pg 21
F-s relation for a single DNA molecule




                                Physics 207: Lecture 13, Pg 22
  Measurement technique: optical tweezers
http://phet.colorado.edu/sims/optical-tweezers/stretching-dna.jnlp




                                                Physics 207: Lecture 13, Pg 23
       Force vs. Energy for a Hooke’s Law spring
   F = - k (x – xequilibrium)
   F = ma = m dv/dt
             = m (dv/dx dx/dt)                                                   m
             = m dv/dx v
             = mv dv/dx                          xf                   vf
   So - k (x – xequilibrium) dx = mv dv          ku du   mv dv
   Let u = x – xeq. & du = dx                  xi                   vi

                                           
                                             1
                                             2
                                                 ku | 
                                                      2 xf
                                                        xi
                                                                     1
                                                                     2
                                                                             2 vf
                                                                           mv vi    |
                             
                                 1
                                 2
                                     kx  kx 
                                      2
                                      f
                                             1
                                             2
                                                      2
                                                      i
                                                             1
                                                             2
                                                                    2
                                                                 mv f      
                                                                               1
                                                                               2
                                                                                     2
                                                                                   mvi

1
2
    kx  mv  kx  mv
         2
         i
                1
                2
                        2
                        i
                               1
                               2
                                       2
                                       f
                                                 1
                                                 2
                                                            2
                                                            f
                                                          Physics 207: Lecture 13, Pg 24
               Energy for a Hooke’s Law spring
          1
          2
              kx  mv  kx  mv
               2
               i
                    1
                    2
                            2
                            i
                                  1
                                  2
                                       2
                                       f
                                           1
                                           2
                                                          2
                                                          f

   Associate ½ kx2 with the
    “potential energy” of the spring                                    m




                   U si  K i  U sf  K f
         Hooke’s Law springs are conservative so the
          mechanical energy is constant

                                               Physics 207: Lecture 13, Pg 25
                            Energy diagrams
    In general:

         Ball falling                         Spring/Mass system
               Emech                                Emech
                        K                                            K




                                     Energy
Energy




                                                                             U
                        U


                   y                                   s



                                                   Physics 207: Lecture 13, Pg 26
                 Energy diagrams
  Spring/Mass/Gravity system

        spring
          alone
Force




                       y

-mg                                                                      Ug
                                        Emech                 K



                               Energy
           spring &
                           m                 K
             gravity
                                                                Uspring
                                                 UTotal

                                                 y
                                        Physics 207: Lecture 13, Pg 27
                          Equilibrium
   Example
      Spring: Fx = 0 => dU / dx = 0 for x=0
     The spring is in equilibrium position

   In general: dU / dx = 0  for ANY function establishes
    equilibrium



            U                          U


            stable equilibrium       unstable equilibrium


                                               Physics 207: Lecture 13, Pg 28
           Comment on Energy Conservation
   We have seen that the total kinetic energy of a system
    undergoing an inelastic collision is not conserved.
      Mechanical energy is lost:
         Heat (friction)
         Bending of metal and deformation

   Kinetic energy is not conserved by these non-conservative
    forces occurring during the collision !

   Momentum along a specific direction is conserved when
    there are no external forces acting in this direction.
       In general, easier to satisfy conservation of momentum
       than energy conservation.

                                              Physics 207: Lecture 13, Pg 29
       Physics 207, Lecture 13, Oct. 15

Assignment:
 HW6 due Wednesday
 For Monday: Read all of chapter 11




                                       Physics 207: Lecture 13, Pg 30

								
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