Physics 115 2012

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					Physics 115 2012

   Final Review
Calculus is about “rates of change”.

A TIME RATE is anything divided by time.

CHANGE is expressed by using the Greek letter,
  Delta, D.

 For example: Average SPEED is simply the “RATE at
 which DISTANCE changes”.

For example, if t = 2 seconds, using x(t) = kt3=(1)(2)3= 8 

The derivative, however, tell us how our DISPLACEMENT (x) 
  changes as a function of TIME (t). The rate at which 
  Displacement changes is also called VELOCITY. Thus if we 
  use our derivative we can find out how fast the object is 
  traveling at t = 2 second. Since dx/dt = 3kt2=3(1)(2)2= 12 
Derivative of a power function
Unit Vector Notation
                       The proper terminology is to use the
                       “hat” instead of the arrow. So we
                       have i-hat, j-hat, and k-hat which are
                       used to describe any type of motion
                       in 3D space.

     How would you write vectors J and K in unit 
     vector notation?
A boat moves with a velocity of 15 m/s, N in a river which flows with a 
   velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with 
   respect to due north.

        8.0 m/s, W

                15 m/s, N

      Rv    q

                     The Final Answer :
            Dot Products in Physics
                                  Consider this situation: A force F is applied to a
                                  moving object as it transverses over a
                                  frictionless surface for a displacement, d.

                                 As F is applied to the object it will increase the
                                 object's speed!
                                                  But which part of F really
                                                  causes the object to increase
                                                  in speed?
         It is |F|Cos θ ! Because it is parallel to the displacement d
In fact if you apply the dot product, you get (|F|Cos θ)d, which happens to be
defined as "WORK" (check your equation sheet!)
                                  Work is a type of energy and energy DOES NOT
                                  have a direction, that is why WORK is a scalar or in
                                  this case a SCALAR PRODUCT
                                  (AKA DOT PRODUCT).
Suppose a person moves in a straight line from the lockers( at a position  x = 1.0 m) 
   toward the physics lab(at a position x = 9.0 m).  “To the right” is taken as positive, 
   as shown below

    The answer is positive so the person must have been traveling horizontally in 
    the positive direction.
 Suppose the person turns around!

The answer is negative so the person must have been traveling horizontally in the 
negative direction

            What is the DISPLACEMENT for the entire trip?

            What is the total DISTANCE for the entire trip?
               Instantaneous Velocity
Instantaneous velocity is a
   measure of an object’s
   displacement per unit time at a
   particular point in time.

 Example: A body’s position is defined as:
          Instantaneous Acceleration
Instantaneous velocity is a measure
   of an object’s velocity per unit
   time at a particular point in time.

   If the velocity of an object is defined as:
What do the “signs”( + or -) mean?
  Quantity       Positive               Negative
  Displacement   Your position has      Your position has
                 changed toward the     changed toward the
                 positive.              negative.
  Velocity       You are traveling in   You are traveling in
                 the +x or +y           the –x or –y
                 direction.             direction.
  Acceleration   If moving in the       If moving in the
                 positive direction,    positive direction,
                 you are speeding up.   you are slowing
                 If moving in the       down. . If moving in
                 negative direction,    the negative
                 you are slowing        direction, you are
                 down.                  speeding up.
       The 3 Kinematic equations
There are 3 major kinematic
  equations than can be
  used to describe the
  motion in DETAIL. All are
  used when the
  acceleration is CONSTANT.
     Kinematics for the VERTICAL Direction
All 3 kinematics can be used to analyze one dimensional
   motion in either the X direction OR the y direction.
A stone is dropped at rest from the top of a cliff. It is 
   observed to hit the ground 5.78 s later. How high is the 
 What do I        What do I      Which variable is NOT given and
 know?            want?          NOT asked for?
                                         Final Velocity!
    v = 0 m/s
     oy                y=?
  g = -9.8 m/s2
     yo=0 m
    t = 5.78 s
                                         -163.7 m

                                         H =163.7m
A pitcher throws a fastball with a velocity of 43.5 m/s. It is determined that 
   during the windup and delivery the ball covers a displacement of 2.5 
   meters. This is from the point behind the body when the ball is at rest to 
   the point of release. Calculate the acceleration during his throwing 
 What do I           What do I             Which variable is NOT given and
                                           NOT asked for?
 know?               want?                                     TIME
     vo= 0 m/s             a=?
     x = 2.5 m
   v = 43.5 m/s

                                                378.5 m/s/s
How long does it take a car at rest to cross a 35.0 m intersection after 
  the light turns green, if the acceleration of the car is a constant 2.00 

  What do I         What do I             Which variable is NOT given and
  know?             want?                 NOT asked for? 
                                                              Final Velocity
     vo= 0 m/s            t=?
     x = 35 m
   a = 2.00 m/s/s

                                                  5.92 s
A car accelerates from 12.5 m/s to 25 m/s in 6.0 seconds. 
  What was the acceleration?
 What do I       What do I      Which variable is NOT given and
 know?           want?          NOT asked for? 

  vo= 12.5 m/s       a=?                    DISPLACEMENT

   v = 25 m/s
     t = 6s

                                          2.08 m/s/s
There are 3 types of MOTION graphs
• Displacement(position) vs. Time
• Velocity vs. Time
• Acceleration vs. Time

There are 2 basic graph models
• Slope
• Area
                      v (m/s)                 a (m/s/s)
x (m)      = 



                                                          area = v
                                  area = x

              t (s)                   t (s)                 t (s)
   Comparing and Sketching graphs
One of the more difficult applications of graphs in physics is when given a certain 
type of graph and asked to draw a different type of graph

                           List 2 adjectives to describe the SLOPE or VELOCITY
                           1. The slope is CONSTANT

x (m)                      2. The slope is POSITIVE

               t (s)
                                      v (m/s)
How could you translate what the
SLOPE is doing on the graph
ABOVE to the Y axis on the graph
to the right?
                                                     t (s)
                                              v (m/s)
x (m)

                                                        t (s)
              t (s)

  1st line                      2nd line
  • The slope is constant       • The slope is “0”
  • The slope is “-”

           3rd line
           • The slope is “+”
           • The slope is constant
Example – Graph Matching
          What is the SLOPE(a) doing?            a (m/s/s)
           The slope is increasing
v (m/s)

                                     a (m/s/s)               t (s)

          t (s)

                  a (m/s/s)                      t (s)

                                     t (s)
               Free Body Diagrams
A pictorial representation of forces complete
  with labels.
          FN                   •Weight(mg) – Always
               T               drawn from the center,
                               straight down
Ff                             •Force Normal(FN) – A
                        T      surface force always drawn
                               perpendicular to a surface.
     W1,Fg1                    •Tension(T or FT) – force in
     or m1g                    ropes and always drawn
                               AWAY from object.
                         m2g   •Friction(Ff)- Always drawn
                               opposing the motion.
Free Body Diagrams

   Ff        FN

    New’s 1st Law and Equilibrium
Since the Fnet = 0, a system moving at a constant
  speed or at rest MUST be at EQUILIBRIUM.

TIPS for solving problems
• Draw a FBD
• Resolve anything into COMPONENTS
• Write equations of equilibrium
• Solve for unknowns
10-kg box is being pulled across the table to the right at a
  constant speed with a force of 50N at an angle of 30
  degrees above the horizontal.
a) Calculate the Force of Friction

b) Calculate the Normal Force

              FN    Fa
 Ff                30
           Springs – Hooke’s Law
One of the simplest type of
  simple harmonic motion
  is called Hooke's Law. 
  This is primarily in
  reference to SPRINGS.
                              The negative sign only tells
                              us that “F” is what is called
                              a RESTORING FORCE, in that
                              it works in the OPPOSITE
                              direction of the
     Hooke’s Law from a Graphical Point of View
Suppose we had the following data:
x(m)     Force(N)
     0              0
   0.1           12
   0.2           24
   0.3           36
   0.4           48
   0.5           60
   0.6           72                      k =120 N/m
A load of 50 N attached to a spring hanging vertically stretches the
   spring 5.0 cm. The spring is now placed horizontally on a table and
   stretched 11.0 cm. What force is required to stretch the spring this

                                           110 N
           1000 N/m
          Newton’s Second Law
The acceleration of an object is directly
  proportional to the NET FORCE and inversely
  proportional to the mass.

                      •Draw an FBD
                      •Resolve vectors into components
                      •Write equations of motion by adding and
                      subtracting vectors to find the NET FORCE.
                      Always write larger force – smaller force.
                      •Solve for any unknowns
              Newton’s 2nd Law
A 10-kg box is being pulled across the table to the
  right by a rope with an applied force of 50N.
  Calculate the acceleration of the box if a 12 N
  frictional force acts upon it.
        FN           In which direction,
                Fa   is this object
                     The X direction!
                     So N.S.L. is worked
                     out using the forces
                     in the “x” direction
A mass, m1 = 3.00kg, is resting on a frictionless horizontal table is connected
to a cable that passes over a pulley and then is fastened to a hanging mass,
m2 = 11.0 kg as shown below. Find the acceleration of each mass and the
tension in the cable.



Example (cont.)
        Where does the calculus fit in?
                                          There could be situations where you are given
                                          a displacement function or velocity function.
                                          The derivative will need to be taken once or
                                          twice in order to get the acceleration. Here is
                                          an example.

You are standing on a bathroom scale in an elevator in a tall
building. Your mass is 72-kg. The elevator starts from rest and
travels upward with a speed that varies with time according to:

 When t = 4.0s , what is the reading on the bathroom scale (a.k.a.
 Force Normal)?

                              4.6 m/s/s                      1036.8 N
         TWO types of Friction
• Static – Friction that keeps an object at rest
  and prevents it from moving
• Kinetic – Friction that acts during motion
                Force of Friction
• The Force of Friction is
  directly related to the
  Normal Force.
                                    The coefficient of
                                    friction is a unitless
                                    constant that is
                                    specific to the
                                    material type and
                                    usually less than one.
A 1500 N crate is being pushed
                                        a) What is the coefficient of kinetic
   across a level floor at a constant   friction between the crate and the
   speed by a force F of 600 N at an    floor?
   angle of 20° below the horizontal
   as shown in the figure.

               Fa        FN

If the 600 N force is instead pulling the block
                                                      FN   Fa
   at an angle of 20° above the horizontal                       Fay
   as shown in the figure, what will be the                20
   acceleration of the crate. Assume that                  Fax
   the coefficient of friction is the same as
   found in (a)
Ff                FN                 q


              q          •Rotate Axis
                         •Break weight into components
                         •Write equations of motion or
   Masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a
   frictionless pulley. As shown in the diagram, m1 is held at rest on the floor and m2 rests on a
   fixed incline of angle 40 degrees. The masses are released from rest, and m2 slides 1.00 m down
   the incline in 4 seconds. Determine (a) The acceleration of each mass (b) The coefficient of
   kinetic friction and (c) the tension in the string.

                       T                 FN
m2gcos40          Ff
                                  m 2g
           m1                             40

               m 1g
 Horizontally Launched Projectiles
To analyze a projectile in 2 dimensions we need 2
  equations. One for the “x” direction and one for the
  “y” direction. And for this we use kinematic #2.

     Remember, the velocity is        Remember that since the
     CONSTANT horizontally, so        projectile is launched
     that means the acceleration is   horizontally, the INITIAL
     ZERO!                            VERTICAL VELOCITY is equal to
  Horizontally Launched Projectiles
Example: A plane traveling with a       What do I        What I want to
   horizontal velocity of 100 m/s is    know?            know?
   500 m above the ground. At 
   some point the pilot decides to      vox=100 m/s      t=?
   drop some supplies to 
   designated target below. (a)         y = 500 m        x=?
   How long is the drop in the air? 
   (b) How far away from point          voy= 0 m/s
   where it was launched will it 
   land?                                g = -9.8 m/s/s

                                                                1010 m
                      10.1 seconds
   Vertically Launched Projectiles
There are several
  things you must
  consider when doing
  these types of
  projectiles besides
  using components. If
  it begins and ends at
  ground level, the “y”
  displacement is
  ZERO: y = 0
    Vertically Launched Projectiles
You will still use kinematic #2, but YOU MUST use
  COMPONENTS in the equation.

          vo     voy

A place kicker kicks a football with a velocity of 20.0 m/s and at 
    an angle of 53 degrees.
(a) How long is the ball in the air?
(b) How far away does it land?
(c) How high does it travel?

              0   m

     q = 53
A place kicker kicks a            What I know     What I want
   football with a velocity of                    to know
   20.0 m/s and at an angle       vox=12.04 m/s   t=?
   of 53 degrees.                 voy=15.97 m/s   x=?
(a) How long is the ball in 
                                  y=0             ymax=?
   the air?
                                  g = - 9.8

       3.26 s
A place kicker kicks a          What I know     What I want
   football with a velocity of                  to know
   20.0 m/s and at an angle  vox=12.04 m/s      t = 3.26 s
   of 53 degrees.               voy=15.97 m/s   x=?
(b) How far away does it        y=0             ymax=?
   land?                        g = - 9.8

                                            39.24 m
Example                            What I know     What I want
                                                   to know
A place kicker kicks a football    vox=12.04 m/s   t = 3.26 s
   with a velocity of 20.0 m/s     voy=15.97 m/s   x = 39.24 m
   and at an angle of 53 
   degrees.                        y=0             ymax=?
                                   g = - 9.8
(c) How high does it travel?       m/s/s


                                       13.01 m
                       A special case…
What if the projectile was launched from the ground at an angle and did
  not land at the same level height from where it started? In other words, 
  what if you have a situation where the “y-displacement” DOES NOT 
  equal zero?

       Let's look at the second kinematic closely!

       Assuming it is shot from the ground. We see we have one squared term
       variable, one regular term variable, and a constant number with no variable.
       What is this?
                                A QUADRATIC EQUATION!
Circular Motion and New’s 2nd Law
Recall that according to
  Newton’s Second Law,
  the acceleration is
  directly proportional to
  the Force. If this is true:

                         Since the acceleration and the force are directly
                         related, the force must ALSO point towards the center.
                         This is called CENTRIPETAL FORCE.

                      NOTE: The centripetal force is a NET FORCE. It could be
                      represented by one or more forces. So NEVER draw it in an
Examples          What is the minimum coefficient of static friction 
                  necessary to allow a penny to rotate along a 33 1/3 rpm 
    Top view      record (diameter= 0.300 m), when
                  the penny is placed at the outer edge of the record?



 Side view
The maximum tension that a 0.50 m 
   string can tolerate is 14 N. A 0.25-kg    T   mg
   ball attached to this string is being 
   whirled in a vertical circle. What is 
   the maximum speed the ball can 
   have (a) the top of the circle, (b)at 
   the bottom of the circle?
 At the bottom?


                 A 2-kg sliding puck whose initial velocity magnitude is v1 = 10
                 m/s strikes a wall at a 30 degree angle and bounces off. If it
                 leaves the wall with a velocity magnitude of v2 = 10 m/s, and
                 if the collision takes a total of 0.02 seconds to complete,
                 what was the average force applied to the puck by the

              There is something you need to consider:
                              Momentum is a VECTOR!!!

    Let’s look at this problem using a X-Y axis for reference
Example cont’

  If we did the same thing for the Y direction we would discover 
  that the Force Net is equal to ZERO!

            The temptation is to treat momentum as a SCALAR...DO 
            Momentum is conserved!
The Law of Conservation of Momentum: “In the absence
  of an unbalanced external force, the total momentum
  before the collision is equal to the total momentum
  after the collision.”
       Several Types of collisions
Sometimes objects stick together or blow apart. In this
  case, momentum is ALWAYS conserved.

                                  When 2 objects collide and DON’T stick

                                  When 2 objects collide and stick together

                                  When 1 object breaks into 2 objects

Elastic Collision = Kinetic Energy is Conserved
Inelastic Collision = Kinetic Energy is NOT Conserved
       The VERTICAL component of the force DOES NOT
       cause the block to move the right. The energy imparted to
       the box is evident by its motion to the right. Therefore
       ONLY the HORIZONTAL COMPONENT of the force
       actually does WORK.

       When the FORCE and DISPLACEMENT are in the SAME
       ANGLE between the force and displacement is ZERO
       degrees. What happens when you put this in for the
       When the FORCE and DISPLACEMENT are in the
       OPPOSITE direction, yet still on the same axis, you get a
       NEGATIVE WORK VALUE. This negative doesn't mean
       the direction!!!! IT simply means that the force and
       displacement oppose each other. The ANGLE between the
       force and displacement in this case is 180 degrees. What
       happens when you put this in for the COSINE?
       When the FORCE and DISPLACEMENT are
       PERPENDICULAR, you get NO WORK!!! The ANGLE
       between the force and displacement in this case is 90 degrees.
       What happens when you put this in for the COSINE?
 A box of mass m = 2.0 kg is moving over a
 frictional floor ( uk = 0.3) has a force whose
 magnitude is F = 25 N applied to it at an angle of
 30 degrees, as shown to the left. The box is
 observed to move 16 meters in the horizontal
 direction before falling off the table.

 a) How much work does F do before taking the
                 Example cont’
What if we had done this in UNIT VECTOR notation?
Example cont’ How much work does the FORCE 
              NORMAL do and Why?

                                             There is NO WORK since “F”
                                             and “r” are perpendicular.


                  How much does the internal energy of the system increase?

                                   34.08 J
               Elastic Potential Energy
The graph of F vs.x for a
  spring that is IDEAL in
  nature will always
  produce a line with a
  positive linear slope.
  Thus the area under the
  line will always be
  represented as a

NOTE: Keep in mind that this can be applied to WORK or can be conserved with any
other type of energy.
Elastic potential energy

 Elastic “potential” energy is a fitting term as springs STORE energy when there
 are elongated or compressed.
Energy is CONSERVED!
A 2.0 m pendulum is released from rest when the support
  string is at an angle of 25 degrees with the vertical. What
  is the speed of the bob at the bottom of the string?

                                q   Lcosq        h = L – Lcosq
                                             L   h = 2-2cosq
                                                 h = 0.187 m

                              EB = E A

                          UO      = K
                          mgho    = 1/2mv2
                          gho     = 1/2v2
                          1.83    = v2
                          1.35 m/s = v

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