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 IUPhysicsP201F2009
 Assignment 4a

 Due at 11:00pm on Tuesday, September 23, 2008

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  Details


                                      A Mass on a Turntable: Conceptual
  Description: An object is sitting on a rotating turntable. The student is asked for the direction of velocity,
  acceleration, and net force. Based on Mechanics Baseline Test.
 A small metal cylinder rests on a circular turntable that
 is rotating at a constant rate, as illustrated in the
 diagram.




  Part A
  Which of the following sets of vectors best describes the velocity, acceleration, and net force acting on the cylinder
  at the point indicated in the diagram?




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   Hint A.1 The direction of acceleration can be determined from Newton's second law
   According to Newton's second law, the acceleration of an object has the same direction as the net force acting on
   that object.

  ANSWER:              a
                       b
                       c
                       d
                       e


  Part B
  Let      be the distance between the cylinder and the center of the turntable. Now assume that the cylinder is moved
  to a new location            from the center of the turntable. Which of the following statements accurately describe the
  motion of the cylinder at the new location?

   Part B.1     Find the speed of the cylinder
   Find the speed     of the cylinder at the new location. Assume that the cylinder makes one complete turn in a
   period of time      .

   Express your answer in terms of            and      .

   ANSWER:
                           =


        Now compare your result with the speed of the cylinder before it is moved.

   Part B.2     Find the acceleration of the cylinder
   Find the magnitude of the acceleration         of the cylinder at the new location. Assume that the cylinder makes one
   complete turn in a period of time    .

   Hint B.2.a Centripetal acceleration
   Recall that the acceleration of an object that moves in a circular path of radius      with constant speed    has
   magnitude given by


                                                                     .


   Note that both the velocity and radius of the trajectory change when the cylinder is moved.

   Express your answer in terms of            and      .

   ANSWER:




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                        =



        Now compare your result with the acceleration of the cylinder before it is moved.

  Check all that apply.

  ANSWER:              The speed of the cylinder has decreased.
                       The speed of the cylinder has increased.
                       The magnitude of the acceleration of the cylinder has decreased.
                       The magnitude of the acceleration of the cylinder has increased.
                       The speed and the acceleration of the cylinder have not changed.



                                          Accelerating along a Racetrack
  Description: Determine the direction of acceleration of cars at various points along a race track
 A road race is taking place along the track shown in the figure . All of the cars are moving at constant speeds. The car
 at point F is traveling along a straight section of the
 track, whereas all the other cars are moving along
 curved segments of the track.




  Part A
  Let      be the velocity of the car at point A. What can you say about the acceleration of the car at that point?

   Hint A.1 Acceleration along a curved path
   Since acceleration is a vector quantity, an object moving at constant speed along a curved path has nonzero
   acceleration because the direction of its velocity is changing, even though the magnitude of its velocity (the
   speed) is constant. Moreover, if the speed is constant, the object's acceleration is always perpendicular to the
   velocity vector at each point along the curved path and is directed toward the center of curvature of the path.


  ANSWER:              The acceleration is parallel to   .




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                       The acceleration is perpendicular to     and directed toward the inside of the track.
                       The acceleration is perpendicular to     and directed toward the outside of the track.
                       The acceleration is neither parallel nor perpendicular to    .
                       The acceleration is zero.


  Part B
  Let      be the velocity of the car at point C. What can you say about the acceleration of the car at that point?

   Hint B.1     Acceleration along a curved path
   Since acceleration is a vector quantity, an object moving at constant speed along a curved path has nonzero
   acceleration because the direction of its velocity is changing, even though the magnitude of its velocity (the
   speed) is constant. Moreover, if the speed is constant, the object's acceleration is always perpendicular to the
   velocity vector at each point along the curved path and is directed toward the center of curvature of the path.


  ANSWER:              The acceleration is parallel to   .
                       The acceleration is perpendicular to     and pointed toward the inside of the track.
                       The acceleration is perpendicular to     and pointed toward the outside of the track.
                       The acceleration is neither parallel nor perpendicular to    .
                       The acceleration is zero.


  Part C
  Let      be the velocity of the car at point D. What can you say about the acceleration of the car at that point?

   Hint C.1 Acceleration along a curved path
   Since acceleration is a vector quantity, an object moving at constant speed along a curved path has nonzero
   acceleration because the direction of its velocity is changing, even though the magnitude of its velocity (the
   speed) is constant. Moreover, if the speed is constant, the object's acceleration is always perpendicular to the
   velocity vector at each point along the curved path and is directed toward the center of curvature of the path.


  ANSWER:              The acceleration is parallel to   .
                       The acceleration is perpendicular to     and pointed toward the inside of the track.
                       The acceleration is perpendicular to     and pointed toward the outside of the track.
                       The acceleration is neither parallel nor perpendicular to    .
                       The acceleration is zero.


  Part D
  Let      be the velocity of the car at point F. What can you say about the acceleration of the car at that point?

   Hint D.1 Acceleration along a straight path




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   The velocity of an object that moves along a straight path is always parallel to the direction of the path, and the
   object has a nonzero acceleration only if the magnitude of its velocity changes in time.

  ANSWER:              The acceleration is parallel to   .
                       The acceleration is perpendicular to     and pointed toward the inside of the track.
                       The acceleration is perpendicular to     and pointed toward the outside of the track.
                       The acceleration is neither parallel nor perpendicular to   .
                       The acceleration is zero.


  Part E
  Assuming that all cars have equal speeds, which car has the acceleration of the greatest magnitude, and which one
  has the acceleration of the least magnitude?

   Hint E.1     How to approach the problem
   Recall that the magnitude of the acceleration of an object that moves at constant speed along a curved path is
   inversely proportional to the radius of curvature of the path.

  Use A for the car at point A, B for the car at point B, and so on. Express your answer as the name the car
  that has the greatest magnitude of acceleration followed by the car with the least magnitude of accelation,
  and separate your answers with a comma.

  ANSWER:           D F


  Part F
  Assume that the car at point A and the one at point E are traveling along circular paths that have the same radius. If
  the car at point A now moves twice as fast as the car at point E, how is the magnitude of its acceleration related to
  that of car E.

   Part F.1     Find the acceleration of the car at point E
   Let      be the radius of the two curves along which the cars at points A and E are traveling. What is the magnitude
         of the acceleration of the car at point E?

   Hint F.1.a Uniform circular motion
   The magnitude of the acceleration of an object that moves with constant speed           along a circular path of
   radius is given by


                                                                  .


   Express your answer in terms of the radius of curvature            and the speed    of car E.

   ANSWER:
                         =



   Part F.2     Find the acceleration of the car at point A




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   If             , what is the acceleration    of the car at point A? Let   be the radius of the two curves along which
   the cars at points A and E are traveling.

    Hint F.2.a Uniform circular motion
    The magnitude of the acceleration of an object that moves with constant speed         along a circular path of radius
      is given by


                                                                   .


   Express your answer in terms of the speed           of the car at E and the radius .

   ANSWER:
                          =



        Since the magnitude of the acceleration of an object that moves with constant speed along a circular path is
        proportional to the square of the speed, the acceleration of the car at point A is proportional to
                      .


  ANSWER:              The magnitude of the acceleration of the car at point A is twice that of the car at point E.
                       The magnitude of the acceleration of the car at point A is the same as that of the car at point E.
                       The magnitude of the acceleration of the car at point A is half that of the car at point E.
                       The magnitude of the acceleration of the car at point A is four times that of the car at point E.



                                   Velocity of a Roller Coaster Ranking Task
  Description: Conceptual question on the velocity of different roller-coaster carts going over a semi-circular hill.
  (ranking task)
 Six roller-coaster carts pass over the same semicircular "bump." The mass          of each cart (including passenger) and
 the normal force of the track on the cart at the top of
 each bump are given in the figures.




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  Part A
  Rank the speeds of the different carts as each passes over the top of the bump.

   Hint A.1 Newton's 2nd law
   Newton's 2nd law is valid for any motion, whether the motion is along a straight line or a circular path. The net
   force acting on the cart must be equal to the product of its mass and acceleration,

                                                                      ,


   and since the cart is moving along a circular path, its acceleration (toward the center of the circular path) has
   magnitude


                                                                  .



   Part A.2    Determine the net force on the cart


   Defining the down direction (toward the center of the circular path) as positive, which of the following
   expressions describes the component of the net force acting on the cart in the vertical direction?

   ANSWER:




     Taking the expressions for the net force on the cart and the centripetal acceleration of the cart and
     substituting into Newton's 2nd law,

                                                                      ,


     results in


                                                                                  .


     Since the radius is the same for every cart, we can ignore       and write


                                                                              .


     But          is simply      so


                                                                          ,




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   or


                                                                        .




  Rank from largest to smallest. To rank items as equivalent, overlap them.

  ANSWER:




                 View

    Taking the expressions for the net force on the cart and the centripetal acceleration of the cart and substituting
    into Newton's 2nd law,

                                                                    ,


    results in


                                                                                    .


    Since the radius is the same for every cart, we can ignore    and write


                                                                                .


    But          is simply      so


                                                                            ,


    or


                                                                        .


    Therefore, larger            implies smaller .




http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1144566                                       9/17/2008
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                                              Orbital Speed of a Satellite
  Description: Short conceptual problem comparing the orbital motion (speed, radius, and period) of two different
  satellites. This problem is based on Young/Geller Conceptual Analysis 6.6

  Part A
  Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed at a distance from the
  center of the earth. The second satellite travels at a speed that is less than . At what distance from the center of the
  earth does the second satellite orbit?

   Hint A.1 How to approach the problem
   Satellites orbiting any large object, like the earth or the moon, are in free fall. The only force that acts on a
   satellite that maintains a stable orbit is the gravitational attraction of the object on the satellite. This force is given
   by Newton's law of gravitation


                                                                            ,


   where     is the gravitational constant,     is the mass of the large object (e.g., in this case the earth),    is the
   mass of the satellite, and   is the distance from the center of the object to the satellite.

   According to Newton's 2nd law the force on the satellite is also given by

                                                                        ,


   where             for objects undergoing uniform circular motion with speed .


   Use this information to find a relationship between a satellite's speed and its orbital distance.

      Hint A.2 Working with the equations



   Because the gravitational attraction of the earth on the satellite is the only force acting on the satellite, you know
   that


                                                                                    .


   You also know that the satellite undergoes uniform circular motion, meaning that


                                                                    .


   Combining these expressions yields


                                                                                .




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   Simplify this expression by getting all the distance terms               on one side of the equation.

  ANSWER:              At a distance that is less than .
                       At a distance equal to .
                       At a distance greater than .


    The speed      of a satellite of mass           orbiting a distance     from the center of a larger object of mass        is
    given by the relationship


                                                                                  ,



    where      is the gravitational constant. We can’t choose the radius               of the orbit and the speed    of the
    satellite independently; choosing a value of             automatically determines . Also note that the satellite’s motion
    does not depend on its mass .


  Part B
  Now assume that a satellite of mass        is orbiting the earth at a distance from the center of the earth with speed
    . An identical satellite is orbiting the moon at the same distance with a speed     . How does the time        it takes
  the satellite circling the moon to make one revolution compare to the time                  it takes the satellite orbiting the earth
  to make one revolution?

   Hint B.1    How to approach this problem
   To compare the time it takes the satellite circling the moon to make one revolution to the time it takes the satellite
   orbiting the earth to make one revolution you should first compare the speeds     and of the satellite. Then you
   can relate the speeds to the time it takes the satellites to make one revolution.

   Part B.2    Compare the speeds of the satellites
   How does the speed of the satellite orbiting the moon,                 , compare to the speed of the satellite orbiting the earth,
     ?

   ANSWER:                  is greater than         .
                            is equal to     .
                            is less than        .


   Hint B.3    Relationship between speed and the time for one revolution
   The time it takes a satellite to make one revolution is called the period              . The period depends on the speed        of
   the satellite and its radius   and is given by


                                                                              .




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  ANSWER:                  is less than        .
                           is equal to     .
                           is greater than         .


    The period of a satellite orbiting an object of mass             at a distance         from its center is given by



                                                                                  .



    This relationship is also known as Kepler's 3rd law.



                                                       Weight on a Neutron Star
  Description: Given your weight on earth, calculate your weight on the surface of a neutron star.
 Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but a much
 smaller diameter.

  Part A
  If you weigh 650       on the earth, what would be your weight on the surface of a neutron star that has the same mass
  as our sun and a diameter of 25.0            ?


  Take the mass of the sun to be          = 1.99×1030         , the gravitational constant to be           = 6.67×10−11               ,
  and the acceleration due to gravity at the earth's surface to be         = 9.810              .


   Hint A.1 How to approach the problem
   "Weight" is the same as "the force of gravitational attraction." Use Newton's law of gravitation to calculate the
   gravitational force that would be exerted on you if you were standing on the surface of the star. Be careful when
   determining the values needed for the equation.

   Hint A.2 Law of universal gravitation
   The gravitational force exerted on a mass               by a second mass           is


                                                                              ,



   where       is the distance between the two masses, and 6.67×10−11                               is the universal gravitational
   constant.

      Part A.3 Calculate your mass


   Calculate your mass       if you weigh 650            on earth.




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   Express your answer in kilograms.

   ANSWER:
                       =



      Part A.4 Calculate the distance between you and the star
   Calculate your distance     from the center of the star if you are standing on its surface.

   Express your answer in meters.

   ANSWER:
                      =


  Express your weight           in newtons.

  ANSWER:
                           =



    This is over      times your weight on earth! You probably shouldn't venture there....




  Summary     0 of 5 items complete (0% avg. score)
              0 of 5 points




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