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 IUPhysicsP201F2009
 Assignment09a

 Due at 11:00pm on Tuesday, October 28, 2008

  View Grading
  Details


                                        Torques on a Seesaw: A Tutorial
  Description: Balancing seesaw: intuition, how it relates to torque, and finally balancing using horizontal force.
  (version for algebra-based courses)

  Learning Goal: To make the connection between intuitive understanding of a seesaw and the standard formalism
  for torque.
 This problem deals with the concept of torque, the "twist" that an off-center force applies to a body that tends to make
 it rotate.




  Use your intuition to try to answer the following question. If your intuition fails, work the rest of the problem and
  return here when you feel that you are more comfortable with torques.

  Part A
  Marcel is helping his two children, Jacques and Gilles, to balance on a seesaw so that they will be able to make it
  tilt back and forth without the heavier child, Jacques, simply sinking to the ground. Given that Jacques, whose
  weight is     , is sitting at distance to the left of the pivot, at what distance  should Marcel place Gilles, whose
  weight is   , to the right of the pivot to balance the seesaw?

   Hint A.1 How to approach the problem

   Consider whether      increases or decreases as each of the variables that it depends on,    ,   , and   , is made




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   larger or smaller.

  Express your answer in terms of          ,       , and      .

  ANSWER:
                          =



  Now consider this problem as a more formal introduction to torque. The torque of each child about the pivot point
  is the product of the child's weight and the distance of the child (strictly speaking, the child's center of mass) from
  the pivot. The sign of the torque is taken to be positive by convention if it would cause a counterclockwise rotation
  of the seesaw. The distance is measured perpendicular to the line of force and is called the moment arm.

  Part B
  Find the torque       about the pivot due to the weight          of Gilles on the seesaw.

  Express your answer in terms of              and        .

  ANSWER:                =


  Marcel wants the seesaw to balance, which means that there can be no angular acceleration about the pivot. For the
  angular acceleration to be zero, the sum of the torques about the pivot must equal zero:

                                                                           .



  Part C

  Determine         , the sum of the torques on the seesaw. Consider only the torques exerted by the children.


   Hint C.1 Torque from the weight of the seesaw
   The seesaw is symmetric about the pivot, and so the gravitational force on the seesaw produces no net torque.
   More generally, when determining torques, the gravitational force on an object in a uniform gravitational field
   can be taken to act at the object's center of mass. Here, the center of mass is directly above the pivot, so the
   weight of the seesaw has zero moment arm and produces no torque about the pivot.

  Express your answer in terms of              ,   ,     , and    .

  ANSWER:



    If you did not solve for the distance              required to balance the seesaw in Part A, do so now.


    The equation                 applies to any body that is not rotationally accelerating. Combining this equation with

                 (which applies to any body that is not accelerating linearly) gives a pair of equations that are
    sufficient to form the basis of statics. The art of applying these equations to large or complicated structures




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  constitutes a significant part of mechanical and civil engineering.



  Gilles has an identical twin, Jean, also of weight . The two twins now sit on the same side of the seesaw, with
  Gilles at distance    from the pivot and Jean at distance   .




  Part D
  Where should Marcel position Jacques to balance the seesaw?

   Part D.1   Find the sum of the torques when the seesaw is balanced
   For the seesaw to balance, the sum of the torques must be zero. What is the sum of the torques due to the three
   children?
   Express your answer in terms of      ,   ,    ,   , and    .

   ANSWER:



     Now solve your torque equation for     .


  Express your answer in terms of       ,   ,    , and   .

  ANSWER:
                      =



  Bad news! When Marcel finds the distance      from the previous part, it turns out to be greater than    , the
  distance from the pivot to the end of the seesaw. Hence, even with Jacques at the very end of the seesaw, the twins
  Gilles and Jean exert more torque than Jacques does. Marcel now elects to balance the seesaw by pushing sideways
  on an ornament (shown in red) that is at height above the pivot.




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  Part E
  With what force in the rightward direction,        , should Marcel push? If you think the force should be toward the
  left, your answer     should be negative.

   Hint E.1    Sign conventions
   It is easy to make sign errors in torque problems, and experience shows that it is better to use standard
   conventions (here, that     goes to the right, the direction of positive x) than to change the direction of positive
   torque or displacement to suit your convenience. (You are likely to forget your unconventional choice at a later
   point in the problem.) In this case, your intuition correctly expects that Marcel must push to the left to make
   things balance; the equations will confirm this by giving a negative result for    . (A positive result would mean
   that the force would be directed to the right in the figure.)

   Part E.2    Find the torque due to Marcel's push
   The sum of all four torques (due to each of the three children, plus Marcel's) must be zero.

   What is the torque            due to Marcel's push? Keep in mind that a positive torque will cause counterclockwise
   rotation of the seesaw.

   Express your answer in terms of the unknown force               and the height   at which it is applied.

   ANSWER:                   =


  Express your answer in terms of          ,     ,    ,   ,   , and   .

  ANSWER:
                        =



    This answer will necessarily be negative, because you were told that to balance the seesaw with the twins on
    the right, Jacques had to be "beyond the end" of the seesaw. Therefore, when        indicates Jacque's position,
        will be less than zero. Hence Marcel must push to the left, as you would expect.




                                                     Young's Modulus


  Description: Contains several questions, conceptual and quantitative, leading to better understanding of Young's
  modulus. (version for algebra-based courses)

  Learning Goal: To understand the meaning of Young's modulus and to perform some real-life calculations related
  to the stretching of steel.
 Hooke's law states that for springs and other "elastic" objects




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                                                                       ,


 where      is the magnitude of the stretching force,         is the corresponding elongation of the spring from equilibrium,
 and     is a constant that depends on the geometry and the material of the spring. If the deformations are small enough,
 most materials, in fact, behave like springs: Their deformation is directly proportional to the external force.
 Therefore, it may be useful to operate with an expression that is similar to Hooke's law but describes the properties of
 various materials, as opposed to objects such as springs.


  Consider, for instance, a bar of initial length     and cross-sectional area       stressed by a force of magnitude       . As
  a result, the bar stretches by      .


  Let us define two new terms:

          Tensile stress is the ratio of the stretching force
           to the cross-sectional area:


                                           .


          Tensile strain is the ratio of the elongation of
           the rod to the initial length of the bar:


                                               .


  It turns out that the ratio of the tensile stress to the tensile strain is a constant as long as the tensile stress is not too
  large. That constant, which is an inherent property of a material, is called Young's modulus and is given by




  Part A
  What is the SI unit of Young's modulus?

   Hint A.1 Look at the dimensions
   If you look at the dimensions of Young's modulus, you will see that they are equivalent to the dimension of
   pressure. Use the SI unit of pressure.

  ANSWER:




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  Part B
  Consider a metal bar of initial length       and cross-sectional area    . The Young's modulus of the material of the bar
  is   . Find the "spring constant"   of such a bar for low values of tensile strain.

   Hint B.1    How to approach the problem
   Consider the equation defining     . Then isolate         and compare the result with Hooke's law,            .

  Express your answer in terms of          ,   , and     .

  ANSWER:
                      =



  Part C
  Ten identical steel wires have equal lengths         and equal "spring constants" . The wires are connected end to end,
  so that the resultant wire has length        . What is the "spring constant" of the resulting wire?

   Hint C.1 How to determine the new spring constant
   Use the expression for the spring constant determined in Part B. From the expression derived in Part B, you can
   determine what happens to the spring constant when the length of the spring increases.

  ANSWER:




  Part D
  Ten identical steel wires have equal lengths         and equal "spring constants" . The wires are slightly twisted
  together, so that the resultant wire has length       and is 10 times as thick as each individual wire. What is the "spring
  constant" of the resulting wire?

   Hint D.1 How to determine the new spring constant
   Use the expression for the spring constant determined in Part B. From the expression derived in Part B, you can
   determine what happens to the spring constant when the area of the spring increases.

  ANSWER:




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  Part E
  Consider a steel guitar string of initial length             and cross-sectional area                     . The Young's

  modulus of the steel is                        . How far (   ) would such a string stretch under a tension of 1500     ?

  Express your answer in millimeters using two significant figures.

  ANSWER:                =



                                          A Bar Suspended by Two Wires
  Description: A nonuniform bar is suspended by two strings at different angles. Use balance of forces and torques
  to find the center of mass of the bar. (version for algebra-based courses)
 A nonuniform horizontal bar of mass      is supported by two massless wires against gravity. The left wire makes an
 angle   with the horizontal, and the right wire makes an angle . The bar has length .




  Part A
  What is the position of the center of mass of the bar, measured as distance    from the bar's left end?

   Hint A.1 The nature of the problem
   The bar is at rest. Therefore, both the net force and the net torque acting on the bar have zero magnitude.

   Part A.2    Find the sum of the x components of the forces

   Assume that the tensions in the left and right wires are    and    , respectively. Which expression represents
   accurately the sum           of the x components of the forces? Because the bar is at rest, these forces should sum
   to zero.




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   Use the sign convention indicated in the figure.

   ANSWER:


                                =




   Part A.3   Find the sum of the y components of the forces
   Assuming that the tensions in the left and right wires are    and    , respectively, which expression accurately
   represents the sum         of the y components of the forces? Because the bar is at rest, these components should
   sum to zero.
   Use the sign convention indicated in the figure.

   ANSWER:


                                =




   Part A.4   Find the sum of the torques about the left end of the bar
   The net torque is zero about any axis you select. Here we ask you to find the net torque of the system about the
   left end of the bar (you may, of course, try any other point if you like and check that it gives the same final
   answer for the center of mass ). Assume that the tensions in the left and right wires are        and , respectively.
   The weight of the bar is            . Using the sign convention shown in the figure, which expression accurately
   gives the sum of the torques about the left end of the bar.

   ANSWER:



                                 =




   Part A.5   Eliminate weight from your equations


   You should have three equations by now. It is possible to eliminate two variables and solve for in terms of the
   others. As an intermediate step, solve your torque equation for in terms of     , , , etc. Then solve your y-
   component force equation for        and substitute back into your expression for   . In other words, find an
   expression for   using the torque and       equations.




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   Express your answer in terms of             ,     ,    ,       , and   .

   ANSWER:
                        =


     You now have an expression for            that contains only             and   . Now you must eliminate    and     from
     your expression. Hint: Use what you know about the x components of the forces acting on the bar.

  Express your answer in terms of          ,        , and     .

  ANSWER:

                       =




                                          An Amusement Park Airplane Ride
  Description: Determine the strain on a rod that is attached to an amusement park airplane of known weight. When
  the system rotates about the center, students must determine if the temsion in the rod increases or decreases.
  (version for algebra-based courses)
 An amusement park ride consists of airplane-shaped cars
 suspended by steel rods. Each rod has length 15  and
 cross-sectional area 8.00       . Take Young's modulus

 for steel to be                      .




  Part A

  How much is the rod stretched (change in length of                  ) when the ride is at rest? Assume that each airplane with
  two riders has a total weight of 1900            and that the
  rods are vertical when the ride is at rest.

  Neglect the stretch due to the weight of the rod itself.




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   Part A.1    Calculate the tensile stress
   Calculate the tensile stress on the rod from the car hanging below it.

   Hint A.1.a Definition of tensile stress
   Tensile stress on a rod is given by the equation




   where       is the force directed along the length of the rod and   is the cross-sectional area of the rod.

   Express your answer in pascals.

   ANSWER:                   =


   Part A.2    Calculate the tensile strain
   Calculate the tensile strain given the tensile stress and Young's modulus.

   Hint A.2.a Definition of tensile strain
   The tensile strain on the rod is defined as


                                                                            ,


   where       is the original (unstressed) length of the rod and      is the elongation, that is, the length by which it
   stretches

   Hint A.2.b Definition of Young's modulus
   Young's modulus for a material is given by the equation


                                                                                     .


   Use this and the results for the tensile stress on the rod to determine the strain and thus the extension of the rod.

   ANSWER:                   =


     Now use the definition of tensile strain and solve for the elongation       .


  Express your answer in meters.

  ANSWER:                =




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  Part B
  When the ride starts operating as shown in the diagram, does the tension in the rods increase or decrease?

  ANSWER:               It increases.
                        It decreases.

       Note that the tensile force equals the tension force in the rod. When the ride is spinning, the rod is no longer
       vertical. Applying Newton's 2nd law in the vertical direction, we see that a component of the tension must
       support the weight of the car. This means that the tension is greater than the car's weight.

       Also, we have a confession to make: The mass of the rod in this case is comparable with the mass of the
       airplane so it does add to the total stretch. However, the contribution of the weight of the rod to the total
       stretch is not that easy to find because the mass is distributed over the length of the rod. In this case, we had to
       settle for a simplified model of the situation—something that physicists often have to do.




                                                 Moments around a Rod
  Description: Find the rotational acceleration about a bent rod that is fixed at a corner and has various forces
  applied at different points and angles. Problem is numerical with moment of inertia and forces specified.
 A rod is bent into an L shape and attached at one point to a pivot. The rod sits on a frictionless table and the diagram
 is a view from above. This means that gravity can be
 ignored for this problem. There are three forces that are
 applied to the rod at different points and angles: , ,

 and      . Note that the dimensions of the bent rod are in
 centimeters in the figure, although the answers are
 requested in SI units (kilograms, meters, seconds).




  Part A
  If           and              , what does the magnitude of       have to be for there to be rotational equilibrium?


   Part A.1      Finding torque about pivot from


   What is the magnitude of the torque |       | provided by     around the pivot point?




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   Give your answer numerically in newton-meters to two significant figures.

   ANSWER:          |   |=


  Answer numerically in newtons to two significant figures.

  ANSWER:               =          N


  Part B
  If the L-shaped rod has a moment of inertia                       ,              ,            , and again         , how long a
  time   would it take for the object to move through           (       /4 radians)?


  Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to
  the object.

   Part B.1    Find the net torque about the pivot
   What is the magnitude of the total torque |           | around the pivot point?

   Answer numerically in newton-meters to two significant figures.

   ANSWER:          |        |=


   Part B.2    Calculate

   Given the total torque around the pivot point, what is       , the magnitude of the angular acceleration?

   Hint B.2.a Equation for

   If you know the magnitude of the total torque (            ) and the rotational inertia (        ), you can then find the
   rotational acceleration (      ) from




   Express your answer numerically in radians per second squared to two significant figures.

   ANSWER:              =


   Hint B.3    Description of angular kinematics


   Now that you know the angular acceleration, this is a problem in rotational kinematics; find the time needed to go
   through a given angle .


   For constant acceleration (     ) and starting with          (where         is angular speed) the relation is given by




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   which is analogous to the expression for linear displacement (             ) with constant acceleration ( ) starting from
   rest,


                                                                          .


  Express the time in seconds to two significant figures.

  ANSWER:            =             s


  Part C
  Now consider the situation in which                   and              , but now a force with nonzero magnitude       is acting
  on the rod. What does           have to be to obtain equilibrium?

   Part C.1   Find the required component of

   Only the tangential (perpendicular) component of           (call it        ) provides a torque. What is    ?

   Answer in terms of            . You will need to evaluate any trigonometric functions.

   ANSWER:
                             =


  Give a numerical answer, without trigonometric functions, in newtons, to two significant figures.

  ANSWER:                =             N



  Summary     0 of 5 items complete (0% avg. score)
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Description: Horizontal bar originated in Germany. 18th century Western European countries appear to seize wire acrobatics big loop action, inspired by this 1811 German gymnast Yang in Berlin on the outskirts of Hudson Hyde gymnastics field instead of using a Mugang wire acrobatics show, the first installation the world's first deputy horizontal bar. 1812 will Mugang to iron, later changed to steel, the bar increases flexibility and endurance. 1920s as an independent event. Bar length 2.4 m, diameter 2.8 cm, 2.55 ~ 2.75 m high. Bars at both ends were fixed on the pillar. 1896 is listed as an Olympic sport.