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Homework Help: A 3d Trajectory

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle of mass 1kg is projected in XYZ space, where Gravity (g=10m/s2) acts in -[tex]\hat{k}[/tex] direction. The initial velocity of a particle is [tex]\vec{u}[/tex]=(-3[tex]\hat{i}[/tex]+4[tex]\hat{j}[/tex])m/s.
    x-component of acceleration = 3t/4
    y-component of acceleration = -1 - 3t/4
    If total work done in interval t=0 to t=4 seconds is 90K Joules, then find the value of K.

    [The format of answering requires K to be an integer between 0 and 9 (inclusive)]


    2. Relevant equations
    Basic Kinematic definitions with complicated level of Calculus


    3. The attempt at a solution
     
  2. jcsd
  3. Mar 18, 2010 #2
    You should show some attempt at the solution. Without that we are not permitted to help.
    What about actually citing relevant equations and substituing them?
     
  4. Mar 18, 2010 #3
    What I tried was:

    ax = 3t/4
    vx = -3 + 3t2/4
    Fx = ma = 3t/4
    Px = Fx vx = -9t/4 + 9t3/32

    ay = -1 - 3t/4
    vy = -t + 3t2/8 +4
    Fy = ma = -1 - 3t/4
    Py = Fy vy = 9t3/32 + 9t2/8 - 2t - 4

    az = -10
    vz = -10t
    Fz = ma = -10
    Pz = Fz vz = 100t

    P = Px + Py + Pz = -9t/4 + 9t3/32 + 9t3/32 + 9t2/8 - 2t - 4 + 100t = 9t3/16 + 9t2/8 + 383t/4 - 4

    [tex]W = \int P dt[/tex]

    W = 9t4/64 + 3t3/8 + 383t2/8 - 4t

    Work Done form 0 sec to 4 sec = W(4)-W(0) = 9(4)4/64 + 3(4)3/8 + 383(4)2/8 - 4(4) = 36 + 24 + 766 - 16 = 810 = 90(9)

    Hence K=9[\b]


    Please tell if this is correct...
     
  5. Mar 19, 2010 #4
    you made a typo in vx, however your Px is okay.

    Your calculation is otherwise correct.
     
  6. Mar 19, 2010 #5
    Oh yes, it should be vx = -3 + 3t2/8

    But is there any smarter method which is less vulnerable to calculation errors?
     
  7. Mar 19, 2010 #6
    I used to avoid calculation errors by using a math package and always denoting the units.
    Your example in sympy:
    Code (Text):

    $ isympy
    Python 2.6.4 console for SymPy 0.7.0-git

    These commands were executed:
    >>> from __future__ import division
    >>> from sympy import *
    >>> x, y, z = symbols('xyz')
    >>> k, m, n = symbols('kmn', integer=True)
    >>> f, g, h = map(Function, 'fgh')

    Documentation can be found at http://sympy.org/

    In [1]: s,m,kg=symbols("s,m,kg",real=True,positive=True)

    In [2]: v0=Matrix([-3*m/s,4*m/s,0*m/s])

    In [3]: t=symbols("t",real=True,positive=True)

    In [4]: a=Matrix([3*t/4*m/s**3,-1.0*m/s**2-3*t/4*m/s**3,-10.0*m/s**2])

    In [5]: v=v0+integrate(a,(t,0,t))

    In [6]: F=a*1*kg
    In [7]: a
    Out[7]:
    ⎡   3⋅m⋅t    ⎤
    ⎢   ─────    ⎥
    ⎢       3    ⎥
    ⎢    4⋅s     ⎥
    ⎢            ⎥
    ⎢  3⋅m⋅t   m ⎥
    ⎢- ───── - ──⎥
    ⎢      3    2⎥
    ⎢   4⋅s    s ⎥
    ⎢            ⎥
    ⎢  -10.0⋅m   ⎥
    ⎢  ───────   ⎥
    ⎢      2     ⎥
    ⎣     s      ⎦

    In [8]: v
    Out[8]:
    ⎡               2  ⎤
    ⎢    3⋅m   3⋅m⋅t   ⎥
    ⎢  - ─── + ──────  ⎥
    ⎢     s        3   ⎥
    ⎢           8⋅s    ⎥
    ⎢                  ⎥
    ⎢                 2⎥
    ⎢4⋅m   m⋅t   3⋅m⋅t ⎥
    ⎢─── - ─── - ──────⎥
    ⎢ s      2       3 ⎥
    ⎢       s     8⋅s  ⎥
    ⎢                  ⎥
    ⎢    -10.0⋅m⋅t     ⎥
    ⎢    ─────────     ⎥
    ⎢         2        ⎥
    ⎣        s         ⎦

    In [9]: F
    Out[9]:
    ⎡    3⋅kg⋅m⋅t     ⎤
    ⎢    ────────     ⎥
    ⎢         3       ⎥
    ⎢      4⋅s        ⎥
    ⎢                 ⎥
    ⎢   ⎛  3⋅m⋅t   m ⎞⎥
    ⎢kg⋅⎜- ───── - ──⎟⎥
    ⎢   ⎜      3    2⎟⎥
    ⎢   ⎝   4⋅s    s ⎠⎥
    ⎢                 ⎥
    ⎢   -10.0⋅kg⋅m    ⎥
    ⎢   ──────────    ⎥
    ⎢        2        ⎥
    ⎣       s         ⎦

    In [10]: pp=[]

    In [11]: for i in range(3):  pp.append((F[i]*v[i]).expand())
       ....:

    In [12]: P=Matrix(pp)
    In [13]: P
    Out[13]:
    ⎡                        2         2  3               ⎤
    ⎢                9⋅kg⋅t⋅m    9⋅kg⋅m ⋅t                ⎥
    ⎢              - ───────── + ──────────               ⎥
    ⎢                      4           6                  ⎥
    ⎢                   4⋅s        32⋅s                   ⎥
    ⎢                                                     ⎥
    ⎢          2           2             2  2         2  3⎥
    ⎢  2⋅kg⋅t⋅m    4.0⋅kg⋅m    1.125⋅kg⋅m ⋅t    9⋅kg⋅m ⋅t ⎥
    ⎢- ───────── - ───────── + ────────────── + ──────────⎥
    ⎢       4           3             5               6   ⎥
    ⎢      s           s             s            32⋅s    ⎥
    ⎢                                                     ⎥
    ⎢                                2                    ⎥
    ⎢                    100.0⋅kg⋅t⋅m                     ⎥
    ⎢                    ─────────────                    ⎥
    ⎢                           4                         ⎥
    ⎣                          s                          ⎦

    In [14]: P_sum=P[0]+P[1]+P[2]

    In [15]: P_sum
    Out[15]:
                2           2             2  2         2  3
    95.75⋅kg⋅t⋅m    4.0⋅kg⋅m    1.125⋅kg⋅m ⋅t    9⋅kg⋅m ⋅t
    ───────────── - ───────── + ────────────── + ──────────
           4             3             5               6  
          s             s             s            16⋅s    

    In [16]: W=integrate(P_sum,t)

    In [17]: W
    Out[17]:
                2              2  2             2  3         2  4
      4.0⋅kg⋅t⋅m    47.875⋅kg⋅m ⋅t    0.375⋅kg⋅m ⋅t    9⋅kg⋅m ⋅t
    - ─────────── + ─────────────── + ────────────── + ──────────
            3               4                5               6  
           s               s                s            64⋅s    

    In [18]: Wsum=integrate(P_sum,(t,0*s,4*s))

    In [19]: Wsum
    Out[19]:
              2
    810.0⋅kg⋅m
    ───────────
          2    
         s    


     
     
  8. Mar 19, 2010 #7
    but i cant really use computer during exams!
     
  9. Mar 20, 2010 #8
    You can check units also by hand.
    And you can practice.
     
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