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Homework Help: Classical mechanics problem

  1. Feb 18, 2010 #1
    1. The problem statement, all variables and given/known data
    heres a problem i working on
    a block slides on a horizontal surface which has been lubricated with a heavy oil such that the block experiences a viscous drag
    resistance in the opposite direction to the velocity and that varies with speed according to:
    [tex]F\left( v\right) =-c\,i\,{v}^{n}[/tex]
    if the initial speed is [itex] v_0 [/itex] at [itex] t=0 [/itex]
    a) find [itex]v[/itex] as a function of time [itex]t[/itex]
    which iv found ;
    [tex]{\left( {v_o}^{1-n}+\frac{c\left( n-1\right) \,t}{m}\right) }^{\frac{1}{1-n}}[/tex]
    b) For [itex]n = \frac{1}{2} [/itex] find the displacement x[itex]\vec{i} [/itex] as a function of t: i is the unit vectoer in x direction. .
    which iv found
    [tex] x- x_o = \frac{12\,{m}^{2}\,t\,vo+6\,c\,m\,{t}^{2}\,\sqrt{vo}+{c}^{2}\,{t}^{3}}{12\,{m}^{2}}[/tex]

    c) show that for [itex] n = 1/2 [/itex] , the block will not travel further than [tex] \frac{2mv_o^{\frac{3}{2}} }{3c} [/tex]
    now this is the sucker that will not work out. the answers for a and b are as given in the
    assignment sheet, so they are ok.
    2. Relevant equations



    3. The attempt at a solution
    give me a clue on how to do this
     
  2. jcsd
  3. Feb 18, 2010 #2

    ideasrule

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    I really don't think b) can be right. For one thing, x-x0 increases without bound as t increases, which implies the block will eventually reach infinite speed.
     
  4. Feb 18, 2010 #3

    kuruman

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    Use

    [tex]a=v\frac{dv}{dx}[/tex]

    and find v(x).
     
  5. Feb 18, 2010 #4


    correction
    the second equation that i found has a sign problem on the second term compared to the solution provided
    [tex] x- x_o = \frac{12\,{m}^{2}\,t\,vo-6\,c\,m\,{t}^{2}\,\sqrt{v_o}+{c}^{2}\,{t}^{3}}{12\,{m}^{2}}[/tex]
     
  6. Feb 19, 2010 #5

    kuruman

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    OK, do it your way. Can you find the velocity as a function of time from your expression? Can you find at what time that velocity goes to zero?
     
  7. Feb 20, 2010 #6
    so from the espression for [itex]f(v)= -cv^n [/itex]we can get using[itex] f=ma [/itex] one of newtons laws
    [itex]a= dv/dt=\frac{-cv^n }{m}[/itex]
    and a little separation of variables,
    [tex]v(t)= {\left( {v_o}^{1-n}+\frac{c\left( n-1\right) \,t}{m}\right) }^{\frac{1}{1-n}}[/tex]
    the integration was done much like that of obtaining newtons equations of motion, from 0 to t, and [itex]dv = v -v_o [/itex] but since the force is variable
    the acceleration was not treated as a constant.
    integrating this wrt time again we get
    [tex]x-x_o= \frac{t\,\left( 12\,{m}^{2}\,v_o-6\,c\,m\,t\,\sqrt{v_o}+{c}^{2}\,{t}^{2}\right) }{12\,{m}^{2}}[/tex]
    and we get this expression from v(t) at n=1/2
    [tex] v(t)={\left( -\sqrt{vo}-\frac{c\,t}{2\,m}\right) }^{2}[/tex]
    when the block stops moving v will be =0 hence we can solve for t at v=0
    [tex]t= \pm \frac{2\,m\,\sqrt{v_o}}{c}[/tex]
    inserting this into the equation for displacement x below that we found
    [tex]\frac{t\,\left( 12\,{m}^{2}\,v_o-6\,c\,m\,t\,\sqrt{v_o}+{c}^{2}\,{t}^{2}\right) }{12\,{m}^{2}}[/tex]
    we get after a little algebra (for the negative t, il try the positive t later)
    [tex]-\frac{14\,m\,{v_o}^{\frac{3}{2}}}{3\,c}[/tex]

    regards
    mechdude.
     
    Last edited: Feb 20, 2010
  8. Feb 20, 2010 #7

    kuruman

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    How do you get two times from

    [tex]
    0={\left( -\sqrt{vo}-\frac{c\,t}{2\,m}\right) }^{2}
    [/tex]

    *** Added on Edit ***

    Also, negative times are unphysical. Check your signs along the derivation of the expression above.
     
    Last edited: Feb 20, 2010
  9. Feb 20, 2010 #8
    Sorry, my bad, the expresion on the right is squared, so theres both negative and positive roots, but negative times are unphysical , il consider the positive root.
     
  10. Feb 20, 2010 #9

    kuruman

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    How do you get a positive root out of

    [tex]

    0={\left( -\sqrt{vo}-\frac{c\,t}{2\,m}\right) }^{2}

    [/tex]

    ??

    Check the math that got you to this point.
     
  11. Feb 26, 2010 #10
    starting from the expression for velocity;
    [tex] v(t)= {\left( \sqrt{v_o}-\frac{c\,t}{2\,m}\right) }^{2} [/tex]
    and expanding it [tex] v(t)= v_o-\frac{c\,t\,\sqrt{v_o}}{m}+\frac{{c}^{2}\,{t}^{2}}{4\,{m}^{2}} [/tex]
    now if we set v to zero (when it finally stops)
    [tex]0 = v_o-\frac{c\,t\,\sqrt{v_o}}{m}+\frac{{c}^{2}\,{t}^{2}}{4\,{m}^{2}} [/tex]
    and solve for t: using the quadratic formula:
    [tex] x=-b \pm \frac{\sqrt{{b}^{2}-4\,a\,c}}{2\,a}[/tex]
    [tex] t=\frac{2\,m\,\sqrt{v_o}}{c} [/tex]
    if we substitute this time into the expression for displacement:
    [tex] x-x_o= \frac{12\,{m}^{2}\,t\,v_o-6\,c\,m\,{t}^{2}\,\sqrt{v_o}+{c}^{2}\,{t}^{3}}{12\,{m}^{2}}[/tex]

    we then get [tex] x-x_o = \frac{2\,m\,{v_o}^{\frac{3}{2}}}{3\,c} [/tex]
    which is what was required.
     
  12. Feb 26, 2010 #11

    kuruman

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    That's better. Note that in the previous postings you had a minus sign in front of both terms between parentheses. Also, it is much quicker to take the square root on both sides

    [tex] 0=\pm {\left( \sqrt{v_o}-\frac{c\,t}{2\,m}\right) } [/tex]
    which clearly has one solution

    [tex]
    t=\frac{2\,m\,\sqrt{v_o}}{c}
    [/tex]
     
  13. Feb 26, 2010 #12
    thanks, i was not exactly cautious while working the problem,
    regards.
     
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