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The total travel time along a frictionless path?

  1. Apr 2, 2010 #1
    Hi,
    I have this problem about an object moving along a track due to a uniform gravitational field. It moves from being at rest in the higher position A (xA,yA) to the lower position B (xB,yB). There is no friction at all. What I am supposed to be doing is showing that the total time from A to B is given by:

    [tex] t_t_o_t = \frac{1}{\sqrt{2g}}\int dt \sqrt{\frac{(dx/dt)^2 + (dy/dt)^2}{y}}[/tex]
    where the limits of the integral are given as between (xA,yA) and (xB,yB).

    The limits I do not quite understand, because surely they should really be times rather than positions if we are integrating with respect to time?

    Anyway, what I have attempted is this:
    From Newton's equation of motion, v2 = u2 + 2as, if we set initial velocity u = 0, gravitational acceleration a = g and vertical displacement s = y, then we have an expression for the object's final velocity, [tex]v = \sqrt{2gy}[/tex].

    Also, I said that the instantaneous velocity will be given by [tex] v_i = \sqrt{(dx/dt)^2 + (dy/dt)^2} [/tex].

    Since these two things match the denominator and numerator in the integrand respectively, they must be important. Clearly vi/v = 0 at time t=0 and position A, and increases to vi/v = 1 at time t=ttot and position B.


    This is where I start to hit trouble though. I'm not really sure what to make of these spatial limits and I can't see the way forward. One thing I have tried is to say that

    [tex] (\sqrt{(dx/dt)^2 + (dy/dt)^2})dt = \sqrt{dt^2(dx/dt)^2 + dt^2(dy/dt)^2} = \sqrt{(dx)^2 + (dy)^2} = dy\sqrt{(dx/dy) + 1}[/tex]

    but when I proceed with the integral integrating with respect to y from there, I end up with a quantity that does not have dimensions of time.

    Can anyone steer me in the right direction?
    Many thanks to those who try.
     
  2. jcsd
  3. Apr 2, 2010 #2
    What is the direction of the instantaneous velocity

    [tex]v_i[/tex]

    and compare with your final velocity direction

    [tex]v[/tex]

    If the dt outside the square root is moved inside the square root, as you have shown, what are the units of the integrand? Then determine the units of

    [tex]\frac{1}{\sqrt{2g}}[/tex]

    You should find the units of the expression is time.
     
  4. Apr 4, 2010 #3

    Cyosis

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    Homework Helper

    Start out with ds=vdt. From this it follows that [itex]t=\int \frac{ds}{v} [/itex]. You know that [itex]ds=\sqrt{dx^2+dy^2}[/itex] which you can rewrite to something in terms of dt as you have already shown in your last line of equations. Now you just have to calculate v. Hint: use conservation of energy.
     
  5. Apr 6, 2010 #4
    Thanks guys.

    What I've tried now is this:-
    In the equation ds = vdt, this velocity is not a final or initial velocity as are used in Newton's equations of motion, it is some instantaneous velocity v=v(t) = v(y).

    Also, kinetic energy Ek = 0.5mv2 and potential energy Ep = -mgy.

    Potential energy is at a maximum of zero when the object is at its highest, and as it falls along its path it becomes more negative. At t=0, velocity is also 0, so that the sum of the kinetic and potential energies is zero, ie. Ek + Ep = 0, which is a constant.

    This means that at some downwards displacement y, we have 0.5mv2 - mgy = 0 at all times. This leads directly to v=(2gy)1/2, which then slots into my final expression for t=ttot.

    Am I justified in saying this?
     
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