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Position vectors in different frames of reference

  1. Oct 1, 2012 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    I am doing a problem involving a man dropping a ball from the top of a mast of a ship at [itex] t =0 [/itex] a height [itex] h [/itex] above the origin of a ship's coordinate system.
    In the sea's frame of reference, the ship is moving with velocity [itex] u\hat{i} [/itex]. The origins of these two frames coincide at [itex] t =0 [/itex].
    The question asks to calculate the position, velocity and acceleration vectors (of the ball)and sketch the position of the ball in both frames of reference, as a function of t.

    3. The attempt at a solution
    So I got the position vector in the frame of the sea as [itex] \vec{h_s} = ut\hat{i} - \frac{1}{2}gt^2\hat{k}, [/itex] and that in the frame of the ship as [itex] \vec{h_b} = -\frac{gt^2}{2}\hat{k} [/itex]. How do you go about sketching these as functions of t?

    Many thanks
     
  2. jcsd
  3. Oct 1, 2012 #2
    In the ship's frame, that's simple, you have only one dimension for position so together with time you can do a 2D diagram. In the sea's frame, however, you would have to do a 3D diagram. But I think you could still do that in 2D by eliminating time from the equation.
     
  4. Oct 1, 2012 #3

    CAF123

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    Thanks. I have very roughly sketched the position from the ship's frame in Paint (see attached). Regarding the position from the sea, what do you mean by eliminating t?
     

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  5. Oct 1, 2012 #4
    At t = 0. z = h. Not zero as sketched.

    Regarding the ship's frame, observe that you have three coordinates: x, z and t. Both x and z depend on t. So you could express, for example, t as a function of x, then plug this into equation for z. Then you have z as a function of x. Which you can easily sketch on the xz plane, just like you sketched the motion in the ship's frame in the tz plane.
     
  6. Oct 1, 2012 #5

    CAF123

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    Eliminating t will give [itex] z = -\frac{g}{2u^2}x^2 [/itex], which again is a parabola in the xz plane. When x = 0, the ball is at z = 0, so I should start the graph there this time?

    The question asks for the graphs to be position as functions of time. Is there a way to do this for the sea's frame?
    Thanks
     
  7. Oct 1, 2012 #6
    In both cases z(0) = h. You forgot that constant in the equations.

    Any sketch will be a planar projection. tx, tz and xz are easiest. You could do a dimetric projection but that requires some skill.
     
  8. Oct 1, 2012 #7

    CAF123

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    ok, thanks. The parabola in the xz plane (sea frame) will be the same parabola as that in the tz plane (ship frame), right?

    For the constant, I understand where it comes from in the ships frame;
    [itex] \vec{h_b} = - \frac{gt^2}{2}\hat{k} => h_f - h = -\frac{gt^2}{2}, [/itex] so at t=0, [itex] h_f = h [/itex] .
    How does it arise when we eliminate t to get [itex] z = -\frac{g}{2u^2}x^2? [/itex]
     
  9. Oct 1, 2012 #8
    Well, not exactly. x = ut, so there is a scaling factor AND the units of t and x are arbitrary. So you can make it exactly the same, or you can stretch it in any cardinal direction.

    If you read the problem carefully, you will note "at t=0 a height h above the origin of a ship's coordinate system". So that means your very first equation should be ## \vec{h}_b = (h - \frac {gt^2}{2}) \vec{k} ##. Hence, ## z = h - {gt^2}{2} ## in either frame.
     
  10. Oct 2, 2012 #9

    CAF123

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    What I should really have is;
    [itex] \vec{h_b} = (h-\frac{gt^2}{2})\hat{k} [/itex] for ship's frame and [itex] \vec{h_s} = ut\hat{i} + (h - \frac{gt^2}{2})\hat{k} [/itex] for sea's frame. Thanks for correcting me.
    I was wondering if it was possible to get these equations by integration methods?
     
  11. Oct 2, 2012 #10
    Yes, it should be possible. In the ship's frame, the full integral of motion would be z = A + Bt - gt^2/2, where A and B are to be determined from the initial conditions.
     
  12. Oct 2, 2012 #11

    CAF123

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    I realise it is quite simple for the ship's frame, since an observer at the origin would be stationary relative to this frame. However, for the sea's frame, the ship is moving with some velocity so how do I incorporate that into the integration process, starting with [itex] \vec{a(t)} = -g\hat{k}?[/itex]
     
  13. Oct 2, 2012 #12
    That would be the same. You would simply have two equations: x'' = 0, z'' = -g, and you would end up with three integration constants, again to be determined from the initial conditions. But knowing that z and x are independent, you could use the solution in the ship's frame right away as part of this solution.
     
  14. Oct 2, 2012 #13

    CAF123

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    I tried this way;
    In the ship's frame,[tex] \int_{0}^{v(t)} dv = \int_{0}^{t} -g dt ,[/tex] which gives [tex] \vec{v(t)} - \vec{v(0)} = -gt\hat{k}[/tex] v(0) is 0 so [itex]\vec{v(t)} = -gt\hat{k}. [/itex]Similarly, [tex] \int_{0}^{s(t)} ds= \int_{0}^{t} -gt dt => \vec{s(t)} - \vec{s(0)} = -\frac{gt^2}{2}\hat{k} [/tex] But s(0) = h so [itex] \vec{s(t)} = (h- \frac{gt^2}{2})\hat{k}.[/itex]

    In the sea's frame, considering the x direction [tex] \int_{0}^{v} dv = \int_{0}^{t} dt => \vec{v(t)} - \vec{v(0)} = u\hat{i} => \vec{v(t)} =u\hat{i} [/tex]. Then, [tex] \int_{0}^{s(t)} ds= \int_{0}^{t} u dt => \vec{s(t)} - \vec{s(0)} = ut\hat{i} => \vec{s(t)} = ut\hat{i}[/tex] Similar argument in z direction. So the total expression is found by adding these two expressions together? (the one in x and the one in z)
     
  15. Oct 2, 2012 #14
    This is not correct. It should be [tex] \int_{0}^{v} dv = \int_{0}^{t} 0 \cdot dt = 0 [/tex] That will yield constant velocity, as expected.

    You could start with [tex] \int_{0}^{v} dv = \int_{0}^{t} [0 \cdot \hat{i} - g \hat{k}]dt [/tex]
     
  16. Oct 2, 2012 #15

    CAF123

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    The integral of zero for an indefinite integral is an arbitrary constant. So here we have;
    [tex] \int_{0}^{v(t)}dv = 0 => \vec{v(t)} - \vec{v(0)}= 0 => \vec{v(t)} = \vec{v(0)} = u\hat{i} [/tex] is that fine now?
     
  17. Oct 2, 2012 #16
    Yes, that is correct.
     
  18. Oct 2, 2012 #17

    CAF123

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    Thanks for all your help.
    Just one more thing: why would the result of this question might have proven crucial for Galileo when he theorised the heliocentric model of the solar system, I.e that the earth goes around the sun?
    I thought it could be because of the varying expressions for the position vector => there is no universal frame of reference, they are all equally valid within their own frame.
    Does this mean from our frame on earth, the sun is moving, while a frame from the sun, it is earth that is moving?
     
  19. Oct 2, 2012 #18
    This is from his book "Dialogue Concerning the Two Chief World Systems". One argument pro immobility of the Earth, ascending to Aristotle, is that a stone thrown upward, or dropped from a tower, falls back (or to the bottom of the tower), whereas, on a moving Earth, it would have landed way behind due to the motion of the Earth. So Galileo invokes the argument with the ship.
     
  20. Oct 2, 2012 #19

    CAF123

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    'Does this mean from our frame on earth, the sun is moving, while a frame from the sun, it is earth that is moving?'
    Is this correct? And we know that the earth actually does go around the sun, so this was known because of an external observer, I presume?
     
  21. Oct 2, 2012 #20
    The particular argument was about, in modern language, indistinguishability of inertial frames of reference by mechanical experiments. The more general argument was that the Earth was orbiting the Sun, not the other way around. However, he failed to give a sound proof of the more general argument, so the entire book might be regarded as an exposition of (Galilean) relativity, even though that most certainly was not its purpose.
     
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