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Coriolis force of a stone throw

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data
    A stone is left to fall from a 80 meters high tower on the equator.
    How far in front of the tower it will fall.

    2. Relevant equations
    The angular velocity of the earth: ω=7.27E-5 [rad/sec]
    It reaches the ground in 4 seconds.

    3. The attempt at a solution
    This problem is solved, in a book, using Coriolis acceleration formulas. the result is that the stone falls approximately 1.5 centimeters in front of the tower.
    But there was also a general explanation which says that since the tangential velocity is higher at the top of the tower than it is at the bottom, it will fall some distance in front.
    I used this explanation and solved:
    7.27E-5 [rad/sec]x80 [m]x4[sec]=0.023[m]=2.3[cm]
    This is greater than the result based on the Coriolis calculation.
    How to combine the two methods?
    The direction of the distancing due to the difference in tangential velocities is eastward, and the coriolis force also acts eastward, so, if the two methods should be added, i should have gotten a smaller result.
    What do i need the coriolis force, if the difference in velocities explains the distancing?
     
    Last edited: Apr 30, 2013
  2. jcsd
  3. Apr 30, 2013 #2

    jambaugh

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    The change in tangential velocities as height (above rotation axis) changes is the coriolis effect so you shouldnt add. Rather each done correctly should give the same answer.

    One is adopting the accelerating frame which rotates with the earth, Coriolis is then an effective force; and the other adopts an inertial frame, moving with the surface tangentially and thus has an added velocity atop the tower.

    Can you give the details of the "book" method?
     
  4. May 1, 2013 #3

    TSny

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    As jambaugh said, if you work it out from the perspective of an inertial frame (so, no Coriolis force) then you should get the same result as working it out in the rotating frame of the earth (with the Coriolis force.)

    Suppose you choose your inertial frame so that the base of the tower is at rest in the inertial frame at the instant the stone is dropped. Let the x axis be parallel to the ground and toward the east. Then at the instant the stone is dropped, the stone will have a velocity of ωh in the x-direction in the inertial frame. ω is the angular speed of rotation of the earth and h is the height of the tower.

    In order to get the correct answer for where the stone strikes the earth, you will need to take into account the x-component of the acceleration due to gravity as the stone falls. I find this to be rather amazing!
     
    Last edited: May 1, 2013
  5. May 1, 2013 #4
    Why is the inertial frame moving with the surface? it should be fixed, for example with an origin in the sun
     
  6. May 1, 2013 #5
    There is no x-component to the acceleration, it's directed towards the center of the earth, exactly parallel to the y axis
     
  7. May 1, 2013 #6

    TSny

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    The acceleration of the stone is always toward the center of the earth. In the inertial frame chosen as moving with the base of the tower at t = 0, the center of the earth is moving in the negative x-direction! So, the center of the earth moves off of the y-axis during the 4 seconds of fall and the acceleration of the stone picks up a negative x component. It's amazing to me that this contributes significantly to where the stone lands :smile:
     
  8. May 1, 2013 #7

    TSny

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    You can certainly pick the inertial frame any way you want. For example, when I first worked it out, I chose an inertial frame with origin at the center of the earth. So, in this non-rotating frame, the center of the earth remains at the origin and the earth rotates around the origin.

    There is not need to worry about the orbital motion of the earth around sun.
     
    Last edited: May 1, 2013
  9. May 2, 2013 #8
    Inertial Frame

    Once you tell me the inertial frame is fixed at the base of the tower and moving with it. how can an inertial frame be moving? then you offer the inertial frame fixed at the center of the earth.
    In both cases the calculation:
    7.27E-5 [rad/sec]x80 [m]x4[sec]=0.023[m]=2.3[cm]
    Seems to me correct, and i don't have to take into consideration the radius of the earth since it reduces: either if the inertial frame is fixed at the base and moving, or it is at the center.
    And this result, as i said, is bigger than the one received from the point of view of a rotating frame.
    If it is possible to select the inertial frame as moving or fixed, i want to select it so that the rotation movement will also be considered, not only the differences in tangential velocities.
     
  10. May 2, 2013 #9
    There is no such thing as a "fixed" internal frame. All you can say about internal frames is that they move with constant velocities relative to each other. So it is entirely possible to have an inertial frame whose velocity is equal to that of the towers's base at the instant of the drop. Note that immediately after that the velocity of the tower's base is different, so it does not stay at rest in this inertial frame.
     
  11. May 2, 2013 #10

    jambaugh

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    Adding to what has already been said, an inertial frame is a frame undergoing no acceleration. In this context we use Galilean relativity..."who's to say which observers are 'really' stationary?"

    You could pick a frame moving with the Sun but the Earth's orbital speed is irrelevant to this problem (and its orbital acceleration inconsequential.). If you did, why not the center of the Milky Way? Or our local galactic cluster.

    But understand the difference between the inertial frame and the rotating frame. The road and the wheel so to speak.
     
  12. May 2, 2013 #11
    From the rotating frame

    The method in the book:
    The small notations F and R note in the fixed and rotating frames.
    [tex]\vec{a}_{R}=\vec{a}_{F}-2\vec{\omega}\times\vec{V}_{R}-\vec{\omega}\times\left(\vec{\omega}\times\vec{r}\right)[/tex]
    [tex]\vec{a}_{R}=-g\hat{z}[/tex]
    [tex]\vec{\omega}=-(\omega\cdot\sin\theta)\hat{x}+(\omega\cos\cdot\theta)\hat{z}[/tex]
    [tex]\vec{r}_{R}=x\hat{x}+y\hat{y}+z\hat{z}[/tex]
    And:
    [tex]\vec{V}_{R}=\dot{x}\hat{x}+\dot{y}\hat{y}+\dot{z}\hat{z}[/tex]
    And:
    [tex]\vec{a}_{R}=\ddot{x}\hat{x}+\ddot{y}\hat{y}+\ddot{z}\hat{z}[/tex]

    Because ω is small, we neglect the term:
    [tex]\vec{\omega}\times\left(\vec{\omega}\times\vec{r}\right)[/tex]

    We substitute all in the first equation, with θ=90 at the equator, compare components, take integrals in order to find the velocities and finally get 1.5 cm distance.

    With my calculation at the inertial frame:
    7.27E-5 [rad/sec]x80 [m]x4[sec]=0.023[m]=2.3[cm]

    The result is bigger. where's my mistake?
     
    Last edited: May 2, 2013
  13. May 2, 2013 #12

    TSny

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    The mistake is not taking into account the horizontal component of acceleration of the stone in the inertial frame during the 4 seconds of fall.

    It will help if you state how you set up your inertial frame.
     
    Last edited: May 2, 2013
  14. May 3, 2013 #13

    jambaugh

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    When I integrated this I got the same 2.3cm answer. Let's see:
    I used the following format: [itex] \ddot{r} = -g, \ddot{x} = 2\omega \dot{r}[/itex]
    [tex] \dot{r} = -gt + \dot{r}_0=-gt[/tex]
    [tex] \ddot{x} = -2g\omega t[/tex]
    [tex] \dot{x} = -g\omega t^2 + \dot{x}_0 = -g\omega t^2[/tex]
    [tex]x = -\frac{g}{3}\omega t^3 [/tex]
    Absorb the negative into the direction choice of x and you have the magnitude of lateral displacement.

    I used polar (or cylindrical) coordinates and worked with components then took [itex]x=R\theta[/itex] with fixed R given Earth's radius >> change in r.

    You should get the same working in the vector format you used. I think your error is in your integration, did you get the t^3/3 factor as I did?
     
    Last edited: May 3, 2013
  15. May 3, 2013 #14

    TSny

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    [itex]\frac{g}{3}\omega t^3 [/itex] should evaluate to approximately 1.5 cm.
     
  16. May 3, 2013 #15

    TSny

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    Suppose we choose our inertial frame to be comoving with the center of the earth and origin at the center of the earth. Then, in our inertial frame, the earth is rotating about the origin of our coordinate system.

    At the moment the stone is dropped, suppose the tower is coincident with the y-axis of our inertial frame. At this moment the ball and tower are moving in the x-direction with a velocity of about 1000 mi/h, with the ball moving about 6 mm/s faster than the base of the tower! As the ball is in free fall, it will pick up a negative x-component of acceleration of gravity as shown in the attachment. (Note how the green acceleration of gravity vector has a negative x-component.)

    Thus, the ball does not travel as far in the x-direction as you would expect if you neglect this x-component of acceleration.
     

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  17. May 3, 2013 #16

    jambaugh

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    Hmmm... I should have checked my numbers. Yesterday I got 2.3 but I didn't save my scratch notes so I worked it out again. Let me work it in full generality to see what factor difference we're getting.

    Dropping a height h with no initial velocity means [itex] h = 1/2 g t^2[/itex] so [itex] t = \sqrt{2h/g}[/itex].

    With the inertial frame method, we get [itex]v=\omega r[/itex] so [itex]\Delta v = \omega h[/itex] (linear in r so easy.) The lateral displacement will be...
    [tex]\Delta x=\omega h t =\sqrt{2} \omega h^{3/2}g^{-1/2}[/tex]

    with the Coriolis integration I got...
    [tex]\Delta x = \frac{1}{3}g\omega t^3 = \frac{2\sqrt{2}}{3} g^{-1/2}\omega h^{3/2}[/tex].
    Hmmm... unitally they both "add up" but there's a factor of 2/3 difference.
    Methodologicallly...

    the factor of 2 in the Coriolis force comes from the two identical terms when differentiating the velocity, you get one term from the [itex]r\omega \hat{\theta}[/itex] tangential velocity and the other from the [itex] \dot{r}\hat{r}[/itex] term. The first can be thought of as due to the change in radius within the radial dependency of the tangential velocity expressed in terms of constant [itex]\omega[/itex]. The second term is the rotation of the radial velocity so that over time it acquires a tangential component due to the rotating frame. Both should be accounted for here.

    The answer may lie in the time dependency of the basis in the integration... Let me work on it and get back to you (I have some free time now).
     
  18. May 3, 2013 #17
    Calculating the negative x component

    I don't know how to calculate the x component, since the acceleration isn't steady.
    I will try. according to the attached drawing:
    The accelaration is in the x direction.
    [tex]a=g\cdot\sin\theta=g\sin(\omega t)[/tex]
    The constant acceleration formula:
    [tex]x=\frac{1}{2}a\cdot t^2[/tex]
    [tex]dx=\frac{1}{2}g\sin(\omega dt)\cdot dt^2[/tex]
    [tex]x=\frac{g}{2}\int_{0}^{4} \sin(\omega t)\cdot t^2[/tex]
    I know from integrals tables that:
    [tex]\int x^2 \sin x=2x\sin x-(x^2-2)\cos x[/tex]
    But i don't know to handle the (ωt) and t that appear in the integral. this is a mathematical question.
    But, is the method good?
     

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  19. May 3, 2013 #18

    TSny

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    Since θ will remain small during the time of fall, you may approximate sinθ ≈ θ.
     
  20. May 3, 2013 #19

    jambaugh

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    OK, I'm re-deriving from scratch, using polar coordinates with polar basis the position of a point on the tower is [itex]\vec{r}=r\cdot \hat{r}[/itex] but for an object at a given height but laterally displaced we get [itex]\vec{r}=r \hat{r}+x\hat{\theta}[/itex].

    The basis time rate of change is:[itex]\dot{\hat{r}}=\omega\hat{\theta}, \dot{\hat{\theta}} = -\omega\hat{r}[/itex].

    [tex]\dot{\vec{r}}=\dot{r}\hat{r}+r\dot{\hat{r}}+ \dot{x}\hat{\theta}+x\dot{\hat{\theta}}=
    (\dot{r}-\omega x)\hat{r} +(r\omega + \dot{x})\hat{\theta}[/tex]
    now acceleration:
    [tex]\ddot{\vec{r}}=(\ddot{r}-\omega\dot{x})\hat{r}+(\dot{r}-\omega x)\dot{\hat{r}}+(\dot{r}\omega + \ddot{x})\hat{\theta}+(r\omega+\dot{x})\dot{\hat{\theta}}=[/tex]
    [tex]=(\ddot{r}-2\omega\dot{x}-r\omega^2)\hat{r}+(\ddot{x}+2\omega\dot{r} -x\omega^2 )\hat{\theta}[/tex]
    OK, that's acceleration which must equal gravity from the dropped object to the center of the Earth. As TSny points out this is not wholly in the [itex]-\hat{r}[/itex] direction. To be super exact it will be in the direction:
    [itex] -\frac{1}{\sqrt{r^2 + x^2}}(r\hat{r}+x\hat{\theta})[/itex].
    So far this is exact. Taking the first approximation:[itex]x << r[/itex]
    [tex]\ddot{\vec{r}}\approx -g(\hat{r}+x/r\hat{\theta})[/tex]
    Here's the offset term TSny mentioned. Note that it is not considered in either method so I don't think it enough to account for the difference. I'm a little fuzzy headed right now but this should give the right answer and so far it looks to me like the extra terms are too small to matter. I'm inclined to trust the smaller number at the moment where before I was suspicious of it and trusting the larger. I think my bar napkin arithmetic earlier was conveniently incorrect to get the same numbers but I didn't save my work.

    I WILL work this out to my understanding and post.
     
  21. May 6, 2013 #20

    TSny

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    The x-component of the acceleration of the stone is ##a_x = -gsin(\theta) = -gsin(\omega t) \approx -g\omega t## using the small angle approximation sinθ ≈θ.

    So, the x-component of acceleration is just proportional to ##t##. So you can easily integrate it to find the horizontal distance x traveled by the stone.

    You will need to take into account that the initial velocity of the stone in the x-direction is ω(R+h) where R is the radius of the earth and h is the height of the tower.

    To the level of approximation that you need, the base of the tower moves in the x-direction a distance Rsinθ ≈ Rθ = Rωt.

    You want to find the difference between the x-distance traveled by the stone and the x-distance traveled by the base of the tower.
     
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