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Displacement equals to zero?

  1. Aug 21, 2005 #1
    erm just wanna clarify this.If one were to move in a circular motion,and u complete one revolution thus meaning you come back to the same point to where you started off,is your displacement equals to zero??
     
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  3. Aug 21, 2005 #2

    Tide

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    Yes! Technically, it is the zero vector which is not the same as zero (a scalar) since vectors and scalars are different entities.
     
    Last edited: Aug 21, 2005
  4. Aug 21, 2005 #3

    jtbell

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    Displacement is defined as the difference between the final position and the initial position of the object, regardless of the path the object follows in between. So if the object ends up returning to its initial position, its net displacement is zero.
     
  5. Aug 21, 2005 #4
    Mathematically, let [itex]r[/itex] be the radius of your circle, which defines the motion of your object. Circles are 2D figures, so we can represent the object's motion in a plane; I will use 2D Cartesian for this example. The location of its center is irrelevant (as it remains constant), so let the center be at the origin (0,0).

    Therefore, the object's position can be represented via the position vector
    [tex] \vec P = r\left( {\hat i\cos \theta + \hat j\sin \theta } \right) [/tex]
    where [tex] \hat i [/tex] & [tex] \hat j [/tex] represent unit vectors in the [itex] x [/itex] and [itex] y [/itex] directions, respectively.

    Simply put, you can represent the object position as the head of a position vector whose tail is the center of the circle, which we set at (0,0). It has a magnitude [itex] r [/itex] with an "[itex] x [/itex]" component equivalent to [tex] r \hat i \cos \theta [/tex] and a
    "[itex] y [/itex]" component equivalent to [tex] r \hat j \sin \theta [/tex], where [tex] \theta [/tex] is the angle swept counterclockwise from the positive x-axis, with the vertex being the origin.

    Displacement is a vector quantity associated with a change in position. Where [itex] \vec P [/itex] is the position vector, displacement is defined as [itex] \Delta \vec P [/itex].

    In a "complete revolution", the [tex] \Delta \theta = 360^\circ = 2\pi [/tex]. If we let [tex] \theta _0 [/tex] represent the initial angle relevant to the object's position vector, then in a complete revolution:
    [tex] \Delta \vec P = r\Delta \left( {\hat i\cos \theta + \hat j\sin \theta } \right) = r\left\{ {\left[ {\hat i\cos \left( {\theta _0 + 2\pi } \right) + \hat j\sin \left( {\theta _0 + 2\pi } \right)} \right] - \left( {\hat i\cos \theta _0 + \hat j\sin \theta _0 } \right)} \right\} = 0 [/tex]
    Thus, you will have ZERO displacement.:smile:

    And when you think about it, much of what I wrote is completely unnecessary! Though it does, "sort of" provide an answer from a mathematical perspective...err, though it might have been unnecessary. :frown: sorry...
     
    Last edited: Aug 21, 2005
  6. Aug 21, 2005 #5
    cool i needed a proof like this,thx guys.well what set me pondering me was that my friends said if u could imagine the circle as a string,and now you take it and seperate and extend into a straight line,you will get displacement.This of course doesnt make sense.In addition, one of them pointed out that when the particle comes back to the same point in a circular motion,its tangential velocity is pointing in the same direction as it first started off.Now what he claimed was that when one walked in a straight line,in order to move back to where you started off,you have to change direction,hence your velocity must point in the opposite direction.That was what set me thinking.lol.anyway thanks!
     
  7. Aug 21, 2005 #6

    arildno

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    I believe your friend muddled up the concepts of displacement and TRAVERSED DISTANCE.
     
  8. Aug 21, 2005 #7

    PerennialII

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    Perhaps he was thinking about a displacement which causes something along the way.
     
  9. Aug 25, 2005 #8
    ok i think i need to revist this question.let say its a circular motion with both linear accleleration and a radial acceleration,and the object starts from rest,i'm told to find the angle made by the total acceleration with the radial acceleration after the object completes one round of revolution.

    we know that the radial acceleration is a=v^2/r,but what about the linear acceleration??
    what my tutor did and set me puzzling was that he used kinematic equation:
    v^2=u^2+2as,to solve for the linear accleration.Hence s,which is the displacement,turns out to be 2pi*r which is the circumference of the circle.How can this equation be used if we conclude that the displacment is zero for one revolution???can we treat the linear portion as a particle moving in a straight line?
    can someone enlighten me??
     
  10. Aug 25, 2005 #9

    Doc Al

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    Treat "s" as the distance along the circumference measured from the starting point. All the usual kinematic formulas for uniformly accelerated motion apply (assuming the tangential acceleration is constant).

    Yes, the displacement of the object after making one complete revolution is zero. This means that the average velocity (and acceleration) over such a cycle is zero. This should make sense: the object keeps coming back to where it started.

    But what you want to know is not the average acceleration, but the instantaneous acceleration at some point. This is certainly not zero: Real forces are acting to accelerate that object. To find the instantaneous acceleration, it helps to treat the radial and tangential components separately.
     
  11. Aug 25, 2005 #10

    reilly

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    Displacement is the difference between two points in (any metric) space. Because of the time dependence in dynamics, it's customary to think of the displacement as a function of time. bomba923 got it right.

    The tricky point stems from the coordinates involved. After a complete revolution, no matter what the speed or acceleration or amount of backtracking, the angular displacement will be 2pi (unless you end up going clockwise to the starting point). But, the usual x and y (and you can always set the starting point at x=0, y=0, angle=0) are periodic functions of the angle, and thus are always zero for the
    angle=2*integer*pi, as can be seen with plots of x and y. So, the angular displace ment is not zero, but the displacements in x and y are zero whenever the particle moves with counterclockwise motion and arrives at the starting point.

    Regards,
    Reilly Atkinson
     
  12. Aug 25, 2005 #11

    He's right. With a circle a similar thing happens, if you look at the opposite side of any circle you will note that the tangential velocities point in opposite directions. For example, if you say that on the right side of the circle the up direction for a tangential velocity starts things off, draw a 180 degree line to the other side of the circle and you will find that the tangential velocity there is pointed downward exactly equal and opposite in both direction and magnitude to the one on the right. Take a pencil and draw a circle slowly to see what I mean, while you are on the right side of the circle your hand is moving "upwards" and while on the left side it is moving "downwards." The two motions act to cancel each other, and you end up right back where you started after you travel the full circle.
     
  13. Aug 25, 2005 #12

    lightgrav

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    v^2 means v.v (vector dot product) which is a scalar, as is u^2 = u.u ;
    but since the a vector points inward toward the circle center,
    while the s vector (displacement) is along the path of the circle, a.s = 0 .
    ( ie, No Work is done by the force that causes the centripetal acceleration)

    Reilly: isn't displacement supposed to be a vector, rather than a change in coordinate?
     
  14. Aug 26, 2005 #13
    hmm so if we want to find the linear acceleration,we can apply kinematics equation solve for it?

    But however,can we treat the circle as a straight line??Arent we suppose to take our position with respect to the cirlce?hence the displacement equals to zero??
    is there any other way than using the kinematics equation to solve for the linear accleration?? say can we differentiate the tangential velocity wrt to time to get the linear acceleration??
     
  15. Aug 26, 2005 #14

    Doc Al

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    Assuming the tangential acceleration is constant, the usual kinematic equations apply.


    The tangential velocity at any time is just [itex]v_t = r \omega[/itex]; thus the tangential acceleration is [itex]a_t = r \alpha[/itex].

    Even better, try this. Take bomba923's expression for the position vector:
    [tex] \vec P = r\left( {\hat i\cos \theta + \hat j\sin \theta } \right) [/tex]
    To find the velocity, take the first derivative; to find the acceleration, take the second derivative. The acceleration will have two components: (1) a radial component equal to the centripetal acceleration, and (2) a tangential component giving the tangential acceleration.
     
  16. Aug 26, 2005 #15

    reilly

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    lightgray --Yes displacement is a vector, and, I apologize for my sloppy, but somewhat standard usage.

    Again, the speed, direction of motion, etc. have little to do with the displacement at any given time -- except to tell you how the object got to where it is. The confusion, once again as clearly as i can say it: it all depends on the coordinates you use. For the angular motion, one should do arithmetic modulo 2pi -- normally we do this with the tacit agreement that 10pi is the same is 2pi. This matter is discussed in as much detail as you want in freshman physics texts, some calculus texts,and, in fact, in more advanced mechanics texts dealing with generalized coordinates. Very basic stuff.

    Regards,
    Reilly Atkinson
     
  17. Sep 14, 2005 #16
    hey now, You don't know if the displacement is zero in that question.

    He did not clarify in what sphere (frame) of reference are we talking. The sphere (frame) of reference may have changed while we were travelling the circle.
     
    Last edited: Sep 14, 2005
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