Displacement equals to zero?

Yes! Technically, it is the zero vector which is not the same as zero (a scalar) since vectors and scalars are different entities.

 

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.

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. sorry...

 

I believe your friend muddled up the concepts of displacement and TRAVERSED DISTANCE.

 

Perhaps he was thinking about a displacement which causes something along the way.

 

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

 

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.

 

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?

 

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

 

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.