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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.

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jtbell

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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...

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,

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

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...

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arildno

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

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PerennialII

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

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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??

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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).SVG84R said: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???

Yes, the displacement of the object after making one complete revolution is zero. This means that the

But what you want to know is not the average acceleration, but the

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reilly

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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

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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.

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.

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lightgrav

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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?

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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??

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Doc Al

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Assuming the tangential acceleration is constant, the usual kinematic equations apply.SVG84R said:hmm so if we want to find the linear acceleration,we can apply kinematics equation solve for it?

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].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??

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.

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reilly

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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

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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.

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.

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