If an object rotates 2pi radians is its angular displacement zero?

Question

I understand that visually the object does not appear to have moved. However, there are a few things that make me question that point.

1. Position is measured by how many radians from the reference point the object is. After a rotation it is reasonable to say it is 2pi from the reference point, and not zero. (i think) If this is the case then displacement equals theta2 - theta 1 = 2pi. Directions are measured clockwise and counter clockwise as opposed to absolute distance from the reference point.

2. work = force x displacement. I understand that this is for linear motion but I think it is safe to say the angular version has a similiar relationship. If rotating 2pi = 0 displacement then rotating a wheel one cycle would result in no work done, which is not the case because the wheel is now spinning with angular velocity.

Is position 2pi = to 0? It does not make any sense if it does. If it actually does please provide a good explanation please. I don't want a meaning less answer.

I am aware of distance and dispalcement and their differences. If you travel a distance but your dispalcement is 0, it means you direction of motion has changed at some point. This is not the case for rotation. The object is constantly going in the same direction (clockwise for example)

In rotation motion directions are measured as clockwise and counter clockwise. If you keep going clockwise how can rotations in one direction not count? If you rotated CW pi and the CCW pi then i understand that it is 0. But now its just CW 2pi.

Please read all my points before answering.

If you apply a force to life a chair and it didnt not move then work does is zero (it did not gain any kinetic or gravitational energy)

now if you rotate a wheel with a force for 2pi radians the wheel is spinning when you're done. it has angular velocity, IT HAS ENERGY!!!. how can work done on it be zero??

Also, driving in a circle is a completly different is circular motion, not rotational and are therefore completly different.

I repeat, please don't give me meaningless answers. I am puzzled.

Answer 1

You are confusing distance and displacement. Displacement is a vector with direction. If I go 1 mile north and then 1 mile south, my distance is 2 miles but my displacement is 0. Since 2pi radians make a complete revolution, the rotation of 2pi radians brings you back where you started and your displacement is zero.

Answer 2

If I rotate an object 3/2 pi, it has the same location as if I rotated it by -1/2 pi. However, the associated arc lengths for the two rotations are not the same, nor would any calculation of expended energy from say having the periphery rubbing against a surface during the two rotations.

So the question is semantics, whether angular displacement and rotation are the same. To me they are. So I would not consider 2pi and zero as being the same angular displacement, although they might share common properties and trig function values.

Answer 3

Work is not equal to force times displacement, it is equal to int( force dot dr), where r is a unit vector in the direction of motion. In non-calculus terms, this means it is force times the distance in the direction of the force, not displacement. Thus if i apply a torque to something while it rotates 2pi, the angular displacement may be 2pi, but the work is non-zero because int(torque dot dr) = 2pi * torque. This could be termed "angular distance" although I've never heard that term actually used.

I'm not sure how to answer your first point, because frankly I don't understand it. Perhapse discussing polar cordinates will clear things up. You can label any point in 2-D space with a set of numbers in cartesian coordinates (x, y) or in polar coordinates (r, theta) where r is distance from the origin and theta is probably roughly equivalent to what you call "position" in the statement "position 2pi=to 0". In this case, yes theta=2pi is equal to theta=0. Theta here is an angle equal to arctan(y/x), and you can verify yourself that an addition or subtraction of 2pi from any angle yeilds the same absolute position. If you're still confused feel free to message me.

Answer 4

I would think of it this way:

If you drove your car in a circle, would your odometer register zero? Of course not. Distance traveled is not zero. The displacement vector (where did you end up from where you started) IS zero

You are confusing displacement (distance traveled) with distance. Look at the wikipedia description...the illustration is good.

With regard to work, consider the term in the way that it is offered. If you unsucessfully tried to lift a chair that was nailed to floor with all your might, did you do work? Seems like it, but in the physical sense, ,you did not. You exerted great force, but the displacement was zero, therefore no work. Same for your angular example.

Answer 5

No. As usual, apply conservation of energy and momentum. In order to get ab object to move 2pi radians, you needed to have accelerated an object and then decelerated the same object. Therefore, work was expended, so the new position is not energy neutral.

Answer 6

Hi. Mathematically yes : http://mathforum.org/library/drmath/view/64434.html

In a real object with mass, I'm not sure.