Laymans time and distance canundrum

[Robin07]

Hi everyone,
I have often wondered about the time and distance relationship.
Case in point: If you wind a flat coil, let's say six inches in diameter, and cause a current to run though it. The electrons that travel on the outer most perimeter will need to travel a much longer distance to complete one circumference compared to the inner most electrons to complete one distance around. I understand that each electron will travel the same speed but cover a different distance because the wire is of one piece. This alone took me months to understand. This is much like the hands on a clock. It's now obvious, to me, that the inner part of the hour hand will spend the same amount of time completing one revoluation. What is so confusing to me is that, when a person is holding a rope and two objects are attached at different distances away on that rope and then spun so that the stones are now air borne, the objects are read to be traveling at different speeds according to a radar detector? How is this discrepancy justified? For me this is much like walking with the moon as it rotates around our Earth. A traveler would take 24hrs. to complete one trip around our planet, which would be X amount of time over the distance but the moon also travels within the same time frame. Would the moon be traveling the same speed as the traveler on Earth? I would say not.

Thanks for all your input

 

[CompuChip]

Note that all your examples involve rotational motion of some sort.
When talking about rotations, we discern two velocities: the angular velocity (usually denoted [itex]\omega[/itex] - small omega, some people just write w) and linear velocity (usually denoted v).
The angular velocity tells you how fast something rotates, it is measured in angle per time (degrees/second or radians/second). For example, if omega = 12 degrees/second, it would take the object 30 seconds to get back where it started (do a complete revolution).
The linear velocity is something the velocity at which an object would be released, if you would release the constraint. For example, if you put a stone on a rotating circular platform and suddenly stop the platform, the stone will fly off tangent to the platform with linear velocity.

The angular and linear velocity are related by
[tex]v = \omega r[/tex]
where r (radius) is the distance from the center of rotation to the object. So if I put two stones on my round platform, one near the inside and one at the edge, they will necessarily have the same angular velocity (they rotate with the platform). However, the one with the smaller radius will have a smaller linear velocity; basically: it has to cover less distance in one revolution than the other stone does. (Of course, this is simply a consequence of the fact that the circumference of the circle which they travel is 2*pi*r).

It's a lot to think about probably, but does it clarify a bit?

 

[Robin07]

Thanks, yes it's abit clearer.

You mentioned that "the one with the smaller radius will have a smaller linear velocity; basically: it has to cover less distance in one revolution than the other stone does. (Of course, this is simply a consequence of the fact that the circumference of the circle which they travel is 2*pi*r)."

So from my understanding than, 'Time over distance' does determine speed if your calculating the objects speed in a straight line but when it come to a circular path the angular and linear velocity must be concidered to determine the actual speed.

 

[CompuChip]

Thanks, yes it's abit clearer.

You mentioned that "the one with the smaller radius will have a smaller linear velocity; basically: it has to cover less distance in one revolution than the other stone does. (Of course, this is simply a consequence of the fact that the circumference of the circle which they travel is 2*pi*r)."

Exactly. The linear velocity is distance / time (not the other way around, as you said below). So v = 2 pi r / t. The angular velocity is number of radians / time (360 degrees = 2 pi radians in x units of time means covering 2 pi / x radians in one unit of time). So omega = 2 pi / t. Then automatically, v = omega r.

So from my understanding than, 'Time over distance' does determine speed if your calculating the objects speed in a straight line but when it come to a circular path the angular and linear velocity must be concidered to determine the actual speed.

Indeed. Note that you can actually consider v as the "real" velocity, and omega more as a tool. Newtons second law ("F = m a") tells you that, in absence of force, objects move in a straight line at constant velocity. When an object is constrained to move in a circle, it must undergo a force (for example, friction with the rotating platform or tension in the cable it is tied to), which accelerates it inward so the velocity is always along the circle. As soon as you would lift this force (cut the rope, for example) the force would be zero, the acceleration would be zero and the object would fly on with the velocity it has at that particular moment: velocity v in the direction perpendicular to the radius of the circle, so to speak. What the constant angular velocity expresses is just that, although the linear velocity changes all the time (in direction, not in magnitude), it just changes in such a way as to keep the object going around the circle at a steady rate.