Why Does Angular Velocity Appear Opposite to Linear Velocity?

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The discussion centers on the relationship between angular and linear velocity, particularly in the context of a coin launched with backspin. It highlights that the coin can only come to rest both translationally and rotationally if it has backspin, as this allows the forward motion and backward spin to cancel out effectively. The conversation also touches on the implications of kinetic friction, noting that rolling resistance and air drag are not considered in this scenario. Participants find the problem unusual yet interesting, emphasizing that it simplifies the analysis compared to natural rolling scenarios. Understanding these dynamics is crucial for grasping the mechanics involved in such motion.
Fascheue
I’m not quite sure where to put this post so forgive me if it’s misplaced, but can somebody explain why the angular velocity in the picture here appears to be in the opposite direction as I would expect if the velocity is in the direction that is is.

I’m not looking to solve this problem (that’s why I didn’t post this in the homework thread), I’m just wondering why the angular velocity is counter-clockwise and the velocity is to the right.
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The problem statement talks about the coin coming to rest (both translationally and rotationally). That can only work out if the coin is launched with back-spin. The diagram shows back-spin.

It is amusing to launch a hula hoop, giving it a backward flick of the wrist at release so that it spins backward. With a little practice one can arrange for the forward progress and the backward spin to cancel out at nearly the same time, whereupon the hoop falls to the ground.

Video:
 
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The coin is sliding relative to the surface of the table while it's moving to the right. This is the case when ω > V/R and the coin is given an impulse to the right. The situation is analogous to striking a cue ball with "backward english" meaning that the ball is struck sharply below center at the six o' clock position.
 
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jbriggs444 said:
The problem statement talks about the coin coming to rest (both translationally and rotationally). That can only work out if the coin is launched with back-spin. The diagram shows back-spin.

It is amusing to launch a hula hoop, giving it a backward flick of the wrist at release so that it spins backward. With a little practice one can arrange for the forward progress and the backward spin to cancel out at nearly the same time, whereupon the hoop falls to the ground.

Video:

Okay I thought that might be it, it just seems a bit unusual.

Why is it the case that the coin can only come to rest translationally and rotationally if it is launched with backspin? If you roll a coin forward does it not have forward velocity and clockwise spin that simultaneousley come to rest?
 
Fascheue said:
Why is it the case that the coin can only come to rest translationally and rotationally if it is launched with backspin?
Because the question implicitly only considers kinetic friction.

Fascheue said:
If you roll a coin forward does it not have forward velocity and clockwise spin that simultaneousley come to rest?
That requires rolling resistance and/or air drag, which are to be neglected here, as no parameters on them are given.
 
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Fascheue said:
it just seems a bit unusual.
Yes. it is an unusual problem, which is what makes it interesting. It is actually easier to solve than a problem in which the coin is rolling naturally because the kinetic friction is an easier concept than 'rolling' resistance.
One can worry too much about the motives of the people who set problems. The motives are often questionable and it's often best to get on and just answer the problem as stated. :wink:
 
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For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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