# The pole and the rope

1. Oct 27, 2014

### SpaceExplorer

A puck of mass 'm' on friction-less ice is attached by a horizontal string of
length 'l' to a very thin vertical pole of radius R. The puck is given a kick
and circles around the pole with initial speed V. The string wraps around the
pole, and the puck gets drawn in and eventually hits the pole. What quantity
is conserved during the motion? What is the puck’s speed right before it hits
the pole?
The answer has to be 'V', right? Because the only thing being conserved here is the 'kinetic energy', since no external force is applied nor there's any extra work done by 'friction'. There's a torque acting on 'm' at each point, so 'angular momentum' cant be conserved.

2. Oct 27, 2014

### Simon Bridge

Doesn't the force from the tension in the string cause a change in linear momentum? (Recall: momentum is a vector.)
In order for there to be an unbalanced torque, and a constant velocity, the distance to the center of rotation bust decrease as the angular velocity increased ... you can check to see if this is the case. How do you get the torque anyway?
What is the significance of the pole being "very thin"?

3. Oct 27, 2014

### SpaceExplorer

No, the tension won't cause any change in 'radial' momentum because there is no change in tension as the rope wraps up, because nothing is pulling it except for the pole and no external agent is doing work on the pole. Initially when the puck is driven, all the radial forces are balanced, and it remains so throughout and so the radial velocity remains constant. And the torque appears from the fact that the rope is not directly connected to the axis of the pole, but to the outer surface, so that the force is not radial but along a tangent to the surface. This torque will change the 'w' but not without a change in 'r' so that 'rw' remains constant, and thus the net velocity should remain constant throughout.

4. Oct 27, 2014

### Simon Bridge

momentum is a vector
velocity is a vector

you don't seem to think that the pole being very thin has anything to do with it ... so, since you are sure, there is nothing more to be done here.

5. Oct 27, 2014

### SpaceExplorer

yes and that is why I am taking both the radial and tangential components separately

6. Oct 27, 2014

### SpaceExplorer

If the pole is that thin then there will be no spiraling, the puck will execute a circular motion, since there will be no torque. I don't know why he(the author, 'David Morin', I really like his writing style) mentioned 'thin'. I was just looking for a hole in my argument because I don't know why but it seems incomplete.