Conservation of Angular Momentum: Hopping onto a Board

[The Head]

Homework Statement

A long board is free to slide on a sheet of frictionless ice. A skater skates to the board (laid horizontally relative to the skater's motion) and hops onto one end, causing the board to slide an rotate. In this situation, are angular and linear momentum conserved?

Homework Equations

sum of torques = I* alpha
sum of forces = ma

The Attempt at a Solution

I understand that the reason that the angular momentum and linear momentum are conserved is because there are no outside forces or torques, so mathematically it makes sense. I'm struggling with this conceptually though.

So when this board begins to slide and rotate, are we saying the angular momentum is zero? Because if the board were to rotate, so would the person attached to it. I can see that the board will eventually straighten out, but and then just stay straight, but I'm caught up on the part where it's sliding.

Also, I imagine there is friction of some sort to keep the skater on the board (and get the board moving). But in order for the board to slide, wouldn't the board also have to slide against the skater's shoes in order to straighten out? Otherwise, it seems as if the board would just stay horizontal relative to the skater's line of motion. And if there's friction, there would be some loss of energy. I'm less concerned about this at the moment, and more so about the bit in the second paragraph, but it's just not clicking for me fully and I'd like to completely understand.

 

[Orodruin]

So when this board begins to slide and rotate, are we saying the angular momentum is zero?

Was the angular momentum zero before the collision? (Also note that angular momentum depends on your choice of reference as there is a net linear momentum in the system)

I can see that the board will eventually straighten out, but and then just stay straight, but I'm caught up on the part where it's sliding.

What do you mean by this? If the ice is frictioless the board will rotate indefinitely.

 

[The Head]

So if I choose the reference frame of a point on the ice in front of the board where the skater will hit it, the board isn't rotating wrt the point before contact and neither is the skater, so I would think not, but maybe I've got something wrong.

 

[haruspex]

if I choose the reference frame of a point on the ice in front of the board where the skater will hit it, the board isn't rotating wrt the point before contact and neither is the skater,

Right.

so I would think not

... not what?

 

[The Head]

See above. I was answering Orodruin's question about if there is an intial angular momentum. And I thought not.

 

[haruspex]

See above. I was answering Orodruin's question about if there is an intial angular momentum. And I thought not.

Ok, but what Orodruin actually asked was

Was the angular momentum zero before the collision?

So a "no" to that would mean it was nonzero.

Anyway, yes, with that reference axis there is no angular momentum.

 

[haruspex]

in order for the board to slide, wouldn't the board also have to slide against the skater's shoes in order to straighten out?

No. On landing on the board there is a coalescence, i.e. a completely inelastic collision. Well, it's a bit more complicated than that, but that is the effect.

 

[The Head]

So it seems the sum angular momentum is zero about any given point. That is some points of mass will be rotating clockwise, while others will be rotating counterclockwise. Is that a correct way of reconciling this?

Also, from earlier, Orodruin mentioned the rotation would happen indefinitely. So would this skater/board system be making some larger loop around the rink?

 

[kuruman]

I understand that the reason that the angular momentum and linear momentum are conserved is because there are no outside forces or torques, so mathematically it makes sense. I'm struggling with this conceptually though.

The answer will not be complete without noting that linear momentum in the vertical direction is not conserved because of the normal force exerted by the ice.

 

[haruspex]

The answer will not be complete without noting that linear momentum in the vertical direction is not conserved because of the normal force exerted by the ice.

I think we can take the leap onto the board as essentially horizontal. The vertical impulses will all cancel out.

 

[haruspex]

So it seems the sum angular momentum is zero about any given point.

About any point along the line of the skater's original momentum.

some points of mass will be rotating clockwise, while others will be rotating counterclockwise.

From the point of view of that chosen axis, yes. Imagine standing on the ice facing the skater head on as the leap is made onto the board. Soon after, the skater, and some of the board, will still be approaching you, but a bit to one side of that original line. Meanwhile, the other end of the board will be even further from that line and moving away from you. That is how these angular momenta can cancel.

Orodruin mentioned the rotation would happen indefinitely. So would this skater/board system be making some larger loop around the rink?

Think about what happens to the mass centre. That corresponds to a fixed point on the board, and it will move in a straight line.

 

[The Head]

About any point along the line of the skater's original momentum.

From the point of view of that chosen axis, yes. Imagine standing on the ice facing the skater head on as the leap is made onto the board. Soon after, the skater, and some of the board, will still be approaching you, but a bit to one side of that original line. Meanwhile, the other end of the board will be even further from that line and moving away from you. That is how these angular momenta can cancel.

Think about what happens to the mass centre. That corresponds to a fixed point on the board, and it will move in a straight line.

Great, thank you. That last part really makes sense.