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rotational dynamics question

rotational dynamics question - Physics Forum

rotational dynamics question - Physics Forum. Discuss and ask physics questions, kinematics and other physics problems.


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  #1  
Old 12-22-2004, 08:06 PM
Starton X. Mangrove
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Default rotational dynamics question





In 2D, a spinning (in the manner of an bicycle tire) disk is dropped from a
height onto a completely non-slip surface. Is 100% of the rotational
momentum converted to linear momentum?

If the surface was slightly elastic and caused the disk to bounce, would the
disk stop spinning after it hit (and was in the air after the bounce). My
confusion is, assuming non-slip, non-elastic surface, the disk should move
in a linear fashion after contact with the surface, *and* also roll along
the surface. So is some of the rotational momentum is not transfered? Or is
the linear momentum transfered back to rotational as it is in contact with
the surface instantly after it hits.

Hope this is clear enough, TIA.



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  #2  
Old 12-23-2004, 12:38 AM
N:dlzc D:aol T:com \(dlzc\)
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Default rotational dynamics question

Dear Starton X. Mangrove:

"Starton X. Mangrove" <[Only registered users see links. ]> wrote in message
news:Rskyd.22827$[Only registered users see links. ].rr.com...

Momentum is conserved. Energy is not.


Try it with a superball. The rotation reverses, because the ball (or tire)
is a deformable solid...


It has been a long time since I tried formulating such a problem. I would
think the radius of gyration would need to be doped out for the wheel *at
the point of rolling contact*. The angular momentum about the point of
contact should equal the angular momentum before contact. I THINK.

David A. Smith


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  #3  
Old 12-23-2004, 12:51 AM
tadchem
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Default rotational dynamics question


"Starton X. Mangrove" <[Only registered users see links. ]> wrote in message
news:Rskyd.22827$[Only registered users see links. ].rr.com...
a

No. For that to happen, the disk would have to stop spinning *entirely* and
_slide_ along the surface. The actual limiting condition (determined by
fact that the friction between the disk and the surface is non-zero) is that
the linear velocity v of the disk as it rolls equals the velocity of the rim
relative to the center of the disk as it spins.

the

Interestingly, a sufficiently elastic _tire_ would engage the non-skid
surface, absorb the rotational kinetic energy in a wind-up spring manner,
and then re-release the energy as the disk rebounds in the opposite
direction, so the rebounding disk could actually *reverse* the direction of
its spin. This is easily seen by imparting a spin with a horizontal axis to
an elastic "superball" and then dropping it - the alternating directions of
successive bounces demonstrate that the spin reverses on each bounce.


....eventually.


The situation can become complicated as the superball example demonstrates,
but for the disk to move horizontally along the surface without skidding, it
MUST roll.

HTH


Tom Davidson
Richmond, VA


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  #4  
Old 12-23-2004, 02:22 AM
Starton X. Mangrove
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Default rotational dynamics question




I dont see how this is possible, you say the angular momentum at the point
of contact will be equal to the angular momentum before contact?


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  #5  
Old 12-23-2004, 02:32 AM
Starton X. Mangrove
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Default rotational dynamics question


demonstrates,
it

I wanted to avoid the superball example -- I should have been clearer. I
only meant to introduce elasticity in order to have the disk in the air
after contact to examine what it's behavior would be without the subsequent
contact with the surface. Assuming the disk simply "bounces" with no elastic
deformation (think reversal of its normal force * coefficient of
restitution) When in the 'air', my belief is the disk would not spin after
contact, having transfered all its rotation momentum to linear momentum.

Anyway, Im still confused. Having a disk w/o any elasticity and 100%
friction -- how do I calculate the resulting linear and angular values,
assuming the disk rolls after contact.


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  #6  
Old 12-23-2004, 04:03 AM
N:dlzc D:aol T:com \(dlzc\)
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Default rotational dynamics question

Dear Starton X. Mangrove:

"Starton X. Mangrove" <[Only registered users see links. ]> wrote in message
news:KZpyd.29199$[Only registered users see links. ].rr.com.. .

Yes. "Radius of Gyration" I believe is what needs to be determined for the
wheel, about the point of contact with the ground. Angular momentum *must*
be conserved. You have no external applied torques...

tadchem is always more correct and complete than I, so if he posted
something different (and understandable) then trust him.

David A. Smith


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  #7  
Old 12-23-2004, 08:01 AM
tadchem
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Default rotational dynamics question


"Starton X. Mangrove" <[Only registered users see links. ]> wrote in message
news:f7qyd.29200$[Only registered users see links. ].rr.com.. .

<snip>

after

The amount of spin remaining after the bounce is variable, from 100% of the
original spin (no loss of angular momentum) to -100% of the original spin
(complete reversal of the angular momentum).


You can easily model a disk w/o any elasticity and 100% friction using
modeling clay, uncooked pizza dough, raw liver, or any of a number of other
substances. No elasticity means it doesn't bounce, and 100% friction means
it loses *all* its energy to deformation due to the impact, and the shape is
not elastically restored.

[One of the hardest things to do in the empirical sciences is to learn how
to ask the right questions.]


Tom Davidson
Richmond, VA


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