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Stationary yoyo

Ryan McGuire
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Joined: Feb 18, 2005
Posts: 1013
    
    3
If you put a yoyo on the floor so that it's free to roll and pull the string straight up it will unwind. If you pull the string horizontally (so that string is coming from the bottom of the axel) the yoyo will wind up.

At what angle will it stay in one place (and just unwind if you pull hard enough)?


[ May 02, 2006: Message edited by: Ryan McGuire ]
Jim Yingst
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Joined: Jan 30, 2000
Posts: 18671
cos θ = r/R

where

r = inner radius
R = outer radius
θ = angle between ground and string


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Ryan McGuire
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Joined: Feb 18, 2005
Posts: 1013
    
    3
Originally posted by Jim Yingst:
cos θ = r/R

where

r = inner radius
R = outer radius
θ = angle between ground and string


I agree. Care to explain how you got that answer?
fred rosenberger
lowercase baba
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Joined: Oct 02, 2003
Posts: 11499
    
  16

i got the answer by reading Jim's post.


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Jim Yingst
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Joined: Jan 30, 2000
Posts: 18671
Let:

T = tension in the string
F = force of friction from floor on yo-yo, to the left
W = weight of yo-yo
N = normal (upward) force fo floor on yo-o

(The last two don't really matter for us, but I list them for completeness.)

The sum of all horizontal forces must be zero:

    T cos θ - F = 0

or

    F = T cos θ

The sum or all moments (torques) about the center must be zero:

    rT - RF = 0

Combining these gives

    rT - RT cos θ = 0

    r / R = cos θ
Ryan McGuire
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Joined: Feb 18, 2005
Posts: 1013
    
    3
You can simplify things a little. For the yoyo to be stationary, the sum of the torques around any point, not just the center, needs to be zero. By considering the point where the yoyo touches the floor, F, W, and N all have zero perpendicular distance from that point. Since there are no forces to counter it, the force from the tension in the string must also have a zero perpendicular distance from that point.

So the qualitative answer, "The angle where the string aims at the point where the yoyo touches the table," would have been equally correct.



Draw the right triangle from the point where the yoyo touches the floor to the yoyo center to the point where the string first touches the axle. Some simple geometry will show that the angle between the string and the floor is the same as the angle from that right triangle at the center of the yoyo, which is arccos r/R.
[ May 08, 2006: Message edited by: Ryan McGuire ]
Jim Yingst
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Joined: Jan 30, 2000
Posts: 18671
Indeed, I considered taking moments about the floor contact point, but figured in this case it would be more effort to explain the geometry (being too lazy to draw a picture). Six of one, half a dozen of the other.
[ May 08, 2006: Message edited by: Jim Yingst ]
 
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