Find Tension in String when Yo-Yo is Released

In summary, the question is asking how to find the tension in the string of a yo-yo with a hub radius of r and circular mass radius of R, with each mass having a mass of M and the hub having a mass of m. The tangential acceleration is assumed to be g (9.8 m/s2) and the tension in the string is equal to 2Mg + mg. However, this may not be entirely accurate as the rotation of the yo-yo may offset some of the tension, and the initial tangential acceleration depends on the inertia of the yo-yo.
  • #1
amcavoy
665
0
Let's say you have a yo-yo that is constructed from two circular masses connected by a small hub. If the radius of the hub is r and the radius of the circular masses is R, how can I find the tension in the string when the yo-yo is released? Assume each circular mass has mass M and the hub is m.

I know a bit about angular kinematics and dynamics, but I am having trouble applying that here. I am going to take a guess here and say that the tangential acceleration (the magnitude) is g (9.8 m/s2). The reason I say this is because the yo-yo would have a downward acceleration of g had it not been connected to any string. Because the string is connected, an object on the perimeter of the yo-yo will accelerate in the tangent direction at the same rate as if the yo-yo was in free-fall. Is this reasoning correct? If so, I believe I can make more progess on the problem.

Thank you.
 
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  • #2
The tension, T, in the string is applying a torque, tau = Tr, where r is the radius of the hub.

Then tau = I.alpha
and a = alpha*R

where a is the tangential acceleration and R is the radius of the circular mass.

Edit: I is the inertia of the combined masses.

Edit2: correction to the expression for linear acceln.
 
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  • #3
Isn't the tension in the string T = 2Mg + mg ?
 
  • #4
Am I right about the tangential acceleration though?

Thanks for the response.
 
  • #5
Hmmm.
Not absolutely certain, but it doesn't seem right.

Your argument would apply to every point in the yo-yo, including diametrically opposite points. Your argument would give both those points a downward accln of g, but we know that they would would start to move in opposite dirns.!

I'm not absolutely sure about what I said about the Tension in the string - it being equal to 2Mg + mg.
Somehow or other it feels as though the rotation of the yo-yo would offset some of the string's tension.
However, at the start, there is no rotation, So there won't be any offset to the tension then, so (I think) the initial tension in the string is T = 2Mg + mg.

And, if that assumption is true, then you can quite easily show that the initial tangential accln isn't neccesarily g, but depends on I.
 
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  • #6
Of course points on opposite sides are traveling in opposite directions, but when I say "g," I mean the magnitude of the tangential acceleration (not caring about direction).

Thanks again Fermat.
 

1. How is tension calculated in a string when a yo-yo is released?

Tension in a string can be calculated using the formula T = m x g x h, where T is the tension, m is the mass of the yo-yo, g is the acceleration due to gravity, and h is the height at which the yo-yo is released.

2. Does the length of the string affect the tension when a yo-yo is released?

Yes, the length of the string does affect the tension in a yo-yo. The longer the string, the more tension it will have when the yo-yo is released due to the increased distance it has to travel.

3. What factors can affect the tension in a yo-yo string?

The tension in a yo-yo string can be affected by factors such as the mass of the yo-yo, the length of the string, the angle at which the string is released, and the strength of the string itself.

4. Is the tension in the string constant as the yo-yo is released?

No, the tension in the string is not constant as the yo-yo is released. As the yo-yo falls, the tension in the string decreases due to the decrease in height and acceleration.

5. How does the tension in a yo-yo string affect the yo-yo's movement?

The tension in a yo-yo string plays a crucial role in the yo-yo's movement. It helps to keep the yo-yo in motion and allows it to return to the hand after being released. If the tension is too high or too low, it can affect the yo-yo's movement and performance.

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