Yo-yo & Torque: Force, Linear and Rotational Motion

[igor123d]

1. A yo-yo lies on a frictionless table. If you apply a horizontal force F to the string to the right, how would the yo-yo move (linearly and rotationally)
2. EF=ma ET=I(alpha)
3. Initially I thought that the yo-yo would move to the right and rotate counter-clockwise (because the string lies below the com). On the test in class that was the right answer apparently. But I looked online and one study guide said that there would be no rotation and the yo-yo would be moving to the right. (which is how I had ultimately answered the question). The idea is that the sum of all forces is ma and that the only external force is the pulling force, F. So if it would rotate, some kinetic energy would go to rotational thus linear would be smaller and a would not be F/m. But how could there be no torque if the force is applied at a point that is not the center of mass and has a component perpendicular to the distance?

 

[OlderDan]

I assume this is a qualitative question. Is the string pulling right or left?

 

[turdferguson]

The answer is all of the above, it depends on the angle from horizontal. It has to do with where the torque is applied. If you pick the correct angle theta, the force is on the instantaneous axis of rotation and produces no torque, only translation. If its below (in your case it was horizontal), then your answer is correct. If the angle is greater than the axis, the yoyo will move in the opposite direction from which you pulled it.

 

[igor123d]

I assume this is a qualitative question. Is the string pulling right or left?

Oh sorry forgot to mention, to the right.

 

[igor123d]

The answer is all of the above, it depends on the angle from horizontal. It has to do with where the torque is applied. If you pick the correct angle theta, the force is on the instantaneous axis of rotation and produces no torque, only translation. If its below (in your case it was horizontal), then your answer is correct. If the angle is greater than the axis, the yoyo will move in the opposite direction from which you pulled it.

Well but in this specific example, the force actos horizontally on a point below the center of mass, so there would have to be torque, right?

 

[igor123d]

By the way here is an image of a similar situation (look at b): http://www.fas.harvard.edu/~scdiroff/lds/NewtonianMechanics/Yo-yo/Yo-yo002.gif

 

[OlderDan]

Oh sorry forgot to mention, to the right.

Then you are correct. The only horizontal force acting is F, so there will be an acceleration F/m to the right. If the string is below the CM there will be a torqe CCW of F*r where r is the radius of the inner spool, so there will be a CCW rotation with angular acceleration.

If there were sufficient friction to prevent slipping instead of no friction, the yo-yo would wind itself up on the string as it rolled to the right. Friction would oppose F and produce a greater CW torque than the CCW torque of the string.

 

[igor123d]

Then you are correct. The only horizontal force acting is F, so there will be an acceleration F/m to the right. If the string is below the CM there will be a torqe CCW of F*r where r is the radius of the inner spool, so there will be a CCW rotation.

If there were sufficient friction to prevent slipping instead of no friction, the yo-yo would wind itself up on the string as it rolled to the right. Friction would oppose F and produce a greater CW torque than the CCW torque of the string.

But would this work with conservation of energy? I mean if the only force is F then the total work would have to be Fx. But in this case, you would also have rotational energy, so you would have Fx+K rotational. How could this work?

 

[OlderDan]

But would this work with conservation of energy? I mean if the only force is F then the total work would have to be Fx. But in this case, you would also have rotational energy, so you would have Fx+K rotational. How could this work?

It is a bit tricky to calculate the work. Think of you doing work on the string, and the string being an energy transmitter to the yo-yo. You apply a force over a distance. How does that distance compare to the distance the yo-yo moves?

 

[igor123d]

Well I guess since the string moves around the yo-yo, its distance would be R*theta. Then would it also have a translational part x, so that d=R*theta+x
But since the rotation is not what's casuing the linear motion, how would x be related to r*theta?

 

[Hootenanny]

use physics

Some really useful comments you've been posting recently...

 

[OlderDan]

Well I guess since the string moves around the yo-yo, its distance would be R*theta. Then would it also have a translational part x, so that d=R*theta+x
But since the rotation is not what's casuing the linear motion, how would x be related to r*theta?

The relationship between x and r*θ will depend on the moment of inertia of the yo-yo. What is important is the relationship between W = F*(x + r*θ) and the total kinetic energy. Does F*x equal the translational KE and F*r*θ equal the rotational KE?

 

[igor123d]

The relationship between x and r*θ will depend on the moment of inertia of the yo-yo. What is important is the relationship between W = F*(x + r*θ) and the total kinetic energy. Does F*x equal the translational KE and F*r*θ equal the rotational KE?

Yes, I guess they will. Ok thanks for your help Dan, this makes sense now.