# Yo-yo falling

1. Dec 18, 2011

### guss

I had a problem a few days ago that involved a yo-yo falling. It was a disk with a string wrapped around it, and the string was attached at a certain height.

In the explanation for the problem, it said the equation "weight - tension = ma" could be used to find the linear acceleration of the disk. This confuses me, though. They are acting if all of the force goes to moving the object around, which it clearly doesn't. I am thinking some of the tension force must contribute to the rotational acceleration, so the linear acceleration should be less.

I have tried and tried to understand this, but I cannot. Can anyone help?

Thanks!

2. Dec 18, 2011

### JHamm

The tension does both, rotating it doesn't affect how it effects the center of mass. The center of mass acts as if all the forces run straight through it.

3. Dec 18, 2011

### rcgldr

That's only part of the solution. You need some method to determine tension. Other factors to consider are angular acceleration, angular inertia and torque. You'll need to determine equations that show the relationships between the factors. Assuming it's a solid uniform disk, then angular inertia = 1/2 m r2.

4. Dec 18, 2011

### guss

Okay -- Why, though?

And rcgldr, I know that.

5. Dec 18, 2011

### rcgldr

The linear acceleration of the center of mass corresponds to the vector sum of the linear forces (divided by the mass of the disc) regardless of the point of application of those forces or if those forces also result in angular acceleration.

Last edited: Dec 18, 2011
6. Dec 18, 2011

### guss

Yes, but this doesn't make sense to me. Is there deeper reasoning?

7. Dec 18, 2011

### rcgldr

I can only think of other examples to help explain this.

Imagine you have a 1kg block glued onto the surface of a very long massless treadmill. The treadmill surface accelerates so that the force applied to the block is 1 Newton. In this case both the treadmill surface and the block accelerate at 1 m / s2. Now imagine a solid uniform sphere rolling on an accelerating treadmill, and again with the treadmill surface accelerating so it applies a force of 1 Newton to the surface of the sphere. The linear acceleration of the center of mass of the sphere will still be 1 m / s2, but because of the angular acceleration of the sphere, and the corresponding reduced inertial response to acceleration of the surface of the sphere, the treadmill will be accelerating at 3.5 m / s2 in order to generate that 1 Newton of force, 3.5 times the rate of linear acceleration of the sphere.

For another example, imagine a vector force of 1 Newton and constant direction is applied to the end of a 1 kg rod, initially at rest and perpendicular to the direction of the force. The center of mass of the rod will accelerate at 1 m / s2 and in a straight line, in the direction of force, even though the rod will pivot back and forth as well as the point of application of force at the end of the rod, relative to the center of mass of the rod. To help visualize such a force, imagine a massless charged particle attached to the end of an uncharged rod in a constant electrical field (the field from an infinite plate or between the plates of a very large capacitor with some constant amount of charge per unit area) in space, free of any other external forces.

I don't know if it will help to look at the work done and the energy change for the yo-yo.

tension x distance = change in angular kinetic energy
(weight - tension) x distance = change in linear kinetic energy

weight x distance = change in total kinetic energy

Last edited: Dec 18, 2011
8. Dec 18, 2011

### guss

Ok, thanks a lot, I think I am starting to understand.

So for the rod with the charge on the end, it is going to start off being pushed and spun one way, but its angular momentum is going to carry through on the other side, and it is going to be spun back to its original position. So its rotation is harmonic motion, but it is being pushed one way the whole time. So, when the rod is rotating, the rotational energy is defined by force*distance, but the distance is higher than if the rod wasn't rotating since the charge at the end of the rod is moving more than the center of mass of the rod. This is what gives the rod the extra rotational energy that it wouldn't normally have. Not sure if this makes sense, just typing down some thoughts.

Now, let's say the charge is at the center of mass of the rod and not at one side. The linear acceleration would be the same, correct?

Is this sort of the same thing as the internal forces of something canceling each other out?

9. Dec 18, 2011

### rcgldr

Correct. Note that the center of mass travels in a straight line (as well as being accelerated at 1 m / s2), but the application of force at the end of the rod is moving side to side with respect to the center of mass, but the direction of the force and it's magnitude remain constant.

During each half cycle, the rate of rotation increases and decreases, returning to zero as the direction of rotation changes. During the time that the rate of rotation is increasing, the component of velocity in the direction of force at the end of the rod is greater than the center of mass, so a greater amount of force times distance. During the time that the rate of rotation is decreasing, the component of velocity in the direction of force at the end of the rod is less than the center of mass, so a lesser amount of force times distance.

So while the linear component of acceleration of the center of mass is constant, the angular acceleration increases and decreases. If the rod had sufficient initial angular velocity, it would always rotate in the same direction, but the angular acceleration would still cycle through increasing and decreasing periods, and the linear acceleration of the center of mass would remain constant.

Correct, the linear acceleration of the center of mass is the same regardless of where the massless charged particle is connected to the rod.

I'm not sure about the internal forces. The main point is that since the only external force is of constant magnitude and direction, its point of application doesn't affect the rate of linear acceleration of the center of mass. The "trick" here is the fact that the force vector is defined as constant (direction and magnitude), but the point of application of that vector force is allowed to move with respect to the center of mass of the accelerated object, whch means the point of application may be accelerating faster or slower than the center of mass, depending on the circumstances. In the case of that sphere rolling on an accelerating treadmill, the sphere's surface (and the treadmill) is accelerating 3.5 times as fast as it's center of mass.

Last edited: Dec 18, 2011
10. Dec 18, 2011

### JHamm

Sorry for being somewhat off-topic, but I think that is just plain cool.

11. Dec 18, 2011

### guss

Alright, thanks a lot for all your help. I think I get it now. Lots of cool little things going on in these situations.

Anyone have anything else to add?

12. Dec 19, 2011

### rcgldr

Well no one seems to be wondering how angular momentum is being conserved in this situation, since the yo-yo's angular momentum is changing at it falls.

One answer is to consider that the string is mounted on a frame that is connected to the earth. For torques, both the direction and position of forces is important. From the earth's perpective, the gravitational force results in an upwards force vector on the earth that goes through the center of mass of the yo-yo, while the frame that the string is mounted to effectively results in a downwards force vector on the earth that lines up with the string. This results in a torque applied to the earth equal and opposing to the torque applied to the yo-yo, and the end result of the yo-yo, frame, and earth closed system is angular momentum is conserved (since the earth is massive, any angular velocity due to that torque will be extremely tiny).

Last edited: Dec 19, 2011
13. Dec 19, 2011

### guss

Cool! Conservation of angular momentum was actually something I was wondering.