This is actually been sitting around for a while waiting for me to post it. Here is another short Christmas-toy demo. I am going to pull this yo-yo at different angles and on two different surfaces. Check it out.
What is going on here? Let me look at the first case where I pull the yo-yo and it slides without rolling. Here is a diagram.
Normally, I would just say - "hey - a free body diagram". And this is one, but you have to be careful. Normally, a free body diagram treats an object as though it were a point mass. You can't do that in this case because you have to consider rotation also (points can't really rotate). When I draw a diagram as a point, this is the key thing I am looking at:
Which I could break into 2 or 3 component equations such as:
Since this object can rotate, I must also consider that with:
I can't believe this, but I never really had a post just devoted to torque. Weird. Well, here is a post that basically goes over all the ideas of torque - Friction Demo with a meterstick. In short:
- tau is the torque about some axis (labeled as O). You can think of torque as the rotational equivalent of force.
- I is the moment of inertia of that object about the same axis as the torque. The moment of inertia can be a complicated thing, but in this case it can be thought of as the object's resistance to change in rotational motion. The moment of inertia depends on both the mass of the object and how this mass is distributed about the axis of rotation.
- Alpha is the rotational (angular) acceleration.
Hopefully, you can see how similar this last equation is to the linear version (Newton's second law). Ok, I am proceeding onward. Back to the yo-yo. Really, I have three equations - the x equation, the y-equation and the rotational equation. I need to note a couple of extra things. First, I will call the radius of the inner part of the yo-yo r and the outer radius R. Also, the mass is m, and the coefficient of static and kinetic friction will be mus and muk. This gives the following:
A couple of notes:
- I picked the case of the sliding and not rolling yo-yo because: the acceleration and angular acceleration are zero. The friction is kinetic friction. This means that I can determine its value. For static friction, I can only calculate the maximum friction. (here is a review of friction)
- The acceleration in the y-direction is zero since the yo-yo stays on the table.
- I can use the model for friction to get an expression for Ff (did you notice I changed Ffriction to the shorter Ff?)
- Also, I have shorter notation for the force from the table (FN), tension (FT) and the gravitational force (mg)
- There are 4 forces. However, I only show two torques. The torque from the force the table exerts is zero about the axis since this forces points right through the axis. The torque due to the gravitational force is also zero. This is because gravity pulls on all parts of the yo-yo.
Here is the model for kinetic friction. Note that this an expression for the magnitude of the friction force - it is not a vector equation.
With this, I can replace all the Ff and I get:
Now, I will get an expression for FT from the last equation:
And now I can substitute this in the other two equations. I get:
From the top expression, if FN is not zero, then:
So, this says that the angle needed to pull the yo-yo so it doesn't slip only depends on the ratio of the inner and outer radius. Note that r would be smaller than R so that the ratio would be less than 1. This is good because the cosine function must produce a number less than one.
If you take the video above and analyze it with Tracker Video Analysis, I get that the yo-yo slides at an angle of about 53 degrees. You should notice that I repeated the experiment with the yo-yo on a different surface (WebKinz mouse pad) that was much slicker. The angle of the string was still 53 degrees. Since the coefficient of friction wasn't as much, I didn't have to pull as hard (for constant speed) but it was the same angle.
If you wanted to, you could measure the outer radius of the yo-yo and use this to calculate the inner radius.
The other two motions:
What happens if I increase the angle of the string above 53 degrees? The frictional force will be less. This is because if I pull at a greater angle with the string, then the normal force will be smaller (since it doesn't have to exert as large of a force to make the vertical acceleration zero). This smaller normal force means the frictional force will be smaller and thus a smaller torque from the friction. Both of these together make the torque larger in the direction that makes it roll to the left.
If the angle of the string is too small, the frictional force will be greater (basically because of the opposite of above).
I think the coolest part of this demo is that by pulling at different angles you can make the yo-yo roll right, roll left, or slide (not roll).
Hi Rhett - love this problem. If you're interested in which way it does (or doesn't) roll, why not make the axis the contact with the table? Only one torque then - none if it's not rolling! That makes solving for the critical angle a one-liner, and drawing angles greater and smaller and finding the direction of the cross product show instantly which way it rolls.
Finding the critical angle for this problem was on the Caltech Physics placement test when I went there 20 years ago. I thought it was pretty neat then too.
Can you post how to find the linear acceleration of the yoyo. I think my solution is flawed.