# Frictionproblem! Yoyo on a horizontal plane.

1. Apr 17, 2010

### center o bass

I have just studied a problem regarding a yoyo being pulled along a horzontal plane and where the task is to find which way the frictionforce from the ground on the yoyo is acting. I have seen the solution to it, but I find it confusing and would therefore like to ask some questions about it.

Allright.. the situation is as shown in (3): http://www.uio.no/studier/emner/matnat/fys/FYS-MEK1110/v10/eks/eksoppg02.pdf [Broken] (it's in norwegian but you can see a sketch of the scenario).

As I said, I have seen the solution. There they are using N2L, N2L for rotation and the kinematic condition we have for rolling motion where the relative velocity between the ground and the point in contact with the ground is zero.
To start with they just assume that the friction force acts backwards and from there on finds an expression for the frictionforce and thereby a condition on the relation between the inner and the outer radius which shows what way the frictionforce is really acting.

The problem I have with this solution is that they use the rolling condition! When one uses this condition isn't that equivalent to saying that we have rolling motion troughout the whole motion. And if we have rolling motion must not the frictionforce allways act forward regardless of the relation between the radii?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 4, 2017
2. Apr 17, 2010

### Staff: Mentor

No. Realize that there is an external force acting on the yoyo (the string), so the static friction is not the only force accelerating the yoyo.

In general, the answer is no anyway. Consider a car accelerating forward without slipping. Friction on the tires acts forward. But when braking, the friction acts backward even though the motion is forward.

3. Apr 17, 2010

### center o bass

Ah.. yeah and the car can have negative acceleration without slipping.

But what then when the yoyo has a negative friction acting on it. Where does this force come from? It's easy to see that the positive friction force is a reaction to the wheel pushing on the ground. But if the yoyo is still rotating and rolling on the same direction, how can a negative friction act on it if it's not slipping?

4. Apr 17, 2010

### Staff: Mentor

That's true with the yoyo as well. The force of the ground on the yoyo is a reaction to the force of the yoyo on the ground. (But that observation doesn't help much!)
Again, you must consider all the forces acting on the yoyo. Try this. Imagine there was no friction at all. Which way would the yoyo tend to rotate? Which way must friction act to prevent slipping?

Static friction is a passive force that will be whatever it needs to be to prevent slipping.

5. Apr 17, 2010

### center o bass

If there were no frictional forces the yoyo would tend to rotate to the right -->, since that is the direction where we have a torque. In order to have no slipping we would need that the relative velocity between the contact point and the ground is zero, so if the wheel is rotating too fast the ground must impress a force in the other direction of the tangential velocity of the wheel in at the contact point, but this is in the positive direction -->+ since the tangential velocity of the wheel is in <---- that direction.

However what if the wheel rotated to slow..? Then the ground would have to speed up the rotation wheel in order for us to have rolling motion. That means that the friction force would have to be in the negative direction <---- producing a torque in the same direction as the tension force from the string.

Is this the solution and the why? If the inner radius is too small the force from the string does not produce a big enough torque so the friction from the ground would have to help it in order to obtain rolling motion?

6. Apr 17, 2010

### center o bass

My first tought was that this should also be a function of time, since a constant torque produces a constant angular acceleration and eventually the wheel would rotate fast enough and faster so that the friction would have to become positive again.

But then again... that same tension force also produces a translational acceleration so it is really the relation between how big a transelational acceleration and angular acceleration we have that is the crucial relation here. And that is a function of the radii!

Have I got it? :D

7. Apr 17, 2010

### Staff: Mentor

Consider the torque applied about the center of mass of the yoyo. The way the yoyo would tend to rotate depends on how the pull is applied.

8. Apr 18, 2010

### rl.bhat

1) Frictional force is a self adjusting force, which always opposes the motion of the center of mass.
2) In the problem referred by you, the center of mass is moving toward right, whether the force acts above or bellow the axis of rotation.
3) Therefore the frictional force f is towards left in both cases.
4) The frictional force f depends on the magnitude and the point of application of the applied force F.
5) If f < f(max) = μ*mg, The body will roll without slipping.
6) If f(max) > F, the body will roll as well as slip.

9. Apr 18, 2010

### Staff: Mentor

Friction opposes slipping between surfaces, not necessarily the motion of the center of mass.

10. Apr 18, 2010

### rl.bhat

If the surface is frictionless, the center of mass of the disc will move in the direction of the applied force. It has translational and rotational motion. If moving disc encounters a rough surface, it will slow down until the velocity of point of contact becomes zero due to frictional force only. There is no question of slipping until f crosses f(max), just like a block will not move until applied force is greater than μmg. In both the cases frictional force reduces the velocity of the center of mass. So its direction is opposite to the applied force.

Last edited: Apr 18, 2010
11. Apr 18, 2010

### Staff: Mentor

Depending upon the ratio of r and R, it's possible for the friction to act in the direction of motion thus increasing the acceleration of the disc. But the friction always opposes slipping between surfaces.

12. Apr 18, 2010

### rl.bhat

It is true. The expression for the frictional force is given by
f = F*(2r+R)/(3R).
When r = R, the frictional force f = F.
If f > μmg, reaction force f will act on the disc in the direction of the force. And the net force will produce slip in the disc along with rolling. But in the problem we are discussing the rolling without slip.