Direction of friction in yo-yo where applied force is at top

[issacnewton]

Homework Statement

We have a yo-yo and the force $F$ which is the tension in the string and this is at the top tangent as shown in the figure attached with this problem. I want to figure out the direction of the friction at the point of contact.

Homework Equations

Newton's laws of motion and laws for rotational motion

The Attempt at a Solution

[/B]
Now I tried to see the net torque around the point of contact. Let's assume that the static friction force $f$, is pointed in the same direction as $F$, which is the right direction. So there are 3 forces acting at the point of contact. $N, mg$ and $f$. So all of them will contribute zero torque around the point of contact. So only $F$ will contribute to the torque around the point of contact. And this is clockwise torque around this point of contact. Now consider the net torque around the point where $F$ is acting. Since the lines of action of $N, mg$ pass through this point, they will contribute zero torque around this point. $F$ will also not contribute to the net torque here since $F$ is acting at that point. So only $f$ will contribute to the net torque. And $f$ will cause a counter clockwise torque around this point. But this is not possible since we know from earlier analysis that the net torque has to be clockwise. So the direction of static friction force $f$ must be in the backward direction, so that the net torque will be clockwise.
But I know that in this problem, the friction force $f$, has a forward direction (same direction as $F$). So there must be something wrong in my reasoning.
So what is it ?

Thanks

 

[haruspex]

Now consider the net torque around the point where F is acting.

Need to be careful in choice of axis. If you choose a non-inertial frame about a point which is not the mass centre you may get a wrong result.

But this is not possible since we know from earlier analysis that the net torque has to be clockwise.

Torque is relative to a chosen axis, if you change axis you cannot assume the torque is the same way.

 

[issacnewton]

Ok, makes sense. I did not know that. I thought that if net torque is clock wise around one axis then it will be clock wise around any other axis. So, what is rigorous way to find the frictional force direction here ? I have seen the following explanation in this particular case. Due to the force $F$ at the top, the bottom will start sliding towards left and so there will be a momentary kinetic friction force in the forward direction. To prevent sliding, this forward kinetic friction force will become forward static friction force. This explanation is kind of convincing but I wonder if more rigorous argument could be made here.

 

[haruspex]

Due to the force F at the top, the bottom will start sliding towards left

That is not sufficiently obvious. Imagine this as an object floating in space. F will cause a translation to the right of the mass centre, and a clockwise rotation about the mass centre, but the direction of movement of the point diametrically opposite from F will depend on the ratio of the two.

 

[issacnewton]

So, how can we deduce that the static friction force is in the same direction as $F$ ? I don't think I clearly understood your point here.

 

[haruspex]

So, how can we deduce that the static friction force is in the same direction as $F$ ? I don't think I clearly understood your point here.

Just write out the equations and solve.

It will turn out that the direction will be in the F direction for a uniform solid cylinder. For a hollow cylindrical shell of zero thickness, I believe there would be no frictional force, and if you were to add a flange to that, projecting beyond the two forces, so that the moment of inertia exceeds mr², then the frictional force will go the other way.

 

[issacnewton]

Ok, so we can not get the direction of friction before solving the equations ?

 

[haruspex]

Ok, so we can not get the direction of friction before solving the equations ?

That's my view.

 

[issacnewton]

Ok, thanks haruspex...