# Direction of friction for rolling disk attached to a spring

I want to consider the application of Newton's law for a rolling disk attached to a spring, but I am getting confused.

First, I consider the situation when the disk is moving to the right, just past the equilibrium point (dashed line). This is the image on the left below. Its position is a vector to the right (red), its velocity is also to the right (blue) and the friction force is to the left (green).

Then, I consider the situation when the disk is moving to the left, approaching the equilibrium point. This is the image on the right. Its position is a again vector to the right (red), but now its velocity is to the left (blue) and the friction force is to the right (green).

The problem is that the equation of motion I get is

$$m\ddot{x}=-kx-A$$

in the first case and

$$m\ddot{x}=-kx+A$$

in the second case. So either I made a mistake or the acceleration is different in those two situations (I suppose the absolute value of the friction is the same in both cases).

Which is it?

The direction of the friction force is incorrect in your first diagram. In both cases the force from the spring acts to the left and the friction force acts to the right, creating the same anti-clockwise torque and the same angular (and therefore also linear) accelerations.

As the disk passes through the equilibrium position, the spring force is zero and the friction force is also zero. There is no torque and no acceleration at this point. To the left of this position there is a clockwise torque accelerating the disk to the right, so the friction force must act to the left. To the right of the equilibrium position the torque becomes anti-clockwise, accelerating the disk back towards the equilibrium point, so the friction force must act to the right.

The friction force changes direction at the equilibrium position, not at the extreme points of the motion. It increases in magnitude as the disk moves away from the equilibrium point. It changes direction as the magnitude passes through zero at equilibrium. It does not flip direction suddently at the extremes where the magnitude is maximum. That would create a sudden change of the net force on the disk, causing a sudden change in acceleration, ie a 'jerk'. This does not happen : the motion is smooth, not jerky.

I think you could write your equation in terms of $x$ and $\ddot x$ - then you will get the same equation for both directions, namely

$$m \ddot x = -k~x - \frac{\ddot x}{|\ddot x|} A$$

That second term gives you the direction of the inertial term as a function of the acceleration, and would match the signs of your expression.

From this you can see that it is probably better to compute an "effective inertia" of the system - that is, the rolling disk behaves like an object with larger inertia (the force needed to change the velocity by a certain amount).

See if you can derive what that effective inertia would look like.

• I understand what you write, more or less, but you sidestep my question: is the acceleration different in those two situations? (that would mean the disk takes longer to go than to come back). Or, to put it differently, how do I solve the equation to find $\ddot{x}(t)$? Aug 30 '17 at 23:27
• Assuming the disk rolls without sliding, what torque is needed for a given linear acceleration? This torque is provided by force F acting over a distance equal to the radius, $Fr = I\ddot\theta = \frac12 m r^2 \ddot{\frac{x}{r}}$. That relates $F$ to $\ddot x$. Can you see how to do it now? Aug 31 '17 at 0:14