Choosing Axis of Rotation in Cylinder Oscillation Problem

[BBA Biochemistry]

Homework Statement

Here is a problem we worked in class. I already know the answer, just had a question on the method.

Two cylinders are connected with by a small rod (with presumably negligent mass) through their centers. The cylinders can roll freely. A spring is attached to the small rod and causes the cylinders to oscillate and roll without slipping. The mass of the cylinders (M) and stiffness of the spring (k) are known. What is the angular frequency of the oscillations.

Homework Equations

τ=Iα
τ=FR
F=-kx
I_cyl=0.5MR²
Parallel Axis Theorem: I_p=I_CM+Md²
α=a_CM/R

The Attempt at a Solution

We simplified the problem and said we can look at the two cylinders + rod as one cylinder with mass, M. The force of the spring provides a torque. We said that the cylinder is rotating about an axis at the point where it touches the floor. Is the reason we chose this axis (instead of the center of the cylinders) because the rod that is connected to the spring is going through the center of the cylinder, so there is no F_tan applied to the center mass that can provide torque?

Aside from that, I'm comfortable with the problem. We used the Parallel-Axis theorem to find the moment of inertia for that axis of rotation, and found torque with these two equations:

τ=Iα
τ=FR

We then set them equal to each other and got a differential equation to find ω.

How would this problem change if somehow the spring were attached to the top of the cylinder in a way that the it would apply force tangentially (without following the rolling motion of the ball)? Would we be able to see the cylinder as rotating about the center mass?

Thanks!

 

[haruspex]

In the absence of a diagram you need to provide a clearer description.
I'm guessing that the cylinders and the rod are all coaxial, that this axis is horizontal, that the spring is also horizontal but at right angles to the rod's axis and attached to a fixed point at its other end.

The choice of axis is not critical, but some choices can eliminate forces that are of no interest and thereby simplify the algebra. In the present case, there is a frictional force from the ground. You have no need to determine that force. By taking the point of contact with the ground you avoid involving it in the equations. If you take the mass centre as axis you will need another equation (horizontal linear forces) in order to eliminate it.

(Masses can be negligible; describers of problems can be negligent.)