# Rod rotating about pivot with spring

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1. Jan 24, 2016

### ClassicalMechanist

1. The problem statement, all variables and given/known data
Consider the following classic problem: we have a rod in the vertical position with a pivot at its midpoint and a spring attached to the bottom of the rod, perpendicular to the rod. The is rotated through a small angle theta to the vertical, and released. Find the period of oscillations, for small theta.

This is a relatively easy problem. It is as simple as finding the torque on the rod as a function of the angular displacement theta and the spring constant k, and then computing the angular acceleration alpha. You have to use the small angle approximation sin(theta)~theta, and you get that the motion is simple harmonic.

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Now consider a variation of that problem in which the rod of length L is positioned horizontally with the pivot at one end, and the unstretched spring (of length L_0) attached to the opposite end, parallel to the rod:

O------------------------------------------- /\/\/\/\/\/\/\//\[]
(Pivot) (Rod) (Spring, with right end anchored in place)

The rod is allowed to rotate in the vertical direction. Find the period of oscillations for small angular displacements theta.

This is a much harder problem.

2. Relevant equations

F=-kΔx
T=Iα=Fr*sin(θ)
I=1/3mL^2.

3. The attempt at a solution

Let the small angular displacement of the rod (i.e the angle between the rod and the horizontal) be θ. When the rod is displaced by this angle θ, the spring is pulled upwards and stretched by some amount Δx, and exerts a restoring force F=kΔx.

Let the angle between the spring and the rod be Φ. Then the spring exerts a torque on the rod given by T=Iα=Fr*sin(Φ)=-kΔxLsin(Φ). So the angular acceleration is given by α=-(kL/I)Δxsin(Φ). Now to find hte period of oscillation we need to make some approximations so that we get the equation for simple harmonic motion, i.e the angular acceleration α needs to be proportional to θ. The problem is the math quickly gets very messy, because we can't find sin(Φ) without using the angle addition formula or law of sines (on the triangle formed by the rod, the spring, and the horizontal). Also we can't find Δx without using the Pythagorean theorem. If the spring stretches to a length L from its original length L_0, then

Δx=L-L_0=sqrt{ (Lsin(θ))^2+(L_0^)2}-L_0

I will spare you the math, but in the end I found a terribly messy expression for α and it's not obvious how to make small angle approximations so that I end up with an SHM equation. Here is the expression:

Iα/(kL)=sinθ(Lcosθ-L_0)[1-L_0/sqrt{(Lsin(θ))^2+(L_0^)2}]

How can I use small angle approximations to get the RHS to look like a constant multiple of theta? Or am I doing the problem in a way which is too complicated?

2. Jan 25, 2016

### haruspex

I note that the problem says vertical oscillations, but you have not taken gravity into account.
In the absence of gravity (e.g. horizontal oscillations on a smooth table) it will not be SHM. The equation will be of the general form $\ddot x=-Ax^3$. I've no idea what the solution would be.
So I suggest you include gravity. That will give an equilibrium position significantly below horizontal, and I would expect the first order approximation then to be SHM.

3. Jan 25, 2016

### TSny

For small oscillations, the spring may be considered as remaining horizontal. Δx can be expressed in terms of θ and the length of the rod in a simple way.

Since the rod is pivoted at its center of mass, you do not need to worry about gravity.

Last edited: Jan 25, 2016
4. Jan 25, 2016

### haruspex

No, you need to read the OP from "Now consider...." The pivot is at one end now.

If it were pivoted at its mass centre in this variant, then you would be stuck with an equation of the form $\ddot x=-x^3$, no? I don't see how treating it as though the spring remains horizontal changes that.
There is the solution x=-ci/t, but there's no obvious real solution. I can get it to an integral like $t=\int (1-x^4)^{-\frac 12}dx$.

5. Jan 25, 2016

### TSny

Thanks, I completely overlooked that part of the OP. Sorry for that.

Yes, I agree.

I can see how you get the integral form for t. Mathematica throws out some sort of elliptic function for the integral, which is not very enlightening for me.