I am thinking about the following pendulum-spring system. But something is off with my equation of motion.


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We have a uniform rod of length L with mass m pivoted at one end. We should assume that mass, m, to be a point mass.

The other end is attached to a horizontal spring with spring constant k. The spring is neither stretched nor compressed when the rod is perfectly vertical. We can also assume that the force due to the spring is always horizontal.


I want to derive the system's equation of motion, and then also find the angular frequency.

Original attempt

Looking at it, I think what we have is a physical pendulum with the caveat that a spring is attached.

Distance from pivot to centre of mass is $\frac{1}{2}L$. I set this to equal $l$.

Find expression for torque:

$\tau_{rod}= -mglsin\theta$

When $\theta$ is small, $\sin\theta \approx \theta$

So $\tau_{rod}= -mgl\theta$

For the spring, if we assume the angle will remain small then torque from spring will be:

$\tau_s = -kL^2\theta$ for length of entire rod (not $l$).

Hence $\sum \tau = (-kL^2\theta)+(-mgl\theta) = -(kL^2 + mgl)\theta$

Equation of motion

Now that we have the torque, my instinct is to try to derived the system's equation of motion.

Newton's second law for rotational motion is:

$\alpha = \frac{d^2 \theta}{dt^2} = \frac{\tau}{I}$, where $\tau$ is the torque and $I$ the inertia.

Invoking Newton's 2nd law, I get:

$\sum \tau = I \ddot{\theta}$

We know that the moment of inertia, I, for a thin rod with point mass at end is $1/3 mL^2$


$\sum \tau = I \ddot{\theta} = \frac{1}{3} mL^2 \ddot{\theta}$

Rearranging this, I get:

$\ddot{\theta} = \frac{\tau}{\frac 13 mL^2}$

Sub for $\tau$

$\ddot{\theta} = \frac{-(kL^2 + mgl)\theta}{\frac 13 mL^2}$

$\therefore$ the equation of motion for the system is: $\ddot{\theta} + \frac{(kL^2 + mgl)}{\frac 13 mL^2}\theta = 0$

Angular frequency

When I then go to calculate the angular frequency, I get:

$\omega = 2\pi F = \sqrt{\frac{(kL^2 + mgl)\theta}{\frac 13 mL^2}}$

But something about the denominator appears off to me. I think it is because I am unsure of how to calculate the moment of inertia, and include it in my equation.

  • 3
    $\begingroup$ Given the title of the question, I was expecting to see a Lagrangian... Also, you need to use the moment of inertia of a rod, corrected according to the Huygens-Steiner theorem. $\endgroup$ – Phoenix87 Nov 19 '18 at 21:01
  • 2
    $\begingroup$ Yeah. Why are you using Newtonian mechanics when you say you want to use Lagrangian mechanics? $\endgroup$ – Aaron Stevens Nov 19 '18 at 21:08
  • $\begingroup$ The kinetic energy in terms of $L$, $m$, and $\dot \theta^2$ is the same as w/o the spring. The potential energy picks up term depending on $k$ and $(L\sin{\theta})^2$. That's all you need. $\endgroup$ – JEB Nov 19 '18 at 21:23
  • $\begingroup$ Also $\ddot\theta$ is not the angular frequency $\endgroup$ – Aaron Stevens Nov 20 '18 at 0:27
  • $\begingroup$ The question has been reframed, and language clarified. $\endgroup$ – thegeneralcase Nov 20 '18 at 17:48

I'm not sure what you don't like about your denominator, but your set up seems fine. The issue is when you find your angular frequency.

The angular frequency for SHM is seen in the differential equation of motion in general by $$\frac{\text d^2 x}{\text d t^2}=-\omega^2x$$

So in comparing what you have, you can see you have the wrong expression for $\omega$. I'll leave it to you to see what it is.

  • $\begingroup$ Thanks, Aaron, I will think this through. I did notice one thing, though. I think the assumption for the moment of inertia might be wrong. Instead of $\frac{1}{3} ML^2$ it would be $\frac{1}{12}ML^2$. Also, @Phoenix87 is right, my original solution is missing the inclusion of the parallel axis theorem, so for the total inertial I would instead have 1/12*mL^2 + m (L/2)^2 ? $\endgroup$ – thegeneralcase Nov 20 '18 at 18:46
  • $\begingroup$ @thegeneralcase Well $\frac{1}{12}+\frac14=\frac13$ so... $\endgroup$ – Aaron Stevens Nov 20 '18 at 18:49
  • $\begingroup$ @thegeneralcase In other words, the parallel axis theorem is fine, but you already had the moment of inertia correct to begin with $\endgroup$ – Aaron Stevens Nov 20 '18 at 19:09
  • $\begingroup$ Yes, I think you're right and I've just convinced myself that it was the right assumption. I still need to think through the angular frequency, which I will do now, but I can see what you're saying. I say that understanding the equation of motion is $\frac{d^2 x}{dt^2} = - \frac{k}{m} x$. Thank you very much for your time. $\endgroup$ – thegeneralcase Nov 20 '18 at 19:21
  • $\begingroup$ @thegeneralcase That equation is for a mass oscillating on a spring. I was not saying $\omega^2=\frac{k}{m}$ for your problem. $\endgroup$ – Aaron Stevens Nov 20 '18 at 19:58

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