1
$\begingroup$

It's easy to use torque to analyze a pendulum. But how do we do this only using forces?

Specifically, working through a Morin problem that models a pencil balancing on its tip with a mass $m$ on a massless rod of length $l$. At the bottom of the rod I specified a force $N$ pointing upwards, and $R$ pointing horizontally (as if there was a hinge there; sorry for no diagram).

Applying the radial and angular equations, I get:

$$R\sin{\theta} - (mg-N)\cos{\theta} = -ml\dot{\theta}^2$$

$$R\cos{\theta} + (mg-N)\sin{\theta} = ml\ddot{\theta}. $$

Because $\theta$ is low I can assume $\cos{\theta} = 1$ and that $\sin{\theta} = \theta$, but I still can't compute $R$ and $N$. I need to find $\dot{\theta}$ and $\ddot{\theta}$. How do do this without applying torque and angular momentum?

$\endgroup$
8
  • $\begingroup$ what is the gemoetric relation of the pencil to the mass and rod, where are they connected. Is this just a pendulum? $\endgroup$
    – JMLCarter
    Dec 22, 2016 at 23:53
  • $\begingroup$ Where is the first equation coming from? for a simple pendulum radial force cannot influence angular velocity. $\endgroup$
    – JMLCarter
    Dec 22, 2016 at 23:56
  • $\begingroup$ Mass m on a massless tod of length l, standing upright. Used circular motion equations. $\endgroup$ Dec 23, 2016 at 0:11
  • $\begingroup$ For a simple pendulum, you can write the Lagrangian of the system $\mathcal L = T - V = \frac{1}{2}ml^2\dot\theta^2 + mgl\cos\theta$ for an angle $\theta$ from the vertical and $y$ measured from the attachment point. If you solve the Euler-Lagrange equations, you will find $\ddot \theta = - (g/l)\sin\theta$. Not sure if that is what you're looking for, but I'm happy to elaborate if it helps. Note this has an analytic solution for $\sin\theta \approx \theta$. $\endgroup$
    – zh1
    Dec 23, 2016 at 3:07
  • $\begingroup$ @FarazMasroor If the basis for your first equation is that the tangential acceleration in uniform circular motion is proportional to the square of the angular velocity, that is inapplicable here. A pendulum is not in uniform circular motion, since there is always a varying component of gravity acting tangentially to its locus. $\endgroup$ Dec 23, 2016 at 4:01

1 Answer 1

3
$\begingroup$

In my original comment, I suggested solving your problem using Lagrangian mechanics. However, I realized you can do this using forces if you change into polar coordinates. Newton's second law, in polar coordinates, has the form

\begin{equation} \vec F_{net} = m(\ddot r - r\dot\phi^2)\hat r + m(2\dot r \dot \phi + r\ddot \phi)\hat \phi \end{equation}

where $\hat r$ is unit vector that points along the pendulum's rod and $\hat \phi$ is a unit vector parallel to the bob's velocity vector. By drawing a force diagram, we find that

$$ \vec F_{net} = (F_T - mg\cos\phi)\hat r + (-mg\sin\phi)\hat \phi,$$

where $F_T$ is the force from the rod. Except that for your purposes, we don't really care about the $\hat r$ direction too much. Since our unit vectors are linearly independent, we can ignore the $\hat r$ direction and combine these two equations to find

$$ -mg\sin\phi = m(2\dot r\dot\phi + r\ddot\phi).$$

Knowing that $\dot r$ is zero (the rod length doesn't change), we can rewrite this to find

$$ \ddot\phi = - \frac{g}{r}\sin\phi, $$

where in your notation $l = r$ and $\phi = \theta$. Note that this is exactly what you find by solving the Euler-Lagrange equations. As it stands, this DE cannot be solved analytically. However, we can constrain the pendulum to small angles such that

$$ \ddot \phi = - \frac{g}{r}\phi, $$

a linear, homogeneous differential equation which we can solve to find

$$ \phi(t) = c_1\sin(\sqrt{\frac{g}{r}}t) + c_2\cos(\sqrt{\frac{g}{r}}t),$$

where $\sqrt\frac{g}{r}$ is the angular frequency of the pendulum. You may notice that torque and angular momentum are somewhat implicit in this derivation. However, we solved the problem using only a force diagram, which is what I perceive to be your question.

$\endgroup$
2
  • $\begingroup$ It's not obvious to me why the force from the rod points strictly in the radial direction and doesn't have a tangential component. $\endgroup$ Dec 23, 2016 at 17:49
  • $\begingroup$ I'm defining the $\hat r$ unit vector to always point in the direction of the rod - regardless of the rod's orientation. That is, the $\hat r$ vector is always orthogonal to the bob's velocity vector. This means that its direction changes over time so that it always points toward the rod's attachment point. You might read section 1.7 of John Taylor's classical mechanics for further clarification. qom.ac.ir/portal/File/… $\endgroup$
    – zh1
    Dec 23, 2016 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.