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Consider a typical presentation of a simple pendulum: a point mass $m$ attached to a rigid rod of length $L$, which is free to rotate around a pivot. Newton's equations are

$$\begin{gather} mg \cos\theta - F = -m L \dot{\theta}^2 \\ -mg \sin\theta = m L \ddot{\theta}, \end{gather}$$

where $F$ is the force from the rod. It seems that it's universally assumed that this force points in the radial direction, like it would if the rod was replaced by a string - though of course the force from the rod can point outwards as well as inwards, while a string can only pull and not push.

Why is this? A rod is rigid, so it can transmit shear stress, so in principle the force from the rod could have a tangential component too. This is easily illustrated by considering a situation where the rod is attached to a motor which makes the system turn with a constant angular velocity $\omega$. We are forced to introduce a tangential component $F_\theta$ into the equations

$$\begin{gather} mg \cos\theta - F_r = -mL \omega^2 \\ -mg \sin\theta + F_\theta = 0 \end{gather}$$

because something has to balance the tangential component of the weight. But this only works because we have the constraint that $\dot{\theta}$ - if we do this for the pendulum, the system becomes undetermined since we're introducing one more unknown.

What is the difference between this case and the pendulum? What condition allows us to assume that the force from the rod is radial for the pendulum? And as a bonus question, is there a simple way to explain this to an introductory physics course?

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  • $\begingroup$ only if the rod is elastic, you have to take care about the tangential force ? $\endgroup$
    – Eli
    Apr 6, 2022 at 6:45

2 Answers 2

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What is the difference between this case and the pendulum? What condition allows us to assume that the force from the rod is radial for the pendulum?

The condition that explains both results is that the rod is assumed to be massless. Since the rod is assumed to be massless then both the net force and the net torque acting on the rod must be zero at all times, regardless of the acceleration.

Specifically, consider the torques acting on the rod at the axis of the hinge or motor. Since the rod is massless the torque from gravity is 0. That means that the torque from the mass must be equal and opposite from the torque from the hinge or motor.

In the case of the pendulum the torque from the hinge is zero, so the torque from the mass is also zero and therefore the tangential force from the mass is zero.

In the case of the motor the torque from the motor is non-zero, so the torque from the mass is equal and opposite and therefore the tangential force from the mass is non-zero.

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  • $\begingroup$ This might be something very basic that I'm not seeing, but how can you show that the hinge does not exert any torque for the pendulum? After all, it does exert a force. $\endgroup$
    – Javier
    Apr 6, 2022 at 18:16
  • $\begingroup$ @Javier that is the definition of a hinge. It is a mechanical or structural linkage that allows bending on one axis and exerts forces on other axes. See here for a brief overview: web.mit.edu/4.441/1_lectures/1_lecture13/1_lecture13.html $\endgroup$
    – Dale
    Apr 7, 2022 at 16:59
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Of course you can transmit shear forces through a rigid pendulum. For example, if you hold a pencil fixed at one end, you can spin it about that point however you like, by exerting a torque at that end. In that case the motion is completely undetermined since you can choose what torque to exert.

However, when we set up a physical pendulum, we are usually attaching the end to a so-called simple support, i.e. one which does not provide any torque. An example would be an ideal, frictionless pivot.

Any explanation of a rigid pendulum needs to distinguish between simple supports and supports that can provide torques. In the case of a simple support, there is no external torque on the pendulum except for gravity. You can then derive the usual equation of motion using $\tau_{\mathrm{ext}} = d L/dt$. Alternatively, in the special case of a massless rod, you can show that the rod carries only a tension along it (as otherwise it would have an infinite angular acceleration), and proceed with the usual derivation in terms of forces.

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