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We know that when a pendulum with a rope is at its maximum height, there is no tension in the rope and no force on the pivot.

If we consider the same situation and position, but where the rope is replaced with a solid rod. Is the pivot supporting the mass's weight? If so, does this have any affect on the motion of the pendulum?

user2005848
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2 Answers2

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Having a rod rather than a string affects the possible motions of the pendulum, and in particular the motion if the pendulum ever passes above the horizontal. If you assume the rod is light (which is the usual thing to do), then the difference between the pendulum with a rod and the pendulum with a string is that the tension in the string must never be negative, while the tension in the rod can be (ie the rod can be compressed, the string can't be or the pendulum will stop being a pendulum).

If you think about the angle, $\theta$ of the pendulum, where $\theta = 0$ is the pendulum hanging directly downwards, and take the bob mass to be $m$, then the tension in the rope / rod has two components.

  • $mg\cos\theta$ is the component due to the mass of the bob. Note that $\cos\theta \le 0$ when $\pi/2 \le \theta \le 3\pi/2$, when the bob is above the horizontal.
  • $m\omega^2l$ is the component due to the angular velocity, $\omega$ of the bob ($\omega = \dot\theta$).

So we have an expression for the total tension which is:

$$t = m\left(g\cos\theta + \omega^2 l\right)$$

$t$ can only be negative if $\theta \in (\pi/2, 3\pi/2)$: it's always positive or zero if $\cos\theta \ge 0$ in other words.

For the rod, $t$ is allowed to be negative, and there are thus no constraints on $\omega$ at all: it can drop to zero at any point (or it can not drop to zero at all, in which case the pendulum just rotates).

For the string, $t$ is not allowed to be negative, and there are therefore constraints on $\omega$ when the pendulum is above the horizontal. The constraint is

$$\omega \ge \sqrt{\frac{-g\cos\theta}{l}}\quad\text{(remember $\cos\theta\le 0$)}$$

Now if you think about this a little you'll see that, if the pendulum ever passes above the horizontal, then $\omega$ can never drop to zero until it falls below the horizontal again, and hence the pendulum must pass right through the vertical (pointing straight up, $\theta = \pi$) position! And it follows from this that $\omega$ is never zero anywhere: the pendulum must rotate right round, endlessly.

And it's also easy to see, just by thinking about energy, that $\omega$ will be at its minimum when $\theta=\pi$ -- when the pendulum is pointing straight up in other words. And we can use the expression above to deduce that

$$\omega_{\theta=\pi} \ge \sqrt{\frac{g}{l}}$$

So this is a constraint for a string pendulum, if it ever passes above the horizontal. There is no such constraint for a rod pendulum.

What this means is that, for a string pendulum, there are three cases:

  • if the maximum angle is less than $\pi/2$ (below the horizontal) then the tension at the maximum angle is always positive;
  • if the maximum angle is exactly $\pi/2$ (at the horizontal) then the tension at the maximum angle is zero, and this is the only case where this is allowed;
  • the maximum angle can not be more than $\pi/2$: if the pendulum swings above the horizontal it must has sufficient angular velocity to swing right round.

For a rod pendulum, the third case turns into

  • if the maximum angle is above $\pi/2$ then the tension at the maximum angle, if there is one, is negative.
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Both the rope and the rod behave similarly for low angles of oscillation. They cannot increase their length. However, when the oscillation amplitude increases, such as when the pendulum goes above the horizontal, things change. If You have enough energy so that the pendulum still has some velocity when it reaches the top of the circle, there is still a tension in the string, trying to extend the rope/rod. However, if the energy is not high enough, the pendulum reaches a point where the tension in the rope/rod is zero. Here the behavior is different. The rope cannot stay extended any longer, and the pendulum will fall on a parabolic trajectory, until the distance to the center of rotation equal to the length of the rope. On the other hand, the rod will not change length, so the pendulum will feel a tension in the rod, pushing outwards. The pendulum on the rod will keep moving back and forth along the circle.

Andrei
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