Below is a picture of a simple pendulum clock. Suppose that the bob (a rigid disk) on the end of the pendulum can spin without friction about its geometrical axis and is spinning at an angular velocity $\omega=const$. Will the clock gain or lose time compared to the ordinary pendulum clock. Or it does not have any effect on the time? If it does then what is the percentage? Ignore air drag.

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The clock runs slightly faster when the disk can spin. When the disk is attached rigidly to the pendulum, its center translates in space, but it also spins around its own center (compared to gravity). The disk experiences net acceleration and net angular acceleration.

When the disk can spin freely, the center of the disk still accelerates, but the disk has no angular acceleration because there is no torque on it. (Uniform gravity cannot produce a torque about the center of mass, and the force between the pendulum and the disk has no moment arm.) This slightly reduces the energy change in the disk for a given movement of the pendulum because there is no energy going in and out of the spin of the disk. The net result is a lower effective moment of inertia in the pendulum+disk system. The torque on the pendulum due to gravity is the same, so the clock swings slightly faster when the disk is free to spin.

Variable definitions:

  • $l$ length of pendulum
  • $m$ mass of pendulum
  • $\theta$ angle of pendulum with vertical
  • $L$ distance between center of disk and pivot of pendulum
  • $R$ radius of disk
  • $M$ mass of disk
  • $\phi$ angle the disk has rotated relative to gravity
  • $\omega = \dot{\phi}$
  • $g$ gravitational acceleration

We'll find the Lagrangian of the system and compare the two cases. When the disk is fixed, $\phi = \theta$. When the disk is free, $\phi$ is a new degree of freedom.

We need to find the kinetic and potential energies of the pendulum and the bob in terms of the variables above.

Assume the pendulum is a uniform rod pivoted at one end. Then its moment of inertia is $ml^2/3$. Its kinetic energy is $\frac{1}{2} I \dot{\theta}^2 = ml^2\dot{\theta}^2/6$.

The potential energy of the pendulum is $mg$ times the height of its center of mass above rest, which is $l\cos\theta /2$.

The kinetic energy of the disk comes from the motion of its center of mass and from its rotation. It is $1/2(Mv^2) + 1/2I\omega^2$. The velocity is $v = L\dot{\theta}$. The moment of inertia is $I = 1/2 M R^2$. The total kinetic energy is then $1/2M ( L^2 \dot{\theta}^2 + 1/2 R^2 \dot{\phi}^2)$

The potential energy of the disk is $Mg$ times its height above equilibrium height, or $MgL\cos\theta$.

Putting these together, the Lagrangian for the clock with spinning bob is

$$\mathcal{L} = \dot{\theta}^2\left(\frac{ml^2}{6}+ \frac{ML^2}{2}\right) + \dot{\phi}^2\frac{MR^2}{4} + g\cos\theta\left(\frac{ml}{2} + M L\right)$$

which we abbreviate

$$\mathcal{L} = \dot{\theta}^2 \frac{I_\theta}{2} + \dot{\phi}^2 \frac{I_b}{2} + g\cos\theta M_{eff}$$

There are two different cases to explore. If the bob of the pendulum is fixed, then $\phi = \theta$ because the bottom of the disk points the same way as the pendulum. That gives us $\dot{\phi} = \dot{\theta}$. The Lagrangian becomes

$$\mathcal{L}_{fixed} = \dot{\theta}^2 \frac{I_\theta + I_b}{2} + g\cos\theta M_{eff}$$

or, setting $I_\theta + I_b = I_{fixed}$, it's just

$$\mathcal{L}_{fixed} = \dot{\theta}^2 \frac{I_{fixed}}{2} + g\cos\theta M_{eff}$$

When the disk is free to swing, $\phi$ is a degree of freedom on its own. The Lagrangian has no cross-terms involving both $\theta$ and $\phi$, meaning their evolutions are independent; these degrees of freedom are decoupled. $\phi$ is a cyclic coordinate so the angular momentum of the disk is conserved. The evolution of theta is the same as in a Lagrangian where $\phi$ does not exist.

$$\mathcal{L}_{free} = \dot{\theta}^2 \frac{I_\theta}{2} + g\cos\theta M_{eff}$$

The only difference between $\mathcal{L}_{fixed}$ and $\mathcal{L}_{free}$ is that the moment of inertia is slightly larger in $\mathcal{L}_{fixed}$.

The potential energy terms are the same. If we make a substitution in the free-bob case, introducing a new time coordinate defined by

$$t' = t \sqrt{\frac{I_\theta}{I_{fixed}}}$$

the Lagrangians take exactly the same form, except that they're expressed in different time coordinates.

This means the free-bob pendulum will run in the same way, but faster by the factor

$$\frac{\nu_{fixed}}{\nu_{free}} = \frac{t}{t'} = \sqrt{\frac{I_{fixed}}{I_\theta}} = \sqrt{\frac{ml^2/3 + M(L^2 + R^2/2)}{ml^2/3 + ML^2}}$$

Where $\nu$ indicates a period.

For tractability, we might look at some limit, such as $l = L$, $m = M$, $L>>R$. The factor by which the free-disk clock is faster becomes

$$1 + \frac{1}{3}\left(\frac{R}{L}\right)^2$$

  • 2
    $\begingroup$ Are you sure that the inertial moment of the pendulum is the same at both cases: a) the bob is rigidly attached to the pendulum and b) the bob can without friction spin about its axis $\endgroup$ – Martin Gales Feb 21 '11 at 7:50
  • $\begingroup$ @Martin Good point. I'm going to rethink this. $\endgroup$ – Mark Eichenlaub Feb 21 '11 at 7:59
  • $\begingroup$ @Martin Thanks for pointing that out. Re-worked the answer. $\endgroup$ – Mark Eichenlaub Feb 21 '11 at 9:10
  • $\begingroup$ I think that your answer is correct. Maybe it is worth stating explicitly that the clock period would not depend on $\omega$. $\endgroup$ – gigacyan Feb 21 '11 at 9:42

In the case of the first pendulum, disk has to rotate and this is classical physical pendulum. In the case of the second pendulum, you don't expect disk to rotate (the only force is in the center of mass, so there is no torque), so therefore you simply have only transitional movement - mathematical pendulum.

Using equation for periods of physical pendulum

$$T = 2 \pi \sqrt{\frac{I}{m g l}}$$

and noting, that

$$I = I_\textrm{cm} + m l^2,$$

where $I_\textrm{cm} > 0$ is center-of-mass inertia of the disk, you get

$$T = 2 \pi \sqrt{\frac{l}{g} + \frac{I_\text{cm}}{m g l}}.$$

On the other hand equation for periods of mathematical pendulum

$$T = 2 \pi \sqrt{\frac{l}{g}}.$$

Obviously, period is smaller for the second pendulum.

  • $\begingroup$ My answer obviously differs from the previous, as $\left( \frac{T}{T'} \right)^2= 1 + \frac{1}{2} \left( \frac{R}{l} \right)^2$. I invite everybody to find the mistake in either of the answers. Greets! $\endgroup$ – Pygmalion Apr 15 '12 at 8:39
  • $\begingroup$ The fixed case is not a mathematical pendulum. It is still a physical pendulum, but the moment of inertia is different. The other answer is the correct one. $\endgroup$ – Michael Brown Mar 20 '13 at 2:04
  • $\begingroup$ I don't quite understand your comment. I do agree that fixed case is physical pendulum. I just say that loose case is equivalent to mathematical pendulum. $\endgroup$ – Pygmalion Mar 22 '13 at 9:23

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