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Downloaded 2009-01-16 from Charles H. Henderson & John F. Woodhull (1901) The Elements of Physics, D. Appleton & Co., New York, p.59, fig.21 |
Adding detail to the model of the pendulum.
The "Solution" header and the bold items below should NOT be changed. Only the regular text within the macros should be altered.
Solution
System: A model of a pendulum, simply supported and free to swing without friction about a supporting axis under the torque due to gravity.
Interactions: torque due to gravity and the upward force exerted against gravity by the axis.
Model: Angular Momentum and External Torque about a Single Axis
Approach:
Diagrammatic Representation
We consider first the usual Simple Model of a Pendulum
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And then a slightly more detailed model
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With two variations.
Mathematical Representation
The simple model has the virtue that it is extremely simple to calculate themoment of inertia, I, of the pendulum about the axis of rotation. We assume a massless stick of length L and a point mass m at the end. The moment of inertia is simply
\begin
[ I = mL^
]\end
If we pull the pendulum away from its vertical equilibrium position by an angle θ, then the restoring force Fres is given by
\begin
[ F_
= m g sin(\theta) ]\end
And the natural frequency, as noted in the vocabulary entry on pendulum, is given by
\begin
[ \omega = \sqrt{\frac
{L}} ]\end
Now consider the second diagram above. Now, instead pf a point particle, the mass of the pendulum ihas real extent. It is a disc of radius r. We can calculate the real moment of inertia by using the parallel axis theorem. According to that, we can calculate the moment of inertia about a point other than the center of mass (about an axis parallel to one running through the center of mass) by simply adding md2 to it, where d is the distance between the center of mass and the axis of rotation. For a uniform disc, rotating about its center, the moment of inertia is
\begin
[ I = \frac
mr^
]\end
In this case d is the length of the pendulum L, so the moment of inertia is
\begin
[ I = \frac
mr^
+ mL^
]\end
and the angular frequency is given by
\begin
[ \omega = \sqrt{\frac
{I}} = \sqrt{\frac
{\frac
mr^
+ mL^
}} ]\end
After some algebra, this reduces to:
\begin
[ \omega = \sqrt{\frac
{L + \frac{r^{2}}
}} ]\end
So taking into account the physical size of the mass reduces the frequency. Why is this? The answer is that now we must consider not only the moment associated with moving the center of mass from side to side, we must also consider the moment of inertia associated with rotating that mass as well. In the simple model we simply ignored this, since you can't rotate a point particle. A real physical pendulum would behave the same way if it were mounted to the end of the stick of length L by a perfectly frictionless axle, leaving it free to rotate. Since nothing exerts a torque on the mass that requires it to rotate (only to move its center of mass), , the mass does not have to rotate, and the frequency will be the same as for a point mass. Notice that, in the limit as r approaches zero, the physical pendulum result tends towards the simple pendulum result. For a radius significantly smaller than L, the physical pendulum will behave like a simple pendulum.
What this means is that, if you have two pendulums of identical dimensions, but with one having the mass rigidly mounted and the other mounted on a freely rotatring bearing, the latter will have a higher rate of oscillation. That's a surprising result.
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