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Definition of a Physical Pendulum
In the article, "Simple Pendulum", we define a simple pendulum as an ideal pendulum that has all of the mass concentrated at one point. We can only use the idea of a simple pendulum when an object hanging from a string can be modeled as a point mass, and the mass of the string is negligible. In real life, however, most real pendulums cannot be modeled accurately as point masses on massless strings. We refer to such a real pendulum as a physical pendulum.
A physical pendulum is a pendulum in which an extended object hangs from a pivot point that is displaced from the center of mass, about which the object is free to rotate.
At the beginning of this article, we mentioned a tire swing as an example of a physical pendulum. The way a tire swing spins during its motion is evidence of how it cannot be modeled as a simple pendulum. Modeling the tire swing would be rather complicated considering how can rotate not only about the pivot point, but also the axis parallel to the rope. A more simple example of a physical pendulum is a straight rod with a pivot on one end, as shown below. We cannot ignore the mass of the rod as we did with the string in a simple pendulum, so we must consider the location of the center of mass and the moment of inertia when describing its motion.
Equation for a Physical Pendulum
Using Newton's second law, we can find an equation of motion for a physical pendulum. Consider a nonuniform object of mass \(m\) that hangs from a pivot so that it can rotate freely about that point. The distance from the pivot point to the center of gravity of the object is \(d.\) When the object experiences an angular displacement of \(\theta,\) the restoring torque on the object is given by:
\[\tau=-mgd\sin\theta,\]
where \(g\) is the acceleration due to gravity, \(g=9.8\,\mathrm{\frac{m}{s^2}}.\) There is a minus sign in this equation because the restoring torque points in the clockwise direction when the displacement is counterclockwise.
Since the restoring torque is proportional to \(\sin\theta\) instead of \(\theta,\) the motion is not simply harmonic. However, for small amplitudes of motion, we can approximate that \(\sin\theta\approx\theta.\) In this approximation, the restoring force is:
\[\tau=-mgd\sin\theta\approx-mgd\theta,\]
and the motion is approximately simply harmonic.
Now, we substitute the restoring force into Newton's second law in the rotational form to get an equation of motion for a physical pendulum:
\[\begin{align*}\sum\tau&=I\alpha\\[8pt]-mgd\theta&=I\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}\\[8pt]\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}&=-\frac{mgd}{I}\theta.\end{align*}\]
In this equation, \(I\) is the moment of inertia and \(\alpha\) is the angular acceleration. We recognize the term on the right hand side of the equation, \(\frac{mgd}{I},\) as the square of the angular frequency, \(\omega,\) so that \(\omega=\sqrt{\frac{mgd}{I}}.\) In terms of the angular frequency, the equation of motion is given by:
\[\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}=-\omega^2\theta.\]
Period of a Physical Pendulum
The period of a physical pendulum can be derived from the angular frequency found in the section above. The period in terms of the angular frequency is given by:
\[\begin{align*}T&=\frac{1}{f}\\[8pt]&=\frac{2\pi}{\omega}\\[8pt]&=\frac{2\pi}{\sqrt{\frac{mgd}{I}}}\\[8pt]&=2\pi\sqrt{\frac{I}{mgd}}.\end{align*}\]
Moment of Inertia of a Physical Pendulum
The moment of inertia of an object with a complex shape can be calculated by first finding its period of motion. First, we determine the center of gravity of the object. Then, the object is allowed to pivot about a certain point that is a distance \(d\) away from the center of gravity. We finally measure the period of the oscillation. Solving for the moment of inertia in the equation found in the previous section gives us:
\[\begin{align*}T&=2\pi\sqrt{\frac{I}{mgd}}\\[8pt]\frac{T}{2\pi}&=\sqrt{\frac{I}{mgd}}\\[8pt]\left(\frac{T}{2\pi}\right)^2&=\frac{I}{mgd}\\[8pt]I&=mgd\left(\frac{T}{2\pi}\right)^2.\end{align*}\]
We then substitute in the measured values for the mass, distance from the center of gravity, and period of oscillation to calculate the moment of inertia.
Simple Pendulum vs. Physical Pendulum
A simple pendulum is a special case of physical pendulums that can be modeled as a point mass on a massless string. In the case of a simple pendulum, the period is only dependent on the length of the string \(l\) and the acceleration due to gravity:
\[T=2\pi\sqrt{\frac{l}{g}}.\]
Try to prove this formula using the formula for the period of a physical pendulum and the expression for the moment of inertia of a simple pendulum: \(I=ml^2\).
A simple pendulum is a special case of a physical pendulums that can be modeled as a point mass that is a certain distance from the pivot point.
Though the calculations for a simple pendulum are less complex than those for a physical pendulum, most real-life pendulums cannot be modeled as a simple pendulum and must be treated as a physical pendulum.
Physical Pendulum Examples
Let's do a couple of examples with physical pendulums for practice!
A \(2.0\,\mathrm{m}\) uniform rod, like the one shown in the image at the beginning of this article, is pivoted at one end and allowed to oscillate. The moment of inertia for a uniform rod is given by \(I=\frac{1}{3}ml^2,\) where \(m\) is the mass of the rod and \(l\) is its length. What is its period of motion?
Since the rod is uniform, the center of gravity is located in the center of the rod. Thus, the distance from the pivot point to the center of gravity is \(d=\frac{l}{2}.\) Using the equation for the period of a physical pendulum, we find the period to be:
\[\begin{align*}T&=2\pi\sqrt{\frac{I}{mgd}}\\[8pt]&=2\pi\sqrt{\frac{\frac{1}{3}ml^2}{mg\frac{l}{2}}}\\[8pt]&=2\pi\sqrt{\frac{\frac{1}{3}\bcancel{m}l^{\bcancel{2}}}{\bcancel{m}g\frac{\bcancel{l}}{2}}}\\[8pt]&=2\pi\sqrt{\frac{2l}{3g}}\\[8pt]&=2\pi\sqrt{\frac{2(2.0\,\mathrm{m})}{3\left(9.8\,\mathrm{\frac{m}{s^2}}\right)}}\\[8pt]&=2.3\,\mathrm{s}.\end{align*}\]
The \(2.0\,\mathrm{m}\) uniform rod from the previous example is now pivoted \(0.5\,\mathrm{m}\) from the end of it, as shown in the image below. Now what is the period of motion? The new moment of inertia is \(I=\frac{5}{24}ml^2\).
We will calculate the period of motion the same way that we did in the previous example, but this time the distance from the center of gravity has changed. The center of gravity is still at the center of the rod, which is \(\frac{l}{2}\) away from one end. The pivot position is at \(0.5\,\mathrm{m},\) which is \(\frac{l}{4}\) away from one end. So, the distance from the center of gravity to the pivot point is:
\[\begin{align*}d&=\frac{l}{2}-\frac{l}{4}\\[8pt]&=\frac{l}{4}.\end{align*}\]
Using this new distance, the period is then:
\[\begin{align*}T&=2\pi\sqrt{\frac{I}{mgd}}\\[8pt]&=2\pi\sqrt{\frac{\frac{5}{24}ml^2}{mg\frac{l}{4}}}\\[8pt]&=2\pi\sqrt{\frac{\frac{5}{24}\bcancel{m}l^{\bcancel{2}}}{\bcancel{m}g\frac{\bcancel{l}}{4}}}\\[8pt]&=2\pi\sqrt{\frac{5l}{6g}}\\[8pt]&=2\pi\sqrt{\frac{5(2.0\,\mathrm{m})}{6\left(9.8\,\mathrm{\frac{m}{s^2}}\right)}}\\[8pt]&=2.6\,\mathrm{s}.\end{align*}\]
Physical Pendulum - Key takeaways
- A physical pendulum is a pendulum in which an extended object hangs from a pivot point that is displaced from the center of mass, about which the object is free to rotate.
- The motion of a physical pendulum is found using the rotational form of Newton's second law of motion; it is approximated to be simply harmonic for small amplitudes of oscillation.
- The period of a physical pendulum is dependent on the mass, moment of inertia, acceleration due to gravity, and the distance of the pivot point from the center of gravity: \(T=2\pi\sqrt{\frac{I}{mgd}}.\)
- A simple pendulum is a special case of a physical pendulum that can be modeled as a point mass that is a certain distance from the pivot point.
References
- Fig. 1 - Tire swing (https://pixabay.com/photos/outdoors-kids-playing-son-summer-1392494/) by halfpintohoney (https://pixabay.com/users/halfpintohoney-1284944/) licensed by Pixabay license (https://pixabay.com/service/license/).
- Fig. 2 - Uniform rod as a pendulum, StudySmarter Originals.
- Fig. 3 - A uniform rod with offset pivot, StudySmarter Originals.
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Frequently Asked Questions about Physical Pendulum
What is a physical pendulum in physics?
A physical pendulum is an extended object that hangs from a pivot point that is displaced from the center of mass, about which the object is free to rotate.
What's the difference between a physical pendulum and a simple pendulum?
A simple pendulum is a special case of physical pendula that can be modeled as a point mass on a massless string. Though the calculations for a simple pendulum are less complex than those for a physical pendulum, most real-life pendula cannot be modeled as a simple pendulum and must be treated as a physical pendulum.
What are some examples of physical pendulum?
Some examples of a physical pendulum are clocks, a tire swing, and a uniform rod pivoted at one end.
What is the equation for solving a physical pendulum?
We use the rotational form of Newton's second law of motion to find the equation of motion for a physical pendulum. This equation of motion leads us to an equation for the period.
What is the derivative of the physical pendulum?
For small amplitudes of motion, the period of a physical pendulum is derived from the application of Newton’s second law in rotational form.
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