L7.5 Simple and physical pendulums
A physical pendulum is a rigid body that undergoes oscillations about a fixed axis.
Derivation from differential equation
Consider a physical pendulum as shown in the figure. \[\begin{align} \sum{\tau} &= I \alpha \\ - m g d \sin{\theta} &= I \dfrac{d^2\theta}{dt^2} \label{eq:n2lphysicalpendulum} \end{align}\] For small \(\theta\), \(\sin{\theta}\approx\theta\), so \[\begin{align} - m g d \theta &= I \dfrac{d^2\theta}{dt^2} \\ 0 &= \dfrac{d^2\theta}{dt^2} + \dfrac{m g d}{I} \theta \end{align}\] As before we use the method of assumed solutions / method of undetermined coefficients / educated guess that \(\theta = A \cos{\omega t + \phi}\). Substituting this test solution and simplifying gives \[\begin{align} 0 &= A \cos{\omega t + \phi} (- \omega^2 + \dfrac{m g d}{I}) \\ \omega^2 &= \dfrac{m g d}{I} \\ \omega &= \sqrt{\dfrac{mgd}{I}} \end{align}\] The eigenvalues of \[eq:n2lphysicalpendulum\] are \(\pm j \sqrt{\dfrac{mgd}{I}}\) and they represent an undamped oscillatory response in which the pendulum swings back and forth.
Solution to \[eq:n2lphysicalpendulum\]
The end solution is testable on the AP Physics C exam. For small amplitudes of motion, the period of a physical pendulum is derived from the application of Newton’s second law in rotational form. Recalling that \(T = \dfrac{1}{f} = \dfrac{2\pi}{\omega}\): \[\begin{equation} T = 2\pi \sqrt{\dfrac{I}{mgd}} \end{equation}\]
While you don’t have to solve \[eq:n2lphysicalpendulum\] yourself, you should know where it comes from. When displaced from equilibrium, the gravitational force exerted on a physical pendulum’s center of mass provides a restoring torque. \[\begin{equation} \tau = -mgd \sin{\theta} \end{equation}\] For small amplitudes of motion, the small angle approximation can be applied to the restoring torque. \[\begin{align} \sin{\theta} &\approx \theta \\ \tau &\approx -mgd\theta = I \alpha \end{align}\] This gives a linear, second order, constant coefficient differential equation for \[eq:n2lphysicalpendulum\] in the area near the equilibrium point. We have linearized the system by assuming small displacements from equilibrium... sound familiar? The small-angle approximation and Newton’s second law in rotational form yield a second order differential equation that describes simple harmonic motion, namely: \[\begin{equation} \dfrac{d^2\theta}{dt^2} = -\omega^2 \theta \end{equation}\]
Simple pendulum as a special case of a physical pendulum
A simple pendulum is a special case of physical pendulums in which the hanging object can be modeled as a point mass at a distance, \(l\) from the pivot point. For this case, \(I=ml^2\), so that \[\begin{align} T &= 2\pi \sqrt{\dfrac{ml^2}{mgl}} \\ &= 2\pi \sqrt{\dfrac{l}{g}} \end{align}\]
Torsional pendulum
A torsion pendulum is a case of simple harmonic motion where the restoring torque is proportional to the angular displacement of a rotating system. For example, a horizontal disk that is suspended from a wire attached to its center of mass may undergo rotational oscillations about the wire in the horizontal plane. \[\begin{equation} I \alpha = - k \Delta\theta \end{equation}\] It is the rotational analog of a mass-spring oscillator, and is solved similarly by assuming a test solution of the form \(A\cos{\omega t+\phi}\) \[\begin{align} I \dfrac{d^2\theta}{dt^2} &= - k \theta \\ 0 &= I \dfrac{d^2\theta}{dt^2} + k \theta \\ &= A \cos{\omega t +\phi} (-I \omega^2 + k) \\ \omega^2 &= \dfrac{k}{I} \\ \omega &= \sqrt{\dfrac{k}{I}} T &= 2\pi \sqrt{\dfrac{I}{k}} \end{align}\]
Bonus: derivation of equations of motion using the Lagrangian
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See also
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