AP Physics C Mechanics- 7.5 Simple and Physical Pendulums- Study Notes- New Syllabus
AP Physics C Mechanics- 7.5 Simple and Physical Pendulums – Study Notes
AP Physics C Mechanics- 7.5 Simple and Physical Pendulums – Study Notes – per latest Syllabus.
Key Concepts:
- Physical Pendulum
- Period of a Physical Pendulum
- Simple Pendulum as a Special Case
- Torsion Pendulum
Physical Pendulum
A physical pendulum is any rigid body that oscillates about a fixed horizontal axis under the influence of gravity. Unlike a simple pendulum (which assumes all mass concentrated at one point), a physical pendulum’s mass is distributed, and its motion depends on the moment of inertia about the pivot and the distance to its center of mass.
Period of a Physical Pendulum
For small angular displacements, the motion of a physical pendulum is simple harmonic. Its period is derived using Newton’s second law in rotational form:
\( \mathrm{T_{phys} = 2\pi\sqrt{\dfrac{I}{Mgd}}} \)
- \( \mathrm{I} \): moment of inertia about the pivot
- \( \mathrm{M} \): mass of the pendulum
- \( \mathrm{g} \): acceleration due to gravity
- \( \mathrm{d} \): distance from pivot to center of mass
Restoring Torque on a Physical Pendulum
When displaced by an angle \( \mathrm{\theta} \) from equilibrium, the gravitational force acting at the center of mass produces a restoring torque:
\( \mathrm{\tau = -Mgd\sin\theta} \)
This torque acts to return the pendulum toward its equilibrium position.
Small-Angle Approximation
For small angular displacements (\( \mathrm{\theta < 15°} \)), we can use the approximation \( \mathrm{\sin\theta \approx \theta} \). Substituting into the torque equation:
\( \mathrm{\tau = -Mgd\theta = I\alpha} \)
or
\( \mathrm{\alpha = -\dfrac{Mgd}{I}\theta} \)
This equation matches the form of simple harmonic motion.
Differential Equation of SHM for a Physical Pendulum
Applying Newton’s second law for rotation with the small-angle approximation yields:
\( \mathrm{\dfrac{d^2\theta}{dt^2} = -\dfrac{Mgd}{I}\theta} \)
This is the standard SHM equation, where \( \mathrm{\omega^2 = \dfrac{Mgd}{I}} \).
Hence, the angular frequency
\( \mathrm{\omega = \sqrt{\dfrac{Mgd}{I}}} \) and period \( \mathrm{T = 2\pi\sqrt{\dfrac{I}{Mgd}}} \).
Example
A uniform rod of length \( \mathrm{L = 1.0\,m} \) and mass \( \mathrm{M = 0.5\,kg} \) is pivoted about one end. Find its period of oscillation.
▶️ Answer / Explanation
Step 1: \( \mathrm{I = \tfrac{1}{3}ML^2,\ d = \tfrac{L}{2}} \)
Step 2: \( \mathrm{T = 2\pi\sqrt{\dfrac{I}{Mgd}} = 2\pi\sqrt{\dfrac{(1/3)ML^2}{Mg(L/2)}} = 2\pi\sqrt{\dfrac{2L}{3g}}} \)
Step 3: Substitute \( \mathrm{L = 1.0\,m,\ g = 9.8\,m/s^2} \): \( \mathrm{T = 2\pi\sqrt{\dfrac{2(1)}{3(9.8)}} = 2\pi(0.26) = 1.63\,s.} \)
Result: The period of the physical pendulum (uniform rod) is \( \mathrm{1.63\,s.} \)
Simple Pendulum as a Special Case
A simple pendulum is a special case of a physical pendulum in which all the mass is concentrated at a single point at distance \( \mathrm{L} \) from the pivot. It behaves as an idealized system with a massless string and a point mass bob.
Relevant equation:
\( \mathrm{T_p = 2\pi\sqrt{\dfrac{L}{g}}} \)
Key Idea: For small angles, the simple pendulum exhibits simple harmonic motion where the restoring torque is \( \mathrm{\tau = -mgL\sin\theta \approx -mgL\theta} \).
Example
Find the period of a simple pendulum of length \( \mathrm{0.75\,m} \) oscillating at small amplitude near Earth’s surface.
▶️ Answer / Explanation
Step 1: Use the period equation:
\( \mathrm{T = 2\pi\sqrt{\dfrac{L}{g}} = 2\pi\sqrt{\dfrac{0.75}{9.8}} = 2\pi(0.276) = 1.73\,s.} \)
Result: The simple pendulum completes one oscillation in \( \mathrm{1.73\,s.} \)
Torsion Pendulum
A torsion pendulum consists of a rigid body (e.g., a disk) suspended by a wire. When twisted, the wire exerts a restoring torque proportional to the angular displacement:
\( \mathrm{\tau = -k\Delta\theta} \)
- \( \mathrm{k} \): torsional constant of the wire (torque per unit angular displacement)
- \( \mathrm{\Delta\theta} \): angular displacement from equilibrium
Applying Newton’s second law in rotational form:
\( \mathrm{I\alpha = -k\theta} \Rightarrow \dfrac{d^2\theta}{dt^2} = -\dfrac{k}{I}\theta \)
This shows the motion is simple harmonic with angular frequency:
\( \mathrm{\omega = \sqrt{\dfrac{k}{I}}} \quad \text{and} \quad \mathrm{T = 2\pi\sqrt{\dfrac{I}{k}}} \)
Example
A torsion pendulum with moment of inertia \( \mathrm{I = 2.0\times10^{-4}\,kg·m^2} \) is suspended by a wire with torsion constant \( \mathrm{k = 0.02\,N·m/rad} \). Find its period of oscillation.
▶️ Answer / Explanation
Step 1: Use torsion pendulum formula:
\( \mathrm{T = 2\pi\sqrt{\dfrac{I}{k}} = 2\pi\sqrt{\dfrac{2.0\times10^{-4}}{0.02}} = 2\pi\sqrt{0.01} = 2\pi(0.1) = 0.63\,s.} \)
Result: The torsion pendulum oscillates with a period of \( \mathrm{0.63\,s.} \)