7.5 Simple and Physical Pendulums

AP Physics C 7.5 Simple vs Physical Pendulum Summary

A physical pendulum is any rigid body that swings about a pivot and undergoes simple harmonic motion for small angles, with period $T_{\text{phys}}=2\pi\sqrt{\frac{I}{mgd}}$. A simple pendulum is the special case where all the mass can be modeled as a point a distance $\ell$ from the pivot, giving $T_p=2\pi\sqrt{\frac{\ell}{g}}$. A torsion pendulum oscillates because a twisted wire supplies a restoring torque proportional to angular displacement.

Why This Matters for the AP Physics C: Mechanics Exam

This topic pulls together rotational dynamics, torque, moment of inertia, and SHM into one model, so it tests whether you can connect ideas across units. You will use Newton's second law in rotational form to set up the differential equation that describes the motion, then read off the angular frequency and period.

Because pendulums are easy to set up and time, they show up in experimental reasoning. You might design a procedure to measure $g$ from a pendulum's period, or determine an unknown moment of inertia from a measured period. Free-response work often asks you to derive an expression, sketch a graph of angular displacement versus time, or justify a claim using the period equation and its functional dependence.

Key Takeaways

• A physical pendulum is a rigid body oscillating about a pivot; the restoring torque comes from gravity acting at the center of mass: $\tau=-mgd\sin\theta$.
• The small-angle approximation $\sin\theta\approx\theta$ turns the torque equation into the SHM differential equation $\frac{d^2\theta}{dt^2}=-\omega^2\theta$ with $\omega=\sqrt{\frac{mgd}{I}}$.
• The physical pendulum period is $T_{\text{phys}}=2\pi\sqrt{\frac{I}{mgd}}$, where $I$ is the moment of inertia about the pivot and $d$ is the pivot-to-center-of-mass distance.
• A simple pendulum is the special case with $I=m\ell^2$ and $d=\ell$, giving $T_p=2\pi\sqrt{\frac{\ell}{g}}$ with the mass canceling out.
• A torsion pendulum follows $I\alpha=-\kappa\Delta\theta$, so $T=2\pi\sqrt{\frac{I}{\kappa}}$, with the torsion constant playing the role of a spring constant.
• The small-angle approximation only holds for small displacements (roughly under about 10 degrees); larger angles break the SHM model.

Properties of Physical Pendulums

Rigid Body Oscillation

A physical pendulum is a rigid body that swings about a fixed pivot point. When you displace it from equilibrium and release it, it undergoes simple harmonic motion for small angular displacements. The restoring torque comes from gravity acting at the center of mass. The motion is oscillatory, and under the small-angle approximation that oscillation is SHM.

• The body can have any shape (rod, disk, irregular object)
• The mass is distributed throughout the body rather than concentrated at a point
• The pivot point can be located anywhere relative to the center of mass
• The restoring torque is provided by gravity acting on the center of mass

Unlike a simple pendulum, which models the mass as a point at the end of a massless string, a physical pendulum accounts for the actual distribution of mass throughout the object.

Period Derivation for Small Amplitudes

When a physical pendulum is displaced by a small angle from equilibrium, it experiences a restoring torque that drives the oscillation. You analyze this with rotational dynamics.

The relevant equation for the period is:

$T_{\text{phys}}=2 \pi \sqrt{\frac{I}{m g d}}$

where $I$ is the moment of inertia about the pivot point, $m$ is the mass, $g$ is the acceleration due to gravity, and $d$ is the distance between the pivot and the center of mass.

The derivation follows from analyzing the torque on the pendulum:

1. The restoring torque is:

   $\tau=-m g d \sin \theta$

   where $\theta$ is the angular displacement from equilibrium.

2. For small angles, use the approximation $\sin \theta \approx \theta$, which simplifies the torque to:

   $\tau=-m g d \theta$

3. Apply Newton's second law in rotational form:

   $\tau=I \alpha$

   where $\alpha$ is the angular acceleration.

4. Substitute and rearrange:

   $-m g d \theta=I \alpha$

   $\alpha = -\frac{mgd}{I}\theta$

5. Since angular acceleration is the second derivative of angular displacement:

   $\frac{d^{2} \theta}{d t^{2}}=-\frac{m g d}{I} \theta=-\omega^{2} \theta$

   where $\omega$ is the angular frequency.

6. This differential equation describes simple harmonic motion with angular frequency:

   $\omega = \sqrt{\frac{mgd}{I}}$

7. The period is related to the angular frequency by:

   $T=\frac{2\pi}{\omega}$

8. Therefore:

   $T_{\text{phys}}=2 \pi \sqrt{\frac{I}{m g d}}$

This shows that the period depends on the moment of inertia, which captures the mass distribution of the object, rather than treating it as a point mass.

Simple Pendulum vs Physical Pendulum

A simple pendulum is a special case of a physical pendulum in which the bob is modeled as a point mass $m$ located a distance $\ell$ from the pivot on a massless string or rod. Because all the mass is concentrated at a single point, the analysis simplifies.

For small angular displacements, the restoring torque is:

$\tau = -mg\ell \sin\theta \approx -mg\ell\theta$

Using $I = m\ell^2$ for a point mass at distance $\ell$ from the pivot, Newton's second law for rotation gives:

$m\ell^2 \frac{d^2\theta}{dt^2} = -mg\ell\theta$

Dividing both sides by $m\ell^2$:

$\frac{d^2\theta}{dt^2} = -\frac{g}{\ell}\theta$

This is the SHM differential equation with $\omega = \sqrt{\frac{g}{\ell}}$. The period is therefore:

$T_{p}=2 \pi \sqrt{\frac{\ell}{g}}$

Notice that the mass cancels out entirely. The period of a simple pendulum depends only on its length and the acceleration due to gravity. You can also recover this result directly from the physical pendulum formula by substituting $I = m\ell^2$ and $d = \ell$:

$T = 2\pi\sqrt{\frac{m\ell^2}{mg\ell}} = 2\pi\sqrt{\frac{\ell}{g}}$

This confirms that the simple pendulum is just a specific instance of the more general physical pendulum model.

Torsion Pendulum

A torsion pendulum is another rotational SHM system. Instead of gravity providing the restoring torque, a twisted wire or rod provides a restoring torque proportional to the angular displacement. A common example is a horizontal disk suspended from a wire attached to its center of mass. When you rotate the disk and release it, it oscillates back and forth about the wire's axis in the horizontal plane.

The equation of motion for a torsion pendulum is:

$I\alpha = -\kappa \Delta\theta$

where $I$ is the rotational inertia about the axis of rotation, $\kappa$ is the torsion constant of the wire (analogous to a spring constant), and $\Delta\theta$ is the angular displacement from equilibrium.

This can be rewritten as:

$\frac{d^2\theta}{dt^2} = -\frac{\kappa}{I}\theta$

This has the standard SHM form, so the angular frequency and period are:

$\omega = \sqrt{\frac{\kappa}{I}} \qquad T = 2\pi\sqrt{\frac{I}{\kappa}}$

Notice the structural similarity to a mass-spring system: the torsion constant $\kappa$ plays the role of the spring constant $k$, and the rotational inertia $I$ plays the role of mass $m$.

How to Use This on the AP Physics C: Mechanics Exam

Problem Solving

When you set up a physical pendulum problem, get these three pieces right before plugging in:

• $I$ is the moment of inertia about the pivot, not about the center of mass. If you only know the center-of-mass value, use the parallel-axis theorem to shift to the pivot.
• $d$ is the straight-line distance from the pivot to the center of mass.
• The small-angle approximation must apply, or the period equation does not hold.

Free Response

Common tasks include deriving $T_{\text{phys}}=2\pi\sqrt{\frac{I}{mgd}}$ from $\tau=-mgd\sin\theta$ and Newton's second law in rotational form, sketching angular displacement versus time, and predicting how the period changes when you move the pivot, change the length, or change the mass distribution. When asked how period responds to a change, reason from the functional dependence in the equation rather than recomputing from scratch.

Experimental Reasoning

Pendulums are good for lab questions. You might plan a procedure to measure $g$ from the period of a simple pendulum, or determine an unknown moment of inertia from the measured period of a physical pendulum. Plotting period-related data with appropriate axes can let you extract a slope tied to $g$, $I$, or the torsion constant.

Practice Problem 1: Physical Pendulum Period

Solution

Use the period formula for a physical pendulum:

$T_{\text{phys}}=2 \pi \sqrt{\frac{I}{m g d}}$

For a uniform rod pivoted at one end:

• The moment of inertia about the pivot is $I = \frac{1}{3}ML^2 = \frac{1}{3} \times 2.0 \text{ kg} \times (1.0 \text{ m})^2 = \frac{2.0}{3} \text{ kg·m}^2$
• The center of mass is at the middle of the rod, so $d = \frac{L}{2} = 0.5 \text{ m}$
• The mass is $m = 2.0 \text{ kg}$
• Acceleration due to gravity is $g = 9.8 \text{ m/s}^2$

Substituting:

$T_{\text{phys}}=2 \pi \sqrt{\frac{\frac{2.0}{3} \text{ kg·m}^2}{2.0 \text{ kg} \times 9.8 \text{ m/s}^2 \times 0.5 \text{ m}}}$

$T_{\text{phys}}=2 \pi \sqrt{\frac{\frac{2.0}{3}}{2.0 \times 9.8 \times 0.5}} = 2 \pi \sqrt{\frac{1}{3 \times 9.8 \times 0.5}}$

$T_{\text{phys}}=2 \pi \sqrt{\frac{1}{14.7}} = 2 \pi \times 0.261$

$T_{\text{phys}} = 1.64 \text{ seconds}$

The period of small oscillations for this physical pendulum is about 1.64 seconds.

Common Misconceptions

• A heavier pendulum does not swing with a different period just because it is heavier. For a simple pendulum the mass cancels, and for a physical pendulum the period depends on how the mass is distributed (through $I$ and $d$), not on the total mass alone.
• The $d$ in $T_{\text{phys}}=2\pi\sqrt{\frac{I}{mgd}}$ is the distance from the pivot to the center of mass, not the full length of the object.
• The moment of inertia in the period formula must be taken about the pivot. Using the center-of-mass value without applying the parallel-axis theorem gives the wrong period.
• The simple pendulum period $T_p=2\pi\sqrt{\frac{\ell}{g}}$ is not exact for all angles. It only works under the small-angle approximation, so large swings change the period.
• Increasing the amplitude (within the small-angle range) does not change the period. Amplitude and period are independent in SHM.
• A torsion pendulum's restoring torque comes from the twisted wire, not from gravity, so its period uses the torsion constant rather than $g$ and $d$.

Related AP Physics C: Mechanics Guides

• 7.1 Defining Simple Harmonic Motion (SHM)
• 7.3 Representing and Analyzing SHM
• 7.2 Frequency and Period of SHM
• 7.4 Energy of Simple Harmonic Oscillators