Topic 7.5 Simple and Physical Pendulums: derive the small-angle period of the simple pendulum and the physical pendulum using the rotational form of Newton's second law and the small-angle approximation.

What this topic is asking

The College Board (Topic 7.5) wants you to treat both the simple pendulum and the physical pendulum (an extended body swinging about a pivot), deriving their small-angle periods from the rotational form of Newton's second law and the small-angle approximation. The physical pendulum, an extended body rather than a point mass, is a distinctive AP Physics C addition that ties together rotational inertia, torque and SHM.

The simple pendulum from rotational dynamics

Treat the pendulum as a rotation about the pivot. The gravity force $mg$ acts at the bob, a distance $L$ from the pivot, producing a restoring torque $\tau = -mgL\sin\theta$ (the minus sign because it acts to reduce $\theta$). The rotational form of Newton's second law, $\tau = I\ddot{\theta}$ with $I = mL^2$ for the point mass, gives

This is nonlinear because of the $\sin\theta$, so it is not yet SHM.

The small-angle approximation

The motion becomes simple harmonic only when the angle is small, so that $\sin\theta \approx \theta$ (with $\theta$ in radians). This linearises the equation:

which is the SHM form $\ddot{\theta} = -\omega^2\theta$ with $\omega^2 = g/L$. Hence $\omega = \sqrt{g/L}$ and $T = 2\pi\sqrt{L/g}$, independent of the bob's mass and (within the approximation) of the amplitude. The small-angle step is essential and frequently scored: without it the restoring torque is not proportional to $\theta$, the motion is periodic but not simple harmonic, and the period grows with amplitude.

The physical pendulum

A physical pendulum is an extended body, a rod, a hoop, a swinging sign, pivoted at some point. Its weight $Mg$ acts at the center of mass, a distance $d$ from the pivot, giving a restoring torque $\tau = -Mgd\sin\theta$. With the body's rotational inertia $I$ about the pivot (use the parallel-axis theorem if needed), the rotational second law gives

So $\omega = \sqrt{\dfrac{Mgd}{I}}$ and the period is

The simple pendulum is the special case $I = ML^2$, $d = L$, which reduces to $T = 2\pi\sqrt{L/g}$. The physical pendulum is more general: its period depends on how the mass is distributed (through $I$) and on the distance to the center of mass ($d$), not just on its overall size.

Try this

Q1. A simple pendulum has period $2.0$ s on Earth. Calculate its length ($g = 9.8$ m/s squared). [2 points]

• Cue. $T = 2\pi\sqrt{L/g} \Rightarrow L = gT^2/(4\pi^2) = (9.8)(4.0)/39.5 = 0.99$ m.

Q2. State the period of a physical pendulum in terms of its rotational inertia $I$ about the pivot, mass $M$, and the distance $d$ from pivot to center of mass. [2 points]

• Cue. $T = 2\pi\sqrt{I/(Mgd)}$.