Topic 7.5 Notes – Simple and Physical Pendulums

1. What a Physical Pendulum Is

A physical pendulum is any rigid body that swings about a fixed pivot under gravity.

Key features:

• The object is a rigid body (rod, disk, irregular shape).
• It rotates about a fixed axis.
• Its center of mass (CM) is a distance $d$ from the pivot.
• Its rotational inertia is $I$, taken about the pivot.

Here’s the geometry you should picture as you read the torque expression below:

Physical pendulum displaced by angle $\theta$

When displaced by an angle $\theta$, gravity acts at the CM and produces a torque about the pivot.

Restoring Torque from Gravity

The torque due to gravity is

$$\tau(\theta) = -mgd\sin\theta$$

Why this form?

• Lever arm is $d$.
• Component perpendicular to the rod gives $\sin\theta$.
• The negative sign means the torque restores the pendulum toward equilibrium.

This plays the same role as $F = -kx$ in linear SHM.

2. Small-Angle Approximation and Why It Leads to SHM

For small angles (in radians, usually less than about 0.2 rad):

$$\sin\theta \approx \theta$$

Then the torque becomes

$$\tau \approx -mgd\,\theta$$

Now apply rotational Newton’s second law:

$$\sum \tau = I\alpha$$

So,

$$I\alpha = -mgd\,\theta$$

Since $\alpha = \dfrac{d^{2}\theta}{dt^{2}}$,

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

That is the exact form of SHM:

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

So,

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

and the period is

$$T_{\text{phys}} = 2\pi \sqrt{\frac{I}{mgd}}$$

What controls the period?

• Larger $I$ → more rotational inertia → longer period
• Larger $d$ → stronger restoring torque → shorter period
• Mass only matters through how it affects $I$

On FRQs, they often want you to derive this from torque and show the SHM differential equation. Make sure you explicitly write $I\alpha = \tau$ and substitute the small-angle approximation.

3. Types of Pendulums

a. Physical Pendulum (General Case)

This is the full formula:

$$T = 2\pi \sqrt{\frac{I}{mgd}}$$

Steps you usually need:

1. Find $d$ (distance from pivot to CM).
2. Find $I$ about the pivot.
   • If given $I_{CM}$, use the parallel-axis theorem: $$I_{\text{pivot}} = I_{CM} + md^{2}$$

For example, a uniform rod pivoted at one end:

• $I_{CM} = \frac{1}{12}mL^{2}$
• $d = L/2$
• Use parallel-axis to get $I_{\text{end}} = \frac{1}{3}mL^{2}$

Students often forget that $I$ must be about the pivot, not the CM. That mistake costs easy points.

b. Simple Pendulum

A simple pendulum is a special case of a physical pendulum:

• Point mass $m$
• Massless string
• Length $l$

Here:

• $I = ml^{2}$
• $d = l$

Plug into the physical pendulum formula:

$$T = 2\pi \sqrt{\frac{l}{g}}$$

Important properties:

• Independent of mass.
• Depends only on $l$ and $g$.
• Valid only for small angles.

If the problem says “small oscillations” and gives a point mass on a string, this is your go-to result.

c. Torsion Pendulum

A torsion pendulum oscillates because a twisted wire provides restoring torque.

Instead of gravity,

$$\tau = -\kappa \theta$$

Apply rotational Newton’s second law:

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

So,

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

Here:

• $I$ plays the role of mass.
• $\kappa$ is the rotational analog of spring constant $k$.

On conceptual questions, connect:

• Linear SHM: $m\ddot{x} = -kx$
• Rotational SHM: $I\ddot{\theta} = -(\text{constant})\theta$

Same structure. Different physical quantities.