Overview#

• A torsion pendulum is the angular analog of a spring–block oscillator: a disk or object suspended by a wire that twists.

• Stiffness comes from the wire’s resistance to twisting; inertia comes from the object’s rotational inertia \(I\).

• The same SHM structure applies: restoring torque \(\tau \propto -\theta\), with period \(T = 2\pi\sqrt{I/\kappa}\).

The torsion pendulum#

A torsion pendulum is an object (e.g., a disk) suspended from a wire. When the wire is twisted and released, the object oscillates in angular simple harmonic motion.

Restoring torque. Twisting the wire produces a torque that opposes displacement:

\(\kappa\) (kappa) is the torsion constant—a property of the wire (length, diameter, material). It plays the role of spring constant \(k\).

Energy (primary derivation):

• Rotational kinetic energy: \(K = \frac{1}{2}I(d\theta/dt)^2\)

• Potential energy in the twisted wire: \(U = \frac{1}{2}\kappa\theta^2\)

• Total energy: \(E = K + U\) → inertia = \(I\), stiffness = \(\kappa\)

By section 15-2, Eq. (59):

Torque method (alternative):

• Newton’s second law for rotation: \(\tau = I\alpha\), with \(\alpha = d^2\theta/dt^2\)

• Restoring torque: \(\tau = -\kappa\theta\)

• So \(I\alpha = -\kappa\theta\), giving

• Comparing with the SHM hallmark \(\alpha = -\omega^2\theta\) yields the same \(\omega\) and \(T\).

Physical interpretation:

• Larger \(I\) → longer period (slower oscillations)

• Stiffer wire (larger \(\kappa\)) → shorter period (faster oscillations)

Comparing two torsion pendulums#

The torsion constant \(\kappa\) depends only on the wire. If you hang different objects from the same wire, one at a time, \(\kappa\) stays the same; only \(I\) changes. So \(T \propto \sqrt{I}\)—measuring periods lets you compare rotational inertias.

Same wire (same \(\kappa\)), different objects \(a\) and \(b\):

If \(T_b > T_a\), then \(I_b > I_a\)—the object with the longer period has the larger rotational inertia.

Poll: Period and inertia

Three objects—a thin rod, a flat disk, and a ball—are each hung in turn from the same torsion wire (one at a time). All have the same mass and the same characteristic size (e.g., rod length = 2\(R\), disk and ball radius = \(R\)). When oscillating, which has the longest period?

• (A) Rod

• (B) Disk

• (C) Ball

• (D) All the same

Summary#

• Torsion pendulum: object on a twisted wire; restoring torque \(\tau = -\kappa\theta\).

• Energy method (section 15-2): \(E = \frac{1}{2}I(d\theta/dt)^2 + \frac{1}{2}\kappa\theta^2\) → inertia \(I\), stiffness \(\kappa\) → \(\omega = \sqrt{\kappa/I}\), \(T = 2\pi\sqrt{I/\kappa}\). Torque method (\(\tau = I\alpha\)) gives the same result.

• Moment of inertia \(I\): defined from \(K = \frac{1}{2}I(d\theta/dt)^2\); depends on mass distribution around the axis.

• Same wire, different objects: \(\kappa\) unchanged; \(I_b/I_a = (T_b/T_a)^2\)—measure periods to compare rotational inertias.

Moment of inertia: definition and computation#

Definition from energy: The rotational kinetic energy \(K = \frac{1}{2}I(d\theta/dt)^2\) defines the moment of inertia \(I\): it is the coefficient that makes \(K\) quadratic in angular speed. \(I\) measures resistance to angular acceleration; mass farther from the axis contributes more (the \(r^2\) factor).

Two forms:

• Discrete: \(I = \sum_i m_i r_i^2\)

• Continuous: \(I = \int \rho\,r^2\,dV\)

Here \(r\) is the perpendicular distance from the mass element to the rotation axis.

How to compute:

1. Choose the rotation axis.

2. For each mass element, find \(r\) (shortest distance to the axis).

3. Sum \(m_i r_i^2\) (discrete) or integrate \(\rho\,r^2\,dV\) (continuous).

Parallel-axis theorem: If the axis is parallel to one through the center of mass, distance \(d\) apart, then \(I = I_{\text{CM}} + Md^2\). Use this to find \(I\) about an end (e.g., rod about end) from \(I\) about the center.

Examples table#

Worked examples#