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Pendulums are systems that swing or twist because a restoring torque pulls them back toward equilibrium. This cheat sheet compares the simple pendulum, physical pendulum, and torsional pendulum so students can choose the correct model quickly. It is useful for solving oscillation problems, identifying small-angle conditions, and connecting torque, inertia, and period.

These ideas appear often in mechanics, rotational motion, and simple harmonic motion units.

The core idea is that a pendulum behaves like simple harmonic motion when its restoring torque is proportional to displacement. A simple pendulum uses T=2πLgT = 2\pi\sqrt{\frac{L}{g}}, a physical pendulum uses T=2πImgdT = 2\pi\sqrt{\frac{I}{mgd}}, and a torsional pendulum uses T=2πIκT = 2\pi\sqrt{\frac{I}{\kappa}}. The period depends on inertia and restoring strength, not on mass alone.

Large amplitudes, damping, and driving forces require corrections beyond the ideal formulas.

Key Facts

  • For a simple pendulum at small angles, the period is T=2πLgT = 2\pi\sqrt{\frac{L}{g}}, where LL is the length from pivot to bob center.
  • The small-angle approximation is sinθθ\sin\theta \approx \theta when θ\theta is measured in radians and is usually accurate for small amplitudes.
  • The angular frequency of a simple pendulum is ω=gL\omega = \sqrt{\frac{g}{L}}, so T=2πωT = \frac{2\pi}{\omega}.
  • For a physical pendulum, the period is T=2πImgdT = 2\pi\sqrt{\frac{I}{mgd}}, where II is rotational inertia about the pivot and dd is the distance from pivot to center of mass.
  • The parallel-axis theorem gives I=Icm+md2I = I_{\mathrm{cm}} + md^2 when converting rotational inertia from the center of mass to the pivot.
  • For a torsional pendulum, the restoring torque is τ=κθ\tau = -\kappa\theta and the period is T=2πIκT = 2\pi\sqrt{\frac{I}{\kappa}}.
  • Frequency and period are related by f=1Tf = \frac{1}{T}, and angular frequency is related by ω=2πf=2πT\omega = 2\pi f = \frac{2\pi}{T}.
  • Ideal pendulum period is independent of amplitude only when the oscillations are small enough for simple harmonic motion.

Vocabulary

Simple pendulum
A point mass on a light string or rod that swings under gravity with period T=2πLgT = 2\pi\sqrt{\frac{L}{g}} for small angles.
Physical pendulum
An extended rigid body that swings about a pivot, with its mass distribution included through rotational inertia II.
Torsional pendulum
A rotating system that twists back and forth because a wire or support produces restoring torque τ=κθ\tau = -\kappa\theta.
Restoring torque
A torque that acts opposite the displacement and tends to return the system to equilibrium.
Rotational inertia
A measure of resistance to angular acceleration, written II, that depends on both mass and how far the mass is from the axis.
Small-angle approximation
The rule sinθθ\sin\theta \approx \theta for θ\theta in radians, which lets many pendulums be modeled as simple harmonic oscillators.

Common Mistakes to Avoid

  • Using degrees inside sinθθ\sin\theta \approx \theta is wrong because the approximation only works when θ\theta is measured in radians.
  • Using the bob mass in T=2πLgT = 2\pi\sqrt{\frac{L}{g}} is wrong because the ideal simple pendulum period does not depend on mass.
  • Using the total object length as dd for a physical pendulum is wrong because dd must be the distance from the pivot to the center of mass.
  • Forgetting the parallel-axis theorem is wrong when the given II is about the center of mass, because the physical pendulum formula needs II about the pivot.
  • Using τ=mgLθ\tau = -mgL\theta for every pendulum is wrong because torsional pendulums use τ=κθ\tau = -\kappa\theta and physical pendulums use τmgdθ\tau \approx -mgd\theta.

Practice Questions

  1. 1 A simple pendulum has length L=0.80mL = 0.80\,\mathrm{m}. Using g=9.8m/s2g = 9.8\,\mathrm{m/s^2}, find its period TT.
  2. 2 A physical pendulum has I=0.36kgm2I = 0.36\,\mathrm{kg\,m^2} about its pivot, mass m=2.0kgm = 2.0\,\mathrm{kg}, and center of mass distance d=0.30md = 0.30\,\mathrm{m}. Find its small-angle period.
  3. 3 A torsional pendulum has rotational inertia I=0.050kgm2I = 0.050\,\mathrm{kg\,m^2} and torsion constant κ=0.20Nm/rad\kappa = 0.20\,\mathrm{N\,m/rad}. Calculate TT and ff.
  4. 4 Explain why increasing the mass of the bob does not change the ideal simple pendulum period, but changing the mass distribution of a physical pendulum can change its period.