Simple and Physical Pendulums - AP Physics C Mechanics

  1. For a simple pendulum, what is LL?
    Length of the string or rod. Distance from pivot to center of mass.
  2. What is the definition of a physical pendulum?
    A pendulum with an extended mass. Unlike simple pendulum's point mass assumption.
  3. What is the simple pendulum approximation for small angles?
    sin⁡(θ)≈θ\sin(\theta) \approx \theta. Valid for angles less than about 15 degrees.
  4. Determine the moment of inertia for a rod pendulum pivoted at the end.
    I=13mL2I = \frac{1}{3}mL^2. Standard formula for rod rotating about end.
  5. Calculate the angular displacement in radians for 30∘30^{\circ}.
    θ=π6 rad\theta = \frac{\pi}{6} \text{ rad}. Converting degrees to radians using π/180\pi/180.
  6. What variables influence the period of a simple pendulum?
    Length LL and gravitational acceleration gg. Period is independent of mass and initial angle.
  7. State the simple harmonic motion condition for pendulums.
    Small angle approximation. Enables sin⁡(θ)≈θ\sin(\theta) \approx \theta linearization.
  8. Identify the type of energy at the highest point of a pendulum swing.
    Maximum potential energy. All kinetic energy converts to potential at extremes.
  9. State the formula for the torque in a physical pendulum.
    τ=−mgdsin⁡(θ)\tau = -mgd \sin(\theta). Restoring torque proportional to angular displacement.
  10. What is the angular frequency formula for a physical pendulum?
    ω=mgdI\omega = \sqrt{\frac{mgd}{I}}. Physical pendulum frequency depends on inertia distribution.
  11. State the expression for the angular frequency of a simple pendulum.
    ω=gL\omega = \sqrt{\frac{g}{L}}. Angular frequency is ω=2πT\omega = \frac{2\pi}{T}.
  12. What is the kinetic energy at the lowest point of a simple pendulum?
    Maximum kinetic energy. All potential energy converts to kinetic at bottom.
  13. Calculate the period of a simple pendulum with L=1 mL = 1 \text{ m} on Earth.
    T≈2π19.8≈2.01 sT \approx 2\pi \sqrt{\frac{1}{9.8}} \approx 2.01 \text{ s}. Substituting L=1 mL = 1 \text{ m} and g=9.8 m/s2g = 9.8 \text{ m/s}^2 into period formula.
  14. Identify the restoring force in a simple pendulum.
    The component of gravitational force. Gravity's tangential component provides restoring force.
  15. What is the condition for small angle approximation in pendulums?
    θmax≈0.1 rad\theta_{max} \approx 0.1 \text{ rad} or 5∘5^{\circ}. Beyond this, sin⁡(θ)≠θ\sin(\theta) \neq \theta approximation fails.
  16. What is the phase of a pendulum at its maximum displacement?
    ϕ=π2\phi = \frac{\pi}{2} or −π2-\frac{\pi}{2}. Phase at amplitude depends on initial conditions.
  17. Identify the pivot point in a simple pendulum.
    The fixed point from which it swings. Attachment point allowing rotational motion.
  18. What is the effect of increasing the mass on the period of a simple pendulum?
    No effect. Mass cancels out in period equation derivation.
  19. Calculate the torque for a 5 kg5 \text{ kg} pendulum with d=0.5 md = 0.5 \text{ m} and θ=30∘\theta = 30^{\circ}.
    τ=−5×9.8×0.5×sin⁡(30∘)≈−12.25 Nm\tau = -5 \times 9.8 \times 0.5 \times \sin(30^{\circ}) \approx -12.25 \text{ Nm}. Using torque formula with sin⁡(30°)=0.5\sin(30°) = 0.5.
  20. What is the amplitude in a pendulum?
    Maximum angular displacement. Maximum angular distance from equilibrium position.
  21. Identify the linear displacement in a pendulum swing.
    Arc length from equilibrium. Arc length equals LθL\theta for small angles.
  22. Determine the frequency of a simple pendulum with T=2 sT = 2 \text{ s}.
    f=1T=0.5 Hzf = \frac{1}{T} = 0.5 \text{ Hz}. Frequency is reciprocal of period.
  23. Calculate the period of a physical pendulum with I=0.02 kg m2I = 0.02 \text{ kg m}^2, m=0.5 kgm = 0.5 \text{ kg}, d=0.1 md = 0.1 \text{ m}.
    T=2π0.020.5×9.8×0.1≈0.9 sT = 2\pi \sqrt{\frac{0.02}{0.5 \times 9.8 \times 0.1}} \approx 0.9 \text{ s}. Substituting given values into physical pendulum period formula.
  24. Which type of pendulum can have a non-uniform mass distribution?
    Physical pendulum. Simple pendulum assumes point mass only.
  25. What is the equilibrium position of a pendulum?
    Vertical position where θ=0\theta = 0. Lowest potential energy configuration.
  26. State the formula for the maximum speed of a pendulum.
    v_{max} = \sqrt{2gL(1 - \cos(\theta_{max})}}. From energy conservation between highest and lowest points.
  27. What does the period of a simple pendulum NOT depend on?
    Mass of the bob. Period formula shows no mass dependence.
  28. State the condition for the pendulum to oscillate as a simple harmonic oscillator.
    Small angle approximation. Required for linear restoring force relationship.
  29. Calculate the angular frequency for a 2 m2 \text{ m} pendulum on Earth.
    ω=9.82≈2.21 rad/s\omega = \sqrt{\frac{9.8}{2}} \approx 2.21 \text{ rad/s}. Using ω=g/L\omega = \sqrt{g/L} with given values.
  30. Calculate the period change if gg is doubled for a simple pendulum.
    T′=T2T' = \frac{T}{\sqrt{2}}; period decreases. Period inversely proportional to square root of gg.
  31. Identify the moment of inertia in a physical pendulum.
    II, the rotational inertia about the pivot. Determines rotational resistance to angular acceleration.
  32. State the restoring torque in a physical pendulum.
    τ=−mgdsin⁡(θ)\tau = -mgd \sin(\theta). Negative sign indicates restoring nature of torque.
  33. Calculate the period of a pendulum with L=2 mL = 2 \text{ m} on the Moon (g=1.6 m/s2g=1.6 \text{ m/s}^2).
    T≈2π21.6≈7.02 sT \approx 2\pi \sqrt{\frac{2}{1.6}} \approx 7.02 \text{ s}. Lower gravity increases period significantly.
  34. What is the relationship between period and frequency?
    f=1Tf = \frac{1}{T}. Inverse relationship between period and frequency.
  35. State the formula for the period of a physical pendulum.
    T=2πImgdT = 2\pi \sqrt{\frac{I}{mgd}}. Accounts for distributed mass via moment of inertia.
  36. Determine the change in gravitational potential energy at the bottom of the swing.
    Zero; all energy is kinetic. Reference level chosen at lowest swing point.
  37. Determine the maximum kinetic energy of a pendulum given m=2 kgm = 2 \text{ kg}, g=9.8 m/s2g = 9.8 \text{ m/s}^2, and h=0.1 mh = 0.1 \text{ m}.
    KEmax=mgh=1.96 JKE_{max} = mgh = 1.96 \text{ J}. Energy conservation: KEmax=PEmax=mghKE_{max} = PE_{max} = mgh.

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