Questions: Physical Pendulum and Rotational Oscillations

Question 1

A uniform rod of mass m and length L is pivoted at one end and swings as a physical pendulum. How does its period compare to a simple pendulum of the same length L?

• A. Same period: T = 2π√(L/g), because the total length is the same
• B. Longer period, because the distributed mass increases the effective pendulum length
• C. Shorter period, because the effective length is L_eq = I/(md) = (2/3)L, which is less than L
• D. Shorter period, because the moment of inertia mL²/3 is smaller than mL²

Answer: (C)

For a uniform rod pivoted at one end: I = mL²/3 and d = L/2 (center of mass is at the midpoint). The equivalent length is L_eq = I/(md) = (mL²/3)/(m·L/2) = 2L/3 < L. Since T = 2π√(L_eq/g), the rod swings faster (shorter period) than a simple pendulum of the same total length. The mass distribution places the effective center of oscillation closer to the pivot than if all mass were at the tip.

Question 2

In the physical pendulum period formula T = 2π√(I/(mgd)), what does d represent?

• A. The total length of the pendulum arm from pivot to the lowest point of the body
• B. The distance from the pivot point to the center of mass of the rigid body
• C. The radius of gyration of the body about its center of mass
• D. The distance from the center of mass to the lowest point of the body

Answer: (B)

d is the distance from the pivot point P to the center of mass of the rigid body. It appears because the restoring torque is τ = −mgd sin(θ) — gravity acts at the center of mass, and d is the moment arm for that force. This is distinct from the total length of the object, and it's also distinct from the radius of gyration. Confusing d with the total length L is the most common error when applying this formula.

Question 3

A physical pendulum's period depends on the moment of inertia about the pivot point, not about the center of mass.

• T. True
• F. False

Answer: True

The equation of motion is Iα = −mgd sin(θ), where I must be computed about the pivot point P (the actual axis of rotation), not about the center of mass. When applying the formula, you often need the parallel-axis theorem: I_pivot = I_cm + md². Using I_cm instead of I_pivot is a common error that gives the wrong period.

Question 4

The period of a physical pendulum can be calculated using T = 2π√(L/g) as long as L is taken to be the total physical length of the object.

• T. True
• F. False

Answer: False

T = 2π√(L/g) applies only to the simple pendulum, where all mass is concentrated at a single point at distance L from the pivot. For a physical pendulum, the correct formula is T = 2π√(I/(mgd)). This reduces to the simple pendulum result only when I = mL² and d = L (all mass at the tip), which is not the case for any real extended object.

Question 5

Explain why a uniform rod pivoted at one end swings faster than a simple pendulum of the same total length, using the concept of equivalent length.

Model answer: The equivalent length of a physical pendulum is L_eq = I/(md). For a uniform rod pivoted at one end, I = mL²/3 and d = L/2, so L_eq = (mL²/3)/(m·L/2) = 2L/3. Since T = 2π√(L_eq/g), and 2L/3 < L, the rod has a shorter period than a simple pendulum of the same length. Physically, the mass is distributed along the rod with much of it closer to the pivot than the tip, so the rod's effective oscillation length is shorter than its total length.

The equivalent length concept lets you map any physical pendulum to a simple pendulum for the purpose of calculating period. The 'effective length' is always less than the total length for uniform objects (center of mass is inside the object), which is why distributed-mass pendulums swing faster than point-mass pendulums of the same size.