1.6.7 · D5Oscillations & Waves
Question bank — Physical pendulum — compound pendulum
Reminders of the symbols used below (all built in the parent note):
- = total mass of the rigid body.
- = acceleration due to gravity (the strength of the downward pull, on Earth).
- = moment of inertia about the pivot (see Moment of inertia).
- = distance from pivot to the center of mass (CM).
- = Radius of gyration about the CM, so .
- = the equivalent Simple pendulum length.
- , valid only for small and only when the CM sits below the pivot.
True or false — justify
The period of a compound pendulum depends on its mass
False — appears in both and in , so it cancels, exactly as for a Simple pendulum; only the mass distribution (through ) and geometry survive.
A real rod swings faster than a point mass placed at its center of mass
False — since , larger means longer period; and exceeds (because ), so the real rod has the longer effective length and swings slower.
Doubling every dimension of the pendulum (keeping shape) leaves the period unchanged
False — and both scale with length , so ; since , a scaled-up copy swings slower.
If you know only the moment of inertia about the pivot, you can find the period
False — you also need and ; alone can't tell you where gravity's lever arm is.
Two differently shaped bodies with the same have the same period
True — the period depends on the body only through , since ; shape is fully collapsed into that one number.
Pivoting at the center of oscillation gives a different period than pivoting at the original point
False — that is the reversibility property: pivot and center of oscillation are interchangeable and give the same , which is what Kater's pendulum exploits to measure .
The small-angle formula slightly under-estimates the true period of a large swing
True — the exact period grows with amplitude, so (the amplitude-independent limit) is a floor that real large swings exceed.
A compound pendulum on the Moon has the same period as on Earth
False — from we see , so weaker Moon gravity gives a longer period; only the shape-dependence is -free, not itself.
Spot the error
"I'll plug straight into for a disk hung from its rim."
Wrong — the body rotates about the pivot, so you need (Parallel axis theorem); the correct value is , not .
"The effective length is just , the pivot-to-CM distance."
The effective length is , strictly bigger than ; using pretends all mass sits at the CM and ignores the spread that measures.
"As I slide the pivot toward the CM, the period keeps shrinking because there's less inertia to swing."
It shrinks only until ; past the minimum the restoring torque dies faster than , so turns around and blows up to infinity at the CM.
"The lever arm of gravity is , so the torque is ."
The perpendicular lever arm is , giving the restoring torque ; is only the coefficient, and dropping the (and the sign) deletes the restoring behaviour entirely.
"Since is just an approximation, the formula is really exact if I'm careful."
No — the SHM form only appears because was linearised; at large amplitude the equation is genuinely nonlinear and the period truly depends on how far you swing.
"A door is not a pendulum because it swings about a vertical axis."
Correct catch — gravity provides no restoring torque about a vertical axis (the CM's height doesn't change), so a door does not oscillate as a gravity pendulum at all.
Why questions
Why must be taken about the pivot and not the CM?
Because the body physically rotates about the fixed pivot axis, and rotational Newton's law uses the moment of inertia about the actual axis of rotation.
Why does the period go to infinity as the pivot approaches the CM?
The restoring torque is , which as , while stays finite; a vanishing restoring "spring" means an infinitely slow return.
Why is torque, not force, the right starting law here?
The motion is a rotation about a fixed axis, so the relevant balance is between the twisting effect (torque) and the rotational sluggishness (), captured by , not .
Why does a minimum period exist at all?
Two competing effects — near the CM the restoring torque dies (period grows), and far from it grows like (period grows) — so a sweet spot in between minimises , occurring at .
Why does mass cancel out of the period?
Every restoring term carries one factor of (gravity ) and every inertial term carries one factor of (mass distribution ); the ratio leaves nowhere, a hallmark of all gravity pendulums.
Why is the equivalent length useful rather than just quoting ?
collapses all the messy mass distribution into a single number, letting you treat any rigid swinger as a Simple pendulum and compare bodies at a glance.
Why does the parallel-axis theorem show up everywhere in this topic?
Tables give , but the physics needs ; the Parallel axis theorem is the bridge between the two, and it directly produces the form of .
Edge cases
What happens to the period if the pivot is placed exactly at the CM ()?
Gravity has zero lever arm, so there is no restoring torque; the body has no preferred orientation and simply doesn't oscillate — .
What happens if the pivot lies below the center of mass (an inverted, top-heavy setup)?
The equilibrium becomes unstable: for a small tilt gravity now produces a torque with the same sign as (i.e. ), pushing the body further from vertical rather than back. There is no oscillation, the formula does not describe real motion (formally ), and the body topples — exactly like an inverted pendulum.
What is the smallest possible period for a body of given radius of gyration ?
, achieved when the pivot is a distance from the CM, giving .
If a body's entire mass really were concentrated at one point (so ), what does the compound formula reduce to?
, giving — exactly the Simple pendulum, confirming the compound result contains the point-mass case.
What limits the accuracy of as amplitude grows?
The approximation breaks down; the true period increases with amplitude, so the formula is only a small-oscillation limit.
If a body has two possible pivot distances giving the same period, how are they related?
They are conjugate points and lying on opposite sides of the CM with ; this pairing (pivot ↔ center of oscillation) underlies Kater's pendulum.
Does the compound-pendulum analysis apply to a Torsional pendulum?
No — a torsional pendulum's restoring torque comes from a twisted wire (), not gravity's lever arm, so its period has no and no .
Recall One-line summary of every trap
Almost every mistake is one of three: (1) using instead of , (2) confusing with , or (3) forgetting that was assumed (and that the CM must hang below the pivot). Keep those straight and the topic is disarmed.