Simple and Physical Pendulums
| English | Chinese | Pinyin |
|---|---|---|
| simple pendulum | 单摆 | dān bǎi |
| physical pendulum | 物理摆 | wù lǐ bǎi |
Every swing keeps time
- A grandfather clock's rod ticks out steady seconds.
- A child on a swing takes the same time whether pushed high or low.
- A swinging body -- thin or bulky -- has its own fixed beat.
- Two formulas time any small swing about a pivot.
The simple pendulum
- A simple pendulum 单摆 is a point mass on a light string of length $L$:
- Gravity provides a restoring torque that grows with the angle.
- For small angles the swing is SHM.
A simple pendulum has $L = 0.25\ \text{m}$ and $g = 9.8\ \text{m/s}^2$. Its period (in s, to 1 decimal)?
$T = 2\pi\sqrt{L/g} = 2\pi\sqrt{0.25/9.8} \approx 1.0\ \text{s}$.
Doubling the bob mass of a simple pendulum changes its period.
$T = 2\pi\sqrt{L/g}$ has no mass -- the bob's mass cancels.
The physical pendulum
- A physical pendulum 物理摆 is any rigid body swinging about a pivot:
- $I$ is the rotational inertia about the pivot, $d$ the pivot-to-centre distance.
- A ruler, a bat, a leg -- each swings with this period.
In the physical pendulum period $T = 2\pi\sqrt{I/(mgd)}$, what is $d$?
$d$ is the pivot-to-centre-of-mass distance -- the lever arm of gravity.
One is a special case of the other
- For a point mass at distance $L$: $I = mL^2$ and $d = L$.
- Plug in: $T = 2\pi\sqrt{mL^2/(mgL)} = 2\pi\sqrt{L/g}$.
- So the simple pendulum is just the physical pendulum's simplest case.
Simple and physical pendulums
Change the length and gravity and see how the pendulum's period responds - amplitude barely matters.
The simple-pendulum formula is a special case of the physical-pendulum formula.
Put $I = mL^2$ and $d = L$ into $T = 2\pi\sqrt{I/(mgd)}$ and you recover $2\pi\sqrt{L/g}$.
A uniform rod of length $L$ pivoted at one end has $I = \tfrac{1}{3}mL^2$ and $d = L/2$.
- $T = 2\pi\sqrt{\dfrac{I}{mgd}} = 2\pi\sqrt{\dfrac{\tfrac{1}{3}mL^2}{mg\cdot L/2}} = 2\pi\sqrt{\dfrac{2L}{3g}}$.
- It swings a little faster than a simple pendulum of the same length.
For a physical pendulum, the rotational inertia $I$ must be measured about the...
The body rotates about the pivot, so $I$ is taken about that axis.
Both pendulum period formulas are valid only when the swing angle is...
Only small angles make the restoring torque proportional to displacement (SHM).
Both formulas assume small angles -- beyond about $15^\circ$ the motion is no longer true SHM. For a physical pendulum, take $I$ about the pivot (use the parallel-axis theorem if you know $I$ about the centre), and let $d$ be the pivot-to-centre-of-mass distance, not the full length. The bob's mass cancels for a simple pendulum but need not for a physical one.
A simple pendulum (point mass on a string) swings with $T = 2\pi\sqrt{L/g}$; a physical pendulum (any rigid body on a pivot) swings with $T = 2\pi\sqrt{I/(mgd)}$, where $I$ is taken about the pivot and $d$ reaches the centre of mass. The simple case falls out of the physical one, and both are SHM only for small angles.