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Rigid bodies swinging under gravity, $\omega = \sqrt{mgd/I}$ with $I$ about the pivot, the equivalent simple pendulum, the best pivot, and torsion pendulums.
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By the end of this lesson you will be able to find the period of any rigid body swinging about a pivot, and of a body twisting on a wire.
From the last lesson you can solve $\ddot{x} = -\omega^2x$ and fit its solution to a start. From the rotation lessons you have torque, rotational inertia, the parallel-axis theorem, and $\tau = I\alpha$. This lesson puts them together for anything that swings: a clock pendulum, a gate, a leg, a disk on a nail.
| Term | What it means |
|---|---|
| Physical pendulum | A rigid body swinging under gravity about a fixed pivot. |
| Pivot distance | $d$, from the pivot to the center of mass. |
| Small-angle approximation | $\sin\theta \approx \theta$ in radians, good to 1 percent below about 14°. |
| Equivalent length | $L_{\text{eq}} = I/(md)$, the simple pendulum with the same period. |
| Torsion pendulum | A body hung on a wire that twists, with restoring torque $-\kappa\theta$. |
| Torsion constant | $\kappa$, the torque per radian of twist, in N·m/rad. |
| Center of percussion | The point, at $L_{\text{eq}}$ from the pivot, where a blow gives no jolt at the pivot. |
Hang a rigid body from a pivot and turn it through an angle $\theta$. Gravity acts at the center of mass, a distance $d$ from the pivot, and gives a torque $-mgd\sin\theta$ back toward hanging straight down. The rotational second law about the pivot is
$$I\ddot{\theta} = -mgd\sin\theta \approx -mgd\,\theta,$$
using $\sin\theta \approx \theta$ for small swings. That is the oscillator equation with
$$\omega = \sqrt{\frac{mgd}{I}}, \qquad T = 2\pi\sqrt{\frac{I}{mgd}}.$$
The period depends on how the mass is spread, through $I$ about the pivot, and on where the center of mass sits, through $d$. A point mass on a string has $I = mL^2$ and $d = L$, giving the familiar $T = 2\pi\sqrt{L/g}$. Any body swings like a simple pendulum of equivalent length $I/(md)$.
Another way: picture
Picture a yardstick hanging from a nail through one end and a small weight on a string of the same length beside it. Start both swinging. The yardstick wins: its mass near the nail is easy to swing, so it behaves like a shorter pendulum, two-thirds as long. Where the mass sits relative to the pivot, not the object's length alone, sets the rhythm.
Another way: steps
The gravitational torque about the pivot is the weight times its lever arm, and the lever arm of a weight at distance $d$ tilted by $\theta$ is $d\sin\theta$. The minus sign says the torque turns the body back toward $\theta = 0$. Without the approximation, the equation $I\ddot{\theta} = -mgd\sin\theta$ has no solution in elementary functions, and the period grows with amplitude.
For small angles, $\sin\theta \approx \theta$ in radians. At $10°$, $\sin\theta$ is $0.1736$ and $\theta$ is $0.1745$, half a percent apart, and the period is only $0.2$ percent longer than the small-angle value. At $30°$ the error in the period is under 2 percent. Clock pendulums swing through a few degrees so that the approximation, and so the period, holds very well.
The $I$ in the period is about the pivot, not about the center of mass. The parallel-axis theorem moves it: $I = I_{\text{cm}} + md^2$. For a uniform rod of length $L$, $I_{\text{cm}} = \tfrac{1}{12}mL^2$; about one end, $d = L/2$ and $I = \tfrac{1}{12}mL^2 + \tfrac{1}{4}mL^2 = \tfrac{1}{3}mL^2$. For a disk on its rim, $I = \tfrac{1}{2}mR^2 + mR^2 = \tfrac{3}{2}mR^2$.
Putting these in $T = 2\pi\sqrt{I/mgd}$, the mass always cancels: the rod about its end has $T = 2\pi\sqrt{2L/3g}$, the disk on its rim $T = 2\pi\sqrt{3R/2g}$. As with a simple pendulum, a heavy body and a light one of the same shape swing together.
With the parallel-axis theorem written in, $\omega^2 = gd/(k^2 + d^2)$, where $k^2 = I_{\text{cm}}/m$ is the square of the radius of gyration. Pivot at the center, $d = 0$, and the body does not swing at all: gravity exerts no torque. Pivot far away, $d \gg k$, and it swings like a point mass, $\omega^2 \approx g/d$.
In between, the frequency is greatest, and the period least, when $d = k$. A uniform rod has $k = L/\sqrt{12} \approx 0.29L$, so a meter stick hung from a hole $29$ cm from its center swings fastest. Pivots on opposite sides at distances $d$ and $k^2/d$ give the same period, a fact Henry Kater used in 1817 to build a pendulum that measured $g$ precisely.
Hang a disk on a thin wire and twist it. The wire resists with a torque proportional to the twist, $\tau = -\kappa\theta$, and this holds for large angles as well as small. The equation is $I\ddot{\theta} = -\kappa\theta$, with $\omega = \sqrt{\kappa/I}$ and no gravity in it at all.
Because $\kappa$ can be tiny for a fine wire, a torsion pendulum can respond to minute torques. Henry Cavendish used one in 1798 to measure the gravitational pull between lead spheres and so to weigh the Earth, and the same instrument measures rotational inertias: time the swing with and without an object of unknown $I$ and compare. Mechanical wristwatches keep time with a balance wheel on a coiled hairspring, a torsion oscillator.
Checking an answer. The mass must cancel for a gravity pendulum. Setting $I = md^2$ must recover $2\pi\sqrt{d/g}$. The equivalent length must lie between $d$ and the far end of the body. And the period must not depend on the amplitude, as long as the swings are small.
Swing a baseball bat and hit a ball at the wrong spot, and the handle stings your hands. There is one point, the center of percussion, where a blow makes the bat rotate about your hands without pushing them. For a body pivoted at a point, it lies at the equivalent length $I/(md)$ from the pivot.
For a uniform rod held at one end, that is $2L/3$ from the hands. Bat makers call the region near it the sweet spot; tennis racquets and hammers are designed with it in mind. The same length that sets the swing period sets where a hit feels clean, because both come from how the mass is distributed about the pivot.
Beyond small angles the period grows with amplitude. For a swing of amplitude $\theta_0$, a good approximation is $T \approx T_0(1 + \theta_0^2/16)$, with $\theta_0$ in radians. At $20°$ that is a 0.8 percent increase; at $90°$, about 15 percent.
This dependence is why pendulum clocks drift if their swing grows or shrinks, and why Christiaan Huygens looked for a curved path that would make the period independent of amplitude. For the lessons that follow, the small-angle result is what matters: a physical pendulum is a harmonic oscillator whose spring is gravity's torque.
A physical pendulum turns a timing into a measurement. Time twenty swings of an irregular object hung from a pin, and with $d$ known you have its rotational inertia, $I = mgdT^2/4\pi^2$. Engineers measure the inertias of car parts and aircraft components this way before they are assembled.
Turned around, a pendulum of known shape measures $g$. Geologists once mapped buried ore bodies and salt domes along the Gulf Coast by timing pendulums: a denser rock underground raises $g$ by a few parts in a million and shortens the period accordingly. Modern gravimeters are more sensitive, but the idea is the same.
In $T = 2\pi\sqrt{I/mgd}$ the mass appears on top, inside $I$, and on the bottom, in the torque. Every piece of a rigid body contributes to both in proportion to its own mass, so doubling the mass everywhere changes nothing. This is the rotational form of Galileo's observation that heavy and light bodies fall together.
The cancellation fails when the mass is not scaled uniformly. Add a small clamp near the bottom of a meter stick and both $I$ and $d$ change, by different fractions, so the period changes. Work such problems by adding the pieces: $I$ is the sum of each part's inertia about the pivot, and $md$ is the sum of each part's mass times its distance. A torsion pendulum never has the cancellation at all, since $\kappa$ belongs to the wire, not the body; a heavier disk on the same wire always turns more slowly.
The tall case clocks made in colonial Philadelphia and New England swung a pendulum about a meter long, with a period of two seconds, so each tick was one second. The pendulum is a physical pendulum: a rod with a heavy bob, whose equivalent length is set by where the bob sits. Turning the nut under the bob moves it, changes $I$ and $d$, and adjusts the rate.
Because a metal rod expands in summer, which lengthens the period and makes the clock lose time, fine clocks used rods of wood or of two metals arranged so that the expansions cancel. The best pendulum clocks, used by the U.S. Naval Observatory until the 1930s, kept time to a few hundredths of a second a day.
Biomechanics researchers model a swinging leg as a physical pendulum pivoted at the hip. A uniform rod $0.9$ m long has a period of about $1.6$ s, and a relaxed walking step, half a swing, takes about $0.8$ s: close to how people naturally walk. Walking at the leg's natural frequency lets gravity do much of the work.
That is why tall people walk with a slower cadence than short ones, and why small animals scurry while large ones stride. Prosthetic legs are designed with this in mind: their mass distribution sets their swing period, and a prosthesis whose period matches the wearer's other leg gives a smoother, less tiring gait.
For a point mass on a string, $T = 2\pi\sqrt{L/g}$, and it is tempting to use the object's length for any swinging body. A rigid body's mass is spread out: some of it sits near the pivot and is easy to swing. The right length is the equivalent length $I/(md)$, which for a rod about its end is two-thirds of the rod.
A related error uses the inertia about the center of mass. The body rotates about the pivot, so the parallel-axis theorem must move $I$ there; otherwise the period comes out too short.
A meter stick swings from a nail through one end. Find its center-of-mass distance.
$d = 0.50\ \text{m}$
The middle of a uniform stick.
Find its inertia about the nail.
$I = \tfrac{1}{3}m(1.0)^2 = 0.333m$
A rod about one end.
Write the angular frequency.
$\omega = \sqrt{\dfrac{m \times 9.8 \times 0.50}{0.333m}} = \sqrt{14.7} = 3.83\ \text{rad/s}$
The mass cancels.
Find the period.
$T = \dfrac{2\pi}{3.83} = 1.64\ \text{s}$
One full swing.
Find the equivalent length.
$L_{\text{eq}} = \dfrac{0.333m}{0.50m} = 0.667\ \text{m}$
Two-thirds of the stick.
A $0.20$ m radius disk hangs from a nail at its rim. Find its inertia about its center.
$I_{\text{cm}} = \tfrac{1}{2}m(0.20)^2 = 0.020m$
A uniform disk.
Move the axis to the nail.
$I = 0.020m + m(0.20)^2 = 0.060m$
Parallel-axis theorem with $d = R$.
Write the period.
$T = 2\pi\sqrt{\dfrac{0.060m}{m \times 9.8 \times 0.20}}$
$T = 2\pi\sqrt{I/mgd}$.
Evaluate the period.
$T = 2\pi\sqrt{0.0306} = 1.10\ \text{s}$
About a second.
Find the equivalent length.
$L_{\text{eq}} = \dfrac{0.060}{0.20} = 0.30\ \text{m}$
One and a half radii.
Check against a simple pendulum at the center.
$L_{\text{eq}} = 0.30 > d = 0.20\ \text{m}$
Spread-out mass makes it slower than a point at the center.
A $1.2$ m uniform beam can be pivoted anywhere along its length. Find its radius of gyration.
$k^2 = \dfrac{L^2}{12} = \dfrac{1.44}{12} = 0.12\ \text{m}^2, \quad k = 0.346\ \text{m}$
$I_{\text{cm}} = mk^2$.
Write the angular frequency squared.
$\omega^2 = \dfrac{gd}{k^2 + d^2}$
Parallel-axis theorem in $mgd/I$.
Set the pivot at the radius of gyration.
$d = k = 0.346\ \text{m}$
Where $\omega^2$ is greatest.
Find the largest angular frequency.
$\omega^2 = \dfrac{9.8 \times 0.346}{0.24} = 14.1, \quad \omega = 3.76\ \text{rad/s}$
Since $k^2 + d^2 = 2k^2$.
Find the shortest period.
$T = \dfrac{2\pi}{3.76} = 1.67\ \text{s}$
No pivot does better.
Compare with a pivot at the end.
$d = 0.60: \ \omega^2 = \dfrac{9.8 \times 0.60}{0.48} = 12.25, \ T = 1.80\ \text{s}$
Slower than the best pivot.
Find the other pivot with the same period as the end.
$d' = \dfrac{k^2}{d} = \dfrac{0.12}{0.60} = 0.20\ \text{m}$
Kater's reversible-pendulum pair.
Find the inertia about the nail.
$I = mR^2 + mR^2 = 2mR^2$
A hoop about its center has $mR^2$.
Write the period with $d = R$.
$T = 2\pi\sqrt{\dfrac{2mR^2}{mgR}} = 2\pi\sqrt{\dfrac{2R}{g}}$
The mass cancels.
Evaluate the period.
A uniform rod $180$ cm long swings from one end. How long a simple pendulum would have the same period?
Complete the worked solution: a uniform rod $0.9$ m long swings from one end. Find $I/m$ about the pivot in m², the center-of-mass distance $d$ in m, and the length of the simple pendulum with the same period, in m.
Divide the rod's inertia by its mass.
$\dfrac{I}{m} = \tfrac{1}{3}L^2 =$ i
A rod about one end, from the parallel-axis theorem.
Locate the center of mass from the pivot.
$d = \tfrac{1}{2}L =$ d
The rod is uniform.
Divide by the center-of-mass distance.
$L_{\text{eq}} = \dfrac{I}{md} =$ e
Matching $\sqrt{mgd/I}$ to $\sqrt{g/L_{\text{eq}}}$.
Check it against the rod's length.
$d < L_{\text{eq}} < L$
A rod swings faster than a point at its end, slower than a point at its middle.
Match each oscillator or quantity to its expression.
| $\sqrt{g/L}$ | $\sqrt{mgd/I}$ | $\sqrt{\kappa/I}$ | $\tfrac{1}{3}mL^2$ | |
|---|---|---|---|---|
| simple pendulum | ||||
| physical pendulum | ||||
| torsion pendulum | ||||
| rod about one end |
A uniform $3$ kg rod $15$ cm long swings from one end. With $g = 10$ m/s², fill in its rotational inertia about the pivot in kg·m², the distance from the pivot to its center of mass in m, and its angular frequency for small swings in rad/s.
| value | |
|---|---|
| rotational inertia (kg·m²) | |
| center-of-mass distance (m) | |
| angular frequency (rad/s) |
A uniform beam $3$ m long hangs from a pivot a distance $d$ m from its center. With $g = 10$ m/s², write $\omega^2$ for small swings, in s⁻², as a formula in $d$.
Answer:
A uniform disk of radius $20$ cm hangs from a nail through a small hole at its rim and swings in its own plane. With $g = 9.8$ m/s², what is its period for small swings, in seconds?
Answer: s for one swing cycle
A gait researcher in Boston models a relaxed leg $0.95$ m long as a uniform rod swinging from the hip. With $g = 9.8$ m/s², what is its natural period for small swings, in seconds?
Answer: s for a full swing
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A uniform beam $9$ m long hangs from a pivot a distance $d$ m from its center. With $g = 10$ m/s², write $\omega^2$ for small swings, in s⁻², as a formula in $d$.
Answer:
You can find physical pendulum periods. Explain to someone why a yardstick swinging from one end keeps faster time than a weight on a string of the same length.
20. Your turn: a thin hoop of radius $0.50$ m hangs on a nail and swings in its plane. Find its period., step 3
$T = 2\pi\sqrt{\dfrac{1.0}{9.8}} = 2.0\ \text{s}$
The same as a simple pendulum one meter long.