Exercises — Physical pendulum — compound pendulum

$T=2\pi \sqrt{\frac{I}{mgd}},I=I_{cm}+m{d}^2=m{k}^2+m{d}^2,L_{\text{eq}}=\frac{I}{md}=\frac{k^2}{d}+d$ Here $I$ is the moment of inertia about the pivot, $d$ is the pivot-to-CM distance, $m$ the mass, and $k$ the radius of gyration about the CM. We use $g=9.8{\text{m/s}}^2$ unless a problem says otherwise.

Everything is already "about the pivot," so just substitute into $T=2\pi \sqrt{I/(mgd)}$: $T=2\pi \sqrt{\frac{0.30}{2.0\cdot 9.8\cdot 0.25}}=2\pi \sqrt{\frac{0.30}{4.9}}=2\pi \sqrt{0.06122}$ $T=2\pi (0.2474)\approx \boxed{1.55\text{s}}$ What we did: dropped the given numbers straight in — no theorem needed because $I$ was already stated about the pivot.

From the parent note, a rod pivoted at one end has $L_{\text{eq}}=\frac{2}{3}\ell$ (because $I_{end}=\frac{1}{3}m{\ell }^2$ and $d=\ell /2$, so $L_{\text{eq}}=I/(md)=\frac{1}{3}{\ell }^2/\frac{1}{2}\ell =\frac{2}{3}\ell$). $L_{\text{eq}}=\frac{2}{3}(1.2)=\boxed{0.80\text{m}}$ What it means: this whole rod swings exactly like a ball on an 0.80 m string.

Step 1 — build $I$ about the pivot. For a disk, $I_{cm}=\frac{1}{2}m{R}^2$. The pivot is at the rim, a distance $d=R$ from the CM, so by the Parallel axis theorem: $I=\frac{1}{2}m{R}^2+m{R}^2=\frac{3}{2}m{R}^2=1.5(0.80)(0.15{)}^2=0.0270{\text{kg⋅m}}^2$ Step 2 — identify $d$. $d=R=0.15\text{m}$. Step 3 — period. $T=2\pi \sqrt{\frac{\frac{3}{2}m{R}^2}{mgR}}=2\pi \sqrt{\frac{3R}{2g}}=2\pi \sqrt{\frac{3(0.15)}{2(9.8)}}=2\pi \sqrt{0.02296}$ $T=2\pi (0.1515)\approx \boxed{0.952\text{s}}$ Notice the mass cancelled — as always for gravity pendulums. See the disk-at-rim geometry below.

Step 1 — $I$ about the pivot. A ring has all its mass at radius $R$, so $I_{cm}=m{R}^2$. Pivot at the edge, $d=R$: $I=m{R}^2+m{R}^2=2m{R}^2$ Step 2 — $d=R$. Step 3 — $T=2\pi \sqrt{\frac{2m{R}^2}{mgR}}=2\pi \sqrt{\frac{2R}{g}}=2\pi \sqrt{\frac{2(0.20)}{9.8}}=2\pi \sqrt{0.04082}$ $T=2\pi (0.2020)\approx \boxed{1.27\text{s}}$ Sanity check: $L_{\text{eq}}=I/(md)=2m{R}^2/(mR)=2R=0.40\text{m}$ — a hooped ring swings like a simple pendulum of length = its diameter. Neat.

(a) The radius of gyration of a rod about its own centre is found from $I_{cm}=\frac{1}{12}m{\ell }^2=m{k}^2$, so $k=\frac{\ell }{\sqrt{12}}=\frac{1.0}{3.464}=0.2887\text{m}.$ The period is minimum when $d=k$ (parent note §4). So $\boxed{d=k=0.289\text{m}}$ (b) At the minimum, $L_{\text{eq}}=2k$, so $T_{\min }=2\pi \sqrt{\frac{2k}{g}}=2\pi \sqrt{\frac{2(0.2887)}{9.8}}=2\pi \sqrt{0.05892}=2\pi (0.2427)\approx \boxed{1.525\text{s}}$ The figure below shows how $T$ dives to this minimum and then climbs back up.

End-pivot period (from the rod formula $T=2\pi \sqrt{2\ell /(3g)}$): $T_{end}=2\pi \sqrt{\frac{2(1.0)}{3(9.8)}}=2\pi \sqrt{0.06803}=2\pi (0.2608)\approx 1.639\text{s}$ Ratio: $\frac{T_{end}}{T_{\min }}=\frac{1.639}{1.525}\approx \boxed{1.075}$ So swinging from the end is only about 7.5% slower than the fastest possible pivot — the minimum is broad and shallow near $d=k$, which is why the exact pivot point is forgiving.

(a) $L_{\text{eq}}=\frac{2}{3}\ell =0.667\text{m}$ (rod at end). (b) The center of oscillation lies a distance $L_{\text{eq}}=0.667\text{m}$ from the pivot, along the line through the CM. Since the CM is at $0.50$ m, the center of oscillation is $0.667-0.500=0.167\text{m}$ beyond the CM. (c) Pivot there. Now the new pivot-to-CM distance is $d^{\prime }=0.167\text{m}$. The radius of gyration $k$ is unchanged ($k=0.2887$ m). Then $L_{\text{eq}}^{\prime }=\frac{k^2}{d^{\prime }}+d^{\prime }=\frac{0.08333}{0.1667}+0.1667=0.5000+0.1667=0.667\text{m}$ Same $L_{\text{eq}}$! Hence the same period. This reversibility is exactly what Kater's pendulum exploits to measure $g$ without knowing $I$ or $m$.

Step 1 — total moment of inertia about the pivot. Add the two contributions (moments of inertia add about the same axis): $I=\underset{\text{rod, end}}{\underbrace{\frac{1}{3}m{\ell }^2}}+\underset{\text{bob at end}}{\underbrace{M{\ell }^2}}=\frac{1}{3}(0.50)(1{)}^2+(0.50)(1{)}^2=0.1667+0.5000=0.6667{\text{kg⋅m}}^2$ Step 2 — locate the combined CM. Rod's CM at $0.50$ m, bob at $1.0$ m: $d=\frac{m(0.50)+M(1.0)}{m+M}=\frac{0.25+0.50}{1.0}=0.75\text{m}$ Step 3 — total mass and period. $m_{tot}=1.0\text{kg}$: $T=2\pi \sqrt{\frac{I}{m_{tot}gd}}=2\pi \sqrt{\frac{0.6667}{1.0\cdot 9.8\cdot 0.75}}=2\pi \sqrt{0.09070}=2\pi (0.3012)\approx \boxed{1.892\text{s}}$ What we synthesised: $I$ and $d$ each came from combining two objects — you cannot use a single table value here.

Use $L_{\text{eq}}=k^2/d+d$ and $T=2\pi \sqrt{L_{\text{eq}}/g}$: $L_{\text{eq}}=\frac{(0.2887{)}^2}{0.010}+0.010=\frac{0.08333}{0.010}+0.010=8.333+0.010=8.343\text{m}$ $T=2\pi \sqrt{\frac{8.343}{9.8}}=2\pi \sqrt{0.8513}=2\pi (0.9227)\approx \boxed{5.80\text{s}}$ The trend: as $d\to 0$, the $k^2/d$ term dominates and $L_{\text{eq}}\to \infty$, so $T\to \infty$. Physically the restoring torque $mgd\sin \theta$ vanishes while $I$ stays finite — the rod, pivoted almost at its balance point, barely wants to return, so it swings ever more slowly. This is the divergent left branch of the U-curve in the L3 figure.

(a) From $T=2\pi \sqrt{L_{\text{eq}}/g}$, solve for $L_{\text{eq}}$: $L_{\text{eq}}=\frac{g{T}^2}{4{\pi }^2}=\frac{9.8(0.85{)}^2}{4{\pi }^2}=\frac{9.8(0.7225)}{39.48}=\frac{7.0805}{39.48}\approx \boxed{0.1794\text{m}}$ (b) Since $L_{\text{eq}}=k^2/d+d$, rearrange for $k$: $k^2=d(L_{\text{eq}}-d)=0.12(0.1794-0.12)=0.12(0.05935)=0.007122$ $k=\sqrt{0.007122}\approx \boxed{0.0844\text{m}}$ What we mastered: running the whole chain backwards — from a stopwatch reading to a geometric property $k$ of the body, exactly the spirit of Kater's pendulum and of measuring Moment of inertia experimentally (since $I_{cm}=m{k}^2$).

For a rod at its end, $T=2\pi \sqrt{2\ell /(3g)}$ — period depends only on $\ell$ (mass cancels). So $\frac{T_{2\ell }}{T_{\ell }}=\sqrt{\frac{2\ell }{\ell }}=\sqrt{2}\approx \boxed{1.414}$ The longer rod swings slower, its period larger by $\sqrt{2}$. Doubling every length multiplies the period by $\sqrt{2}$, not by $2$ — the square-root law again, the same scaling that governs the Simple pendulum ($T\propto \sqrt{L}$).

Cover the page: for a disk hung at its rim, what is $L_{\text{eq}}$? Answer ::: $L_{\text{eq}}=\frac{3}{2}R$ (from $I=\frac{3}{2}m{R}^2$, $d=R$).