Exercises — Parallel axis theorem — I = I_CM + Md² — proof

WHAT the axis is: it passes through the CM (midpoint of a uniform rod). WHY no theorem: if the axis already goes through the CM, then $d=0$, so $M{d}^2=0$. $I=I_{CM}+M(0{)}^2=\frac{1}{12}M{L}^2$ The theorem is not needed — but it still agrees. This is the sanity check: at the CM the correction vanishes.

Identify $d$: the rim is a perpendicular distance $R$ from the centre, so $d=R$. $I=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

$d=R$ again (the ring's edge is $R$ from its centre). $I=M{R}^2+M{R}^2=2M{R}^2$

Find $d$ (look at the figure). The CM is at the midpoint, a distance $L/2$ from either end. The new axis sits at $L/4$ from the end. So the gap between axis and CM is $d=\frac{L}{2}-\frac{L}{4}=\frac{L}{4}$ WHY subtract: $d$ is the CM-to-new-axis distance, and both points are measured from the same end. $I=\frac{1}{12}M{L}^2+M{\left(\frac{L}{4}\right)}^2=\frac{1}{12}M{L}^2+\frac{1}{16}M{L}^2$ Common denominator $48$: $\frac{4}{48}M{L}^2+\frac{3}{48}M{L}^2=\frac{7}{48}M{L}^2$. $\boxed{I=\frac{7}{48}M{L}^2}$

A tangent line touches the surface, so it is exactly one radius from the centre: $d=R$. $I=\frac{2}{5}M{R}^2+M{R}^2=\frac{7}{5}M{R}^2$

Find $d$. The CM sits at the centre of the rectangle. A corner is at distance $d=\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{b}{2}\right)}^2}=\frac{1}{2}\sqrt{a^2+b^2}$ WHY the square root: the corner is offset in both $x$ and $y$, and straight-line distance is the hypotenuse — this is the perpendicular distance to the parallel axis. $I=\frac{1}{12}M(a^2+b^2)+M\cdot \frac{1}{4}(a^2+b^2)=\left(\frac{1}{12}+\frac{3}{12}\right)M(a^2+b^2)=\frac{1}{3}M(a^2+b^2)$

KEY insight (the parent's Example 3 trap): you may not use the separation of A and B directly. Route each through the CM. $I_A=\frac{2}{5}M{R}^2+M(3R{)}^2=\frac{2}{5}M{R}^2+9M{R}^2$ $I_B=\frac{2}{5}M{R}^2+M(2R{)}^2=\frac{2}{5}M{R}^2+4M{R}^2$ The $I_{CM}$ terms cancel on subtraction: $I_A-I_B=9M{R}^2-4M{R}^2=5M{R}^2$ WHY the side doesn't matter: $d$ enters as $d^2$, so being "left" or "right" of the CM is irrelevant — only the magnitude of the distance counts.

Set up both from the CM: $I_1=I_{CM}+M(2{)}^2,I_2=I_{CM}+M(5{)}^2$ Subtract to kill the unknown $I_{CM}$: $I_2-I_1=M(25-4)=21M$ WHY subtract: we don't know $I_{CM}$ and don't need to — differencing eliminates it, exactly the algebra the proof relies on. $21M=84\Rightarrow M=4\text{kg}$

Any parallel axis has $I(d)=I_{CM}+M{d}^2$. Read it as a function of $d$: $M{d}^2\ge 0$ always (mass $M>0$, square $\ge 0$), and it equals $0$ only when $d=0$. Therefore $I(d)\ge I_{CM}$ for every $d$, with equality only at the CM. So the CM-axis is the unique minimum, and any other parallel axis exceeds it by exactly $M{d}^2$. See Radius of gyration: this minimum $I_{CM}$ fixes the smallest possible gyration radius for that direction.

Step 1 — in-plane central axis via Perpendicular axis theorem. For a flat disc, $I_z=I_x+I_y$ where $z$ is perpendicular to the disc and $x,y$ are two in-plane diameters. By symmetry $I_x=I_y$, and $I_z=\frac{1}{2}M{R}^2$. So $2I_x=\frac{1}{2}M{R}^2\Rightarrow I_x=I_{\text{diameter}}=\frac{1}{4}M{R}^2$ Step 2 — shift to the rim (parallel axis). The rim axis is parallel to this diameter, distance $d=R$ from the centre: $I=\frac{1}{4}M{R}^2+M{R}^2=\frac{5}{4}M{R}^2$ WHY both tools: perpendicular axis got us the central diameter inertia (hard to integrate directly); parallel axis then moved it to the rim. One theorem builds $I_{CM}$, the other relocates it.

Rotational kinetic energy is $KE=\frac{1}{2}I{\omega }^2$. Rim axis: $K{E}_{\text{rim}}=\frac{1}{2}\left(\frac{3}{2}M{R}^2\right){\omega }^2=\frac{3}{4}M{R}^2{\omega }^2$. Central axis: $K{E}_{\text{cm}}=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\omega }^2=\frac{1}{4}M{R}^2{\omega }^2$. Ratio: $K{E}_{\text{rim}}/K{E}_{\text{cm}}=3$. Spinning about the rim costs three times the energy at the same $\omega$ — the extra $M{d}^2=M{R}^2$ shows up directly as extra kinetic energy. This is the parent's "pay $M{d}^2$ to move out" made physical.

Step 1 — get $I_{\text{pivot}}$ with the parallel axis theorem. Pivot at the end, CM at the middle, $d=L/2$: $I_{\text{pivot}}=\frac{1}{12}M{L}^2+M{\left(\frac{L}{2}\right)}^2=\frac{1}{3}M{L}^2$ Step 2 — identify $d_{\text{cm}}$. The CM hangs $L/2$ below the pivot, so $d_{\text{cm}}=L/2$. Step 3 — substitute. The mass $M$ cancels (it appears in both numerator and denominator): $T=2\pi \sqrt{\frac{\frac{1}{3}M{L}^2}{Mg\cdot \frac{L}{2}}}=2\pi \sqrt{\frac{2L}{3g}}$ WHY $M$ cancels: inertia grows with mass and so does the restoring torque; they trade off, so the period depends only on geometry. Numbers: $T=2\pi \sqrt{\frac{2(1.2)}{3(9.8)}}=2\pi \sqrt{0.081633}\approx 2\pi (0.28572)\approx 1.795\text{s}$. $\boxed{T\approx 1.80\text{s}}$

Additivity of inertia: total $I$ = sum of each piece's $I$ about the same axis. Centre axis.

• Rod about its centre: $\frac{1}{12}{m}_r{L}^2$.
• Each end mass at distance $L/2$: $m_p(L/2{)}^2$; two of them: $2m_p(L/2{)}^2=\frac{1}{2}{m}_p{L}^2$. $I_{\text{centre}}=\frac{1}{12}{m}_r{L}^2+\frac{1}{2}{m}_p{L}^2$ Numbers: $\frac{1}{12}(2)(4)+\frac{1}{2}(1)(4)=0.6667+2=2.6667{\text{kg⋅m}}^2$. End axis. Now shift by the whole system. First find the total mass and CM.
• Total mass $M=m_r+2m_p=2+2=4\text{kg}$.
• By symmetry the composite CM is still the rod's centre. So the end axis is $d=L/2=1\text{m}$ from the CM of the whole dumbbell.
• The parallel axis theorem uses total mass: $I_{\text{end}}=I_{\text{centre}}+M(L/2{)}^2$. $I_{\text{end}}=2.6667+4(1{)}^2=6.6667{\text{kg⋅m}}^2$ WHY total mass, not a piece: the theorem's $M$ is the mass of everything being moved to the new axis; the whole rigid dumbbell shifts together.