Exercises — Moment of inertia I = Σmᵢrᵢ² — concept

WHAT the axis is: the $y$-axis is the vertical line where $x=0$. Look at figure

The mass lies on the axis, so its perpendicular distance is $r=0$. $I=m{r}^2=4\times 0^2=0{\text{kg m}}^2.$ A mass sitting on the axis contributes nothing — you can't make it swing in a circle about a line it already sits on.

Sum each mass times its squared distance: $I=1(0.2{)}^2+2(0.4{)}^2+3(0.6{)}^2=0.04+0.32+1.08=1.44{\text{kg m}}^2.$ Notice $m_3$ (farthest) supplies $1.08$ — three-quarters of the total — because its distance is squared.

WHAT the geometry is: see figure

WHY we can't just write $\frac{1}{2}I{v}^2$: each corner mass moves in its own circle, so let's add up the ordinary kinetic energy $\tfrac12 m v^2$ of all four particles and watch $I$ appear. Step 1 — speed of each corner. In a rigid body every particle shares the same $\omega$, and angular velocity gives each a speed $v_i=\omega r_i$. Here all four have $r_i=r=0.7071\text{m}$, so each moves at $v=\omega r=3\times 0.7071=2.121\text{m/s}$. Step 2 — sum the linear KE, then factor. $K{E}_{\text{rot}}={\sum }_i\frac{1}{2}{m}_i{v}_i^2={\sum }_i\frac{1}{2}{m}_i(\omega r_i{)}^2=\frac{1}{2}(\underset{=I}{\underbrace{\sum _i{m}_i{r}_i^2}}){\omega }^2=\frac{1}{2}I{\omega }^2.$ The bracket is exactly the moment of inertia — that is why $\frac{1}{2}\sum m(\omega r{)}^2$ collapses to $\frac{1}{2}I{\omega }^2$: $\omega$ is common to all, so it factors out and leaves the pure geometry sum $I$. Step 3 — plug in. $K{E}_{\text{rot}}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times 4\times 3^2=18\text{J}.$

(a) Axis through $m_1$: then $r_1=0$, $r_2=L=2\text{m}$. $I_a=3(0{)}^2+5(2{)}^2=20{\text{kg m}}^2.$ (b) Axis through midpoint: each mass is $L/2=1\text{m}$ from the axis. $I_b=3(1{)}^2+5(1{)}^2=3+5=8{\text{kg m}}^2.$ (c) $I_a=20>I_b=8$. The end axis wins because it forces the whole heavy $5\text{kg}$ mass out to the full distance $2\text{m}$ (squared to $4$), whereas the midpoint keeps everyone at only $1\text{m}$. Distance dominates here: the same masses, just repositioned relative to the axis, change $I$ by a factor of $2.5$.

$I$ is minimised when the axis passes through the centre of mass, because that's the balance point that keeps the mass-weighted spread smallest. Locate it (distance $x$ from $m_1$ along a $2\text{m}$ rod): $x=\frac{m_{2}L}{m_1+m_2}=\frac{5\times 2}{3+5}=1.25\text{m}.$ Then $r_1=1.25$, $r_2=2-1.25=0.75$: $I_{\min }=3(1.25{)}^2+5(0.75{)}^2=4.6875+2.8125=7.5{\text{kg m}}^2.$ Indeed $7.5<8<20$ — the centre-of-mass axis beats both the midpoint and the end. The heavier mass sits closer to this axis (only $0.75\text{m}$), which the $r^2$ weighting rewards.

WHAT the theorem says, geometrically: look at figure

(a) Conservation of $L$: $I_1{\omega }_1=I_2{\omega }_2$, so ${\omega }_2=\frac{I_1{\omega }_1}{I_2}=\frac{6\times 2}{2}=6\text{rad/s}.$ Shrinking $I$ by a factor of 3 speeds her up by a factor of 3. (b) New energy: $K{E}_2=\frac{1}{2}{I}_2{\omega }_2^2=\frac{1}{2}\times 2\times 6^2=36\text{J}.$ Old energy: $K{E}_1=\frac{1}{2}\times 6\times 2^2=12\text{J}$. Energy tripled. Where from? Her muscles did work pulling the arms inward against the outward (centrifugal) tendency. Momentum is conserved (no external torque), but energy is not — the skater actively adds $24\text{J}$.

(a) All six masses share $r=R=0.4\text{m}$: $I=6m{R}^2=6\times 0.5\times 0.4^2=6\times 0.5\times 0.16=0.48{\text{kg m}}^2.$ (b) Newton's second law for rotation, $\tau =I\alpha$: $\alpha =\frac{\tau }{I}=\frac{1.2}{0.48}=2.5{\text{rad/s}}^2.$ (c) From rest with constant $\alpha$: $\omega =\alpha t=2.5\times 5=12.5\text{rad/s}.$ (d) Energy via kinematics: $KE=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times 0.48\times 12.5^2=0.24\times 156.25=37.5\text{J}.$ Energy via work done by torque, with $\theta =\frac{1}{2}\alpha t^2=\frac{1}{2}\times 2.5\times 25=31.25\text{rad}$: $W=\tau \theta =1.2\times 31.25=37.5\text{J}.✓$ Both routes give $37.5\text{J}$ — the work–energy theorem for rotation is internally consistent.

Work backwards through the chain. First the required $I$ from $KE=\frac{1}{2}I{\omega }^2$: $I=\frac{2KE}{{\omega }^2}=\frac{2\times 25}{10^2}=0.5{\text{kg m}}^2.$ The two masses each contribute $m{r}^2$, so $I=2m{r}^2$: $0.5=2\times 1\times r^2\Rightarrow r^2=0.25\Rightarrow r=0.5\text{m}.$ Place each mass $0.5\text{m}$ from the axis.

$\sum m_i{r}_i^2$ to the continuous integral $\int r^{2}dm$ Every exercise here used discrete point masses, so $I={\sum }_i{m}_i{r}_i^2$ was a finite sum. What about a solid object — a rod, a disk, a real skater — where mass is smeared out continuously? The bridge: chop the object into a huge number of tiny chunks. Let the $i$-th chunk have a tiny mass $\Delta m_i$ at perpendicular distance $r_i$. Then $I={\sum }_i{r}_i^2\Delta m_i.$ Now make the chunks smaller and more numerous without limit. The sum becomes an integral — the tool that answers "what is the limit of a sum of infinitely many infinitesimal pieces?": $\boxed{I=\int r^{2}dm.}$ Why an integral and not a bigger sum? Because a continuous body has uncountably many points; no finite list of $m_i$ describes it. The integral $\int r^{2}dm$ is the exact continuous version of the same recipe — still "each bit of mass times its distance-squared, all added up." Sanity check on our L2.2 square: four discrete corners kept the sum finite ($I=4{\text{kg m}}^2$). A real solid square plate of the same mass would use $\int r^{2}dm$ instead, and its mass, being spread closer to the centre on average, would give a smaller $I$ — the same "far mass dominates" logic from the parent note, now in continuous form.