Exercises — Rotational kinetic energy = ½Iω²

WHAT: We want $K_{rot}$, and we are handed $I$ and $\omega$ directly. WHY this formula: $K_{rot}=\frac{1}{2}I{\omega }^2$ is exactly the quantity asked for; nothing needs to be built. $K_{rot}=\frac{1}{2}(4)(10{)}^2=\frac{1}{2}(4)(100)=200\text{J}.$ Answer: $200\text{J}$. Notice we squared $\omega$ first ($10^2=100$), then multiplied.

WHAT: Same $I$, double the $\omega$. WHY: $K$ depends on ${\omega }^2$, so doubling $\omega$ multiplies $K$ by $2^2=4$. $K_{rot}=\frac{1}{2}(4)(20{)}^2=\frac{1}{2}(4)(400)=800\text{J}.$ Answer: $800\text{J}$, which is $4\times$ the value in L1·1. The square is the whole story.

Step 1 — get $I$. A solid disc about its centre has $I=\frac{1}{2}M{R}^2$. $I=\frac{1}{2}(3)(0.2{)}^2=\frac{1}{2}(3)(0.04)=0.06{\text{kg⋅m}}^2.$ Why this shape formula: the disc's mass is spread from $r=0$ to $r=R$; the factor $\frac{1}{2}$ is the average of $r^2$ over that spread (from Moment of inertia). Step 2 — apply the KE formula. $K_{rot}=\frac{1}{2}I{\omega }^2=\frac{1}{2}(0.06)(50{)}^2=\frac{1}{2}(0.06)(2500)=75\text{J}.$ Answer: $I=0.06{\text{kg⋅m}}^2$, $K_{rot}=75\text{J}$.

Hoop: $I_{\text{hoop}}=M{R}^2=(2)(0.3{)}^2=0.18{\text{kg⋅m}}^2$. $K_{\text{hoop}}=\frac{1}{2}(0.18)(8{)}^2=\frac{1}{2}(0.18)(64)=5.76\text{J}.$ Disc: $I_{\text{disc}}=\frac{1}{2}M{R}^2=0.09{\text{kg⋅m}}^2$. $K_{\text{disc}}=\frac{1}{2}(0.09)(64)=2.88\text{J}.$ Answer: hoop $5.76\text{J}$, disc $2.88\text{J}$; the hoop stores exactly twice as much, because all its mass sits at radius $R$ (largest possible $\sum m{r}^2$). Look at Figure 1 — the hoop's mass is on the outer rim (red), the disc's is spread inward (blue).

The new tool — the rolling link $v_{cm}=R\omega$. Why we need it: rolling ties the spin to the forward speed, so a single $v_{cm}$ fixes $\omega =v_{cm}/R$. This lets us kill $R$ entirely. (a) Translational: treat the whole mass moving at $v_{cm}$ (this is the ordinary Kinetic energy of a particle $\frac{1}{2}m{v}^2$). $K_{trans}=\frac{1}{2}M{v}_{cm}^2=\frac{1}{2}(5)(4{)}^2=40\text{J}.$ (b) Rotational: substitute $\omega =v_{cm}/R$ into $\frac{1}{2}I{\omega }^2$:

The new tool — Work-energy theorem (rotational). Why: net work equals the change in KE, $W=\Delta K_{rot}=K_{final}-K_{initial}$. Starting from rest, $K_{initial}=0$, so $W=\frac{1}{2}I{\omega }^2$. Rearrange for $\omega$ — we invert the formula because $\omega$ is the unknown: $\omega =\sqrt{\frac{2W}{I}}=\sqrt{\frac{2(100)}{0.5}}=\sqrt{400}=20\text{rad/s}.$ Answer: $\omega =20\text{rad/s}$. The square root undoes the square in ${\omega }^2$; we take the positive root because speed of spin is a magnitude.

Step 1 — energy bookkeeping. No slipping means friction does no net work, so all the gravitational potential energy becomes kinetic: $Mgh=\underset{\text{translation}}{\underbrace{\frac{1}{2}M{v}^2}}+\underset{\text{rotation}}{\underbrace{\frac{1}{2}{I}_{cm}{\omega }^2}}.$ Step 2 — kill $\omega$ with the rolling link. With $\omega =v/R$: $\frac{1}{2}{I}_{cm}{\omega }^2=\frac{1}{2}\left(\frac{2}{5}M{R}^2\right)(\frac{v}{R}{)}^2=\frac{1}{5}M{v}^2.$ Step 3 — combine and solve. $Mgh=\frac{1}{2}M{v}^2+\frac{1}{5}M{v}^2=\frac{7}{10}M{v}^2\Rightarrow v=\sqrt{\frac{10gh}{7}}.$ Plug in: $v=\sqrt{\frac{10(9.8)(1.4)}{7}}=\sqrt{19.6}=4.427\text{m/s}$ (to 3 d.p.). Sliding block (no spin): all PE goes to translation, $v_{slide}=\sqrt{2gh}=\sqrt{27.44}=5.238\text{m/s}$. Answer: the rolling sphere reaches $\approx 4.43\text{m/s}$, slower than the sliding block's $\approx 5.24\text{m/s}$, because part of its energy is locked in spin (Figure 2).

Step 1 — energy balance. The block's lost PE becomes the block's translational KE plus the flywheel's rotational KE: $mgH=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2.$ Step 2 — link $\omega$ to $v$. The cord unwinds without slipping, so the rim speed equals the block speed: $v=R\omega \Rightarrow \omega =v/R$. $\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right)\frac{v^2}{R^2}=\frac{1}{4}M{v}^2.$ Step 3 — solve. $mgH=\frac{1}{2}m{v}^2+\frac{1}{4}M{v}^2=(\frac{1}{2}m+\frac{1}{4}M)v^2.$ $v=\sqrt{\frac{mgH}{\frac{1}{2}m+\frac{1}{4}M}}=\sqrt{\frac{(2)(9.8)(1.0)}{1+1}}=\sqrt{9.8}=3.130\text{m/s}.$ Answer: $v\approx 3.13\text{m/s}$. If the flywheel were massless the block would reach $\sqrt{2gH}=4.43\text{m/s}$; the flywheel "steals" energy into spin.

Step 1 — get $I$ about the pivot (the end). Two ways: quote the end result $I=\frac{1}{3}M{L}^2$, or build it with Parallel axis theorem from the centre value. Centre: $I_{cm}=\frac{1}{12}M{L}^2$. Shift the axis by $d=L/2$ using $I=I_{cm}+M{d}^2$: $I=\frac{1}{12}M{L}^2+M(\frac{L}{2}{)}^2=\frac{1}{12}M{L}^2+\frac{1}{4}M{L}^2=\frac{1}{3}M{L}^2.$ Why this tool: the energy formula needs $I$ about the actual rotation axis (the pivot), not about the centre; the parallel axis theorem is the bridge. Numerically: $I=\frac{1}{3}(1.2)(0.6{)}^2=\frac{1}{3}(1.2)(0.36)=0.144{\text{kg⋅m}}^2$. Step 2 — apply the KE formula. $K_{rot}=\frac{1}{2}(0.144)(3{)}^2=\frac{1}{2}(0.144)(9)=0.648\text{J}.$ Answer: $I=0.144{\text{kg⋅m}}^2$, $K_{rot}=0.648\text{J}$.

General rolling result. For any rolling body write $I_{cm}=cM{R}^2$ (so $c=\frac{2}{5}$ solid, $c=\frac{2}{3}$ hollow). Energy conservation with $\omega =v/R$:

(a) $c\to 0$: all mass at the axis $\Rightarrow$ negligible moment of inertia $\Rightarrow$ no energy goes into spin. The formula gives $v=\sqrt{2gh}$ — exactly the frictionless sliding block limit. Good: the rolling formula smoothly reduces to the sliding one when there is nothing to spin. (b) Hoop: $c=1$ (all mass at radius $R$). Then $v=\sqrt{\frac{2gh}{2}}=\sqrt{gh}$, and the rotational fraction is $\frac{K_{rot}}{K_{total}}=\frac{c}{1+c}=\frac{1}{2}=50\%.$ A hoop splits its energy exactly half translation, half rotation — the most "spin-heavy" of the common rollers. Answer: (a) sliding-block limit $v=\sqrt{2gh}$; (b) $c=1$, half the energy is rotational.

Doubling $\omega$ multiplies $K$ by what factor? ::: Four (energy carries ${\omega }^2$). Same $M,R,\omega$: hoop or disc stores more KE? ::: Hoop — twice the disc, mass all at radius $R$. Rolling body total KE? ::: $\frac{1}{2}M{v}_{cm}^2+\frac{1}{2}{I}_{cm}{\omega }^2$ with $v_{cm}=R\omega$. Sphere down height $h$ final speed? ::: $v=\sqrt{10gh/7}$, slower than the sliding $\sqrt{2gh}$. Rolling body with $I=cM{R}^2$ final speed from height $h$? ::: $v=\sqrt{2gh/(1+c)}$. Does a fixed-axle flywheel gain/lose PE while spinning? ::: No — its centre does not move, only $K_{rot}$ changes. Rod swung about its end: which $I$? ::: $\frac{1}{3}M{L}^2$ (use parallel axis from $\frac{1}{12}M{L}^2$).