Exercises — Rolling KE = ½mv² + ½Iω²

The total is König's split: $KE=\underset{\text{translation (sliding forward)}}{\underbrace{\frac{1}{2}M{v}_{cm}^2}}+\underset{\text{rotation (spinning)}}{\underbrace{\frac{1}{2}{I}_{cm}{\omega }^2}}$ The first term is the energy the centre of mass carries as it slides; the second is the energy locked in the spin about that centre. Nothing else is present because König's theorem guarantees the cross term vanishes.

By definition $\beta =\frac{I_{cm}}{M{R}^2}=\frac{\frac{1}{2}M{R}^2}{M{R}^2}=\frac{1}{2}$. Rotational fraction $=\frac{\beta }{1+\beta }=\frac{1/2}{3/2}=\frac{1}{3}$. So one third of the disc's kinetic energy is spinning, two thirds is forward motion.

Use $KE=\frac{1}{2}M{v}_{cm}^2(1+\beta )$: $KE=\frac{1}{2}(3)(2^2)\left(1+\frac{2}{5}\right)=\frac{1}{2}(3)(4)(1.4)=8.4\text{J}.$ Why the compact form? Because rolling ties $\omega$ to $v$, so we never need $\omega$ or $R$ separately.

Disc: $\beta =\frac{1}{2}$. (a) $KE=\frac{1}{2}(4)(25)(1.5)=75\text{J}$. (b) Translational $=\frac{1}{2}M{v}^2=\frac{1}{2}(4)(25)=50\text{J}$. (c) Rotational $=$ total $-$ translational $=75-50=25\text{J}$. Check the fraction: $\frac{25}{75}=\frac{1}{3}=\frac{\beta }{1+\beta }$ ✓.

Energy conservation: all PE becomes rolling KE, so $Mgh=\frac{1}{2}M{v}^2(1+\beta )$. $v=\sqrt{\frac{2gh}{1+\beta }}=\sqrt{\frac{2(10)(1.4)}{1+2/5}}=\sqrt{\frac{28}{1.4}}=\sqrt{20}\approx 4.47\text{m/s}.$ Compare: a frictionless slide would give $\sqrt{2gh}=\sqrt{28}\approx 5.29\text{m/s}$ — the sphere is slower because some PE went into spin. ✓

The figure below shows the physical setup: the sphere starts at rest with potential energy $Mgh$ (lavender height arrow) and arrives at the bottom with a forward centre speed $v_{cm}$ (coral arrow). Notice the KE label carries the $(1+\beta )$ factor — that extra $\beta$ is the spin bucket that steals speed.

Rolling constraint: $\omega =\frac{v_{cm}}{R}=\frac{6}{0.2}=30\text{rad/s}$. Ring: $\beta =1$, so rotational/translational $=\beta =1$ — the two energies are equal (50/50 split). Why exactly $\beta$? Write both KEs out: rotational $=\frac{1}{2}{I}_{cm}{\omega }^2=\frac{1}{2}(\beta M{R}^2)(v/R{)}^2=\frac{1}{2}\beta M{v}^2$, translational $=\frac{1}{2}M{v}^2$. Their ratio is $\frac{\frac{1}{2}\beta M{v}^2}{\frac{1}{2}M{v}^2}=\beta$ — everything but $\beta$ cancels because the rolling constraint forced $\omega R=v$.

Start from the bottom-speed formula, which itself came from $Mgh=\frac{1}{2}M{v}^2(1+\beta )$ — the same height $h$ for both bodies, so $Mgh$ is the shared energy budget. $v=\sqrt{\frac{2gh}{1+\beta }}.$ Why does this give a proportionality? For both bodies the numerator $2gh$ is identical (same drop, same $g$); only $(1+\beta )$ differs. So $v$ depends on shape only through $\frac{1}{\sqrt{1+\beta }}$, and dividing the two speeds cancels the common $\sqrt{2gh}$: $\frac{v_{\text{sphere}}}{v_{\text{ring}}}=\frac{\sqrt{2gh}/\sqrt{1+{\beta }_{\text{sphere}}}}{\sqrt{2gh}/\sqrt{1+{\beta }_{\text{ring}}}}=\sqrt{\frac{1+{\beta }_{\text{ring}}}{1+{\beta }_{\text{sphere}}}}=\sqrt{\frac{1+1}{1+2/5}}=\sqrt{\frac{2}{1.4}}=\sqrt{\frac{10}{7}}\approx \mathrm{1.195.}$ The sphere arrives about 20% faster. Mass $M$ cancelled because it multiplies both the PE ($Mgh$) and the KE ($\frac{1}{2}M{v}^2(1+\beta )$); radius $R$ never survived because the rolling constraint already removed it. That is why the result is independent of $M$ and $R$.

The bar chart makes the ordering visible: each shape's final speed (in units of the frictionless $\sqrt{2gh}$, the dashed lavender line) is taller when $\beta$ is smaller. Sphere (mint) beats disc (butter) beats ring (coral).

From $v=\sqrt{\frac{2gh}{1+\beta }}$, square both sides: $v^2=\frac{2gh}{1+\beta }$. Given $v^2=gh$: $gh=\frac{2gh}{1+\beta }\Rightarrow 1+\beta =2\Rightarrow \beta =1$. $\beta =1$ is a ring / hollow cylinder. It is the slowest common shape, exactly matching $v=\sqrt{gh}$.

(a) $I_{cm}=\frac{1}{2}M{R}^2$, $\omega =v/R$: $KE=\frac{1}{2}M{v}^2+\frac{1}{2}\left(\frac{1}{2}M{R}^2\right)\frac{v^2}{R^2}=\frac{1}{2}M{v}^2+\frac{1}{4}M{v}^2=\frac{3}{4}M{v}^2.$ (b) Parallel axis: $I_{\text{contact}}=I_{cm}+M{R}^2=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$. With $\omega =v/R$: $KE=\frac{1}{2}\left(\frac{3}{2}M{R}^2\right)\frac{v^2}{R^2}=\frac{3}{4}M{v}^2.$ Identical ✓. The contact-point view absorbs the translational term into a bigger moment of inertia.

(a) On the flat, all PE has become rolling KE (no slipping ⇒ no energy lost to friction): $Mgh=\frac{1}{2}M{v}^2(1+\beta )\Rightarrow v=\sqrt{\frac{2gh}{1+\beta }}=\sqrt{\frac{2(10)(3)}{1.5}}=\sqrt{40}\approx 6.32\text{m/s}.$ (b) Rotational fraction $=\frac{\beta }{1+\beta }=\frac{1/2}{3/2}=\frac{1}{3}\approx 33.3\%$. It does not change once shape is fixed.

Energy conservation for the whole system. The falling block loses PE; that becomes the block's translational KE plus the sphere's rolling KE. $m_{b}gh=\frac{1}{2}{m}_b{v}^2+\underset{\text{rolling sphere}}{\underbrace{\frac{1}{2}m{v}^2(1+\beta )}}.$ With $\beta =\frac{2}{5}$: sphere factor $=1.4$. $m_{b}gh=\frac{1}{2}{v}^2[m_b+m(1+\beta )]=\frac{1}{2}{v}^2[1+2(1.4)]=\frac{1}{2}{v}^2(3.8).$ Left side: $1(10)(0.7)=7$. So $7=1.9v^2\Rightarrow v^2=\frac{7}{1.9}\approx 3.684\Rightarrow v\approx 1.92\text{m/s}.$

Sign/direction convention first (the subtle part). Set the positive $x$-axis in the direction the top of the wheel is being flung — that is the direction the contact point slides across the floor. Kinetic friction always opposes the sliding of the contact point, so friction points in the $+x$ direction on the wheel's base, i.e. it pushes the centre forward (spins-up into forward motion). Meanwhile that same forward-pointing friction force, acting at the bottom, produces a torque about the centre that opposes the spin — it slows $\omega$ down. So friction simultaneously: increases $v$ from $0$, decreases $\omega$ from ${\omega }_0$. Rolling begins the instant these meet at $v=\omega R$. Getting this sign right is what makes $\omega$ fall to $\frac{1}{2}{\omega }_0$ rather than rise.

(a) Because friction is horizontal at the contact point, it has zero moment arm about that point, so angular momentum about the contact point is conserved (that is why we choose that axis — it kills the only external horizontal force's torque). Initial (pure spin, cm at rest): $L_i=I_{cm}{\omega }_0=M{R}^2{\omega }_0$. Final (rolling): $L_f=I_{cm}{\omega }_f+MR{v}_f=M{R}^2{\omega }_f+MR({\omega }_{f}R)=2M{R}^2{\omega }_f$ (using $v_f={\omega }_{f}R$). Set equal: $M{R}^2{\omega }_0=2M{R}^2{\omega }_f\Rightarrow {\omega }_f=\frac{1}{2}{\omega }_0$, so $v_f={\omega }_{f}R=\frac{1}{2}{\omega }_{0}R=\frac{1}{2}(20)(0.5)=5\text{m/s}.$ (b) Initial KE (pure spin): $\frac{1}{2}{I}_{cm}{\omega }_0^2=\frac{1}{2}M{R}^2{\omega }_0^2$. Final KE (rolling ring): using $KE=\frac{1}{2}M{v}_f^2(1+\beta )$ with $\beta =1$ gives $K{E}_f=\frac{1}{2}M{v}_f^2(2)=M{v}_f^2=M(\frac{1}{2}{\omega }_{0}R{)}^2=\frac{1}{4}M{R}^2{\omega }_0^2$. So $\frac{K{E}_f}{K{E}_i}=\frac{\frac{1}{4}M{R}^2{\omega }_0^2}{\frac{1}{2}M{R}^2{\omega }_0^2}=\frac{1}{2}$. Half the kinetic energy is lost to friction — because here it does slip until rolling starts. Energy conservation fails; angular-momentum conservation is the right tool.

The figure shows the two states: on the left the ring spins in place (${\omega }_0$, centre at rest, contact point slips — coral dot); friction (grey arrow) acts forward at the base; on the right the ring has reached rolling with ${\omega }_f=\frac{1}{2}{\omega }_0$ and $v_f=\frac{1}{2}{\omega }_{0}R$.

Equal final speed means equal $v^2$: $\frac{2g{h}_s}{1+{\beta }_s}=\frac{2g{h}_d}{1+{\beta }_d}$. Cancel $2g$: $\frac{h_s}{1+{\beta }_s}=\frac{h_d}{1+{\beta }_d}$. $h_d=h_s\cdot \frac{1+{\beta }_d}{1+{\beta }_s}=1.0\cdot \frac{1+1/2}{1+2/5}=\frac{1.5}{1.4}\approx 1.071\text{m}.$ The disc needs a slightly higher ramp because it locks more energy into spin, so it needs a bit more PE to match the sphere's speed.

L1 — Q1: $KE=\frac{1}{2}M{v}_{cm}^2+\frac{1}{2}{I}_{cm}{\omega }^2$. Q2: $\beta =\frac{1}{2}$, rotational fraction $\frac{1}{3}$. Q3: $KE=8.4\text{J}$. L2 — Q1: total $75\text{J}$, translational $50\text{J}$, rotational $25\text{J}$. Q2: $v=\sqrt{20}\approx 4.47\text{m/s}$. Q3: $\omega =30\text{rad/s}$, KE ratio $=1$. L3 — Q1: $v_{\text{sphere}}/v_{\text{ring}}=\sqrt{10/7}\approx 1.195$. Q2: $\beta =1$ (ring). Q3: both methods give $\frac{3}{4}M{v}^2$. L4 — Q1: $v=\sqrt{40}\approx 6.32\text{m/s}$, rotational fraction $\frac{1}{3}$. Q2: $v\approx 1.92\text{m/s}$. L5 — Q1: $v_f=5\text{m/s}$, half the KE lost. Q2: $h_d\approx 1.071\text{m}$.