Question bank — Rolling KE = ½mv² + ½Iω²

False. Same $v_{cm}$ but different $\beta$: ring has $\frac{1}{2}M{v}^2(1+1)$, sphere has $\frac{1}{2}M{v}^2(1+\frac{2}{5})$, so the ring carries more KE at equal speed.

True. $KE=\frac{1}{2}M{v}_{cm}^2(1+\beta )$ is linear in $M$, and $\beta$ depends only on shape, not on mass — so mass scales KE directly.

True. $v=\sqrt{2gh/(1+\beta )}$ and $\beta =I_{cm}/M{R}^2$ is $\frac{2}{5}$ for any solid sphere; radius cancels entirely, so both arrive at the same $v_{cm}$.

True — but only for $\beta >1$, i.e. mass concentrated beyond $R$ (rare). For a plain ring $\beta =1$ gives exactly $1/2$; standard solid shapes stay below.

False. König's split is valid for any rigid body in plane motion. Rolling-without-slipping is only needed to link $v_{cm}=\omega R$ and collapse KE into one variable.

False. Translational KE is zero but $\frac{1}{2}{I}_{cm}{\omega }^2$ is large — all the energy is locked in spin because the no-slip link is broken.

False. With zero friction there is no torque about the center, so $\omega$ never changes; the ball slides down and arrives at $\sqrt{2gh}$, faster than if it rolled.

True. The contact point is instantaneously at rest, so the friction force acts through zero displacement per instant — it supplies torque but transfers no energy.

Wrong: it is also spinning. You must add $\frac{1}{2}{I}_{cm}{\omega }^2$; ignoring it undercounts KE and falsely predicts all shapes reach the ramp bottom at equal speed.

Double-counting. Either use $\frac{1}{2}M{v}^2+\frac{1}{2}{I}_{cm}{\omega }^2$ or $\frac{1}{2}{I}_{contact}{\omega }^2$ alone — never both. By the parallel-axis theorem $I_{contact}=I_{cm}+M{R}^2$ already contains the $\frac{1}{2}M{v}^2$ piece.

Mass is irrelevant. Both PE ($Mgh$) and KE ($\frac{1}{2}M{v}^2(1+\beta )$) scale with $M$, so it cancels; only $\beta$ decides the order.

Only true when slipping. For rolling without slipping the constraint $v_{cm}=\omega R$ locks them together — they are not free to choose independently.

No. ${\beta }_{solid}=\frac{2}{5}<{\beta }_{hollow}=\frac{2}{3}$, so the solid sphere ($v=\sqrt{2gh/1.4}$) beats the hollow one ($v=\sqrt{2gh/1.667}$).

No — it vanishes exactly and always, because $\sum m_i{\vec{u}}_i=M{\vec{u}}_{cm,rel}=0$: the center of mass has zero velocity relative to itself, regardless of how fast ${\vec{v}}_{cm}$ is.

Wrong $R$. The rolling constraint uses the geometric radius (the one touching the ground). The radius of gyration $k$ (with $I=M{k}^2$) is a different length used inside $\beta$, not in $v=\omega R$.

The ring's larger $\beta$ diverts a bigger fraction ($\frac{1}{2}$ vs $\frac{2}{7}$) of the released PE into spinning, leaving less for forward motion, so its $v_{cm}$ is smaller.

Because König's cross term dies ($\sum m_i{\vec{u}}_i=0$), the two energies decouple cleanly — nothing is shared or double-counted between the sliding and spinning pictures.

$v\propto 1/\sqrt{1+\beta }$ for every ramp; changing $h$ or angle rescales all speeds by the same factor, so the ordering by $\beta$ never changes.

The contact point has zero relative sliding velocity, which is exactly the condition for static friction; kinetic friction only appears once the surfaces slide against each other.

Parallel-axis: $I_{contact}=I_{cm}+M{R}^2$, so $\frac{1}{2}{I}_{contact}{\omega }^2=\frac{1}{2}{I}_{cm}{\omega }^2+\frac{1}{2}M(\omega R{)}^2=\frac{1}{2}{I}_{cm}{\omega }^2+\frac{1}{2}M{v}^2$ — algebraically identical.

With too little friction the wheel slips, so the contact point moves relative to the surface; $v_{cm}\ne \omega R$ and you must track $v$ and $\omega$ as separate quantities.

Pure translation: $KE=\frac{1}{2}M{v}^2$ with $v=\sqrt{2gh}$. This is the $\beta \to 0$ limit / no-spin case, the fastest possible descent.

Exactly half: $\beta /(1+\beta )=1/2$. Equal energy in sliding and spinning — the extreme among common shapes.

$v=\sqrt{2gh/(1+\beta )}\to 0$: almost all PE goes into spin, so translational descent slows toward a crawl.

If it also isn't spinning ($\omega =0$), no. But if released rolling from rest both are zero at $t=0$; KE only appears once it acquires speed and, by the constraint, spin together.

$\beta =0$, so $KE=\frac{1}{2}M{v}^2$ and $v=\sqrt{2gh}$ — it behaves like a frictionless slider, since there's no moment of inertia to store spin energy.