Mistake Master
Mistake Master
A rolling object has kinetic energy in two places at once: its forward motion ($K_\text{trans}$) and its spin ($K_\text{rot}$). For pure rolling, $v_\text{cm} = R\,\omega$ ties them together, and static friction does no work. Three traps wait here: treating $v_\text{cm}$ and $\omega$ as independent, dropping the $K_\text{rot}$ piece from the energy tally, and assuming friction always drains energy.
6.1 had a wheel on a fixed axle with $K_\text{rot} = \tfrac{1}{2}I\omega^2$. Take the axle away and let the wheel roll along the ground. Now two things happen at once: the wheel translates (the center of mass moves) and the wheel rotates (it spins about that center). The total kinetic energy is the sum:
$$K_\text{tot} = K_\text{trans} + K_\text{rot} = \tfrac{1}{2}Mv_\text{cm}^2 + \tfrac{1}{2}I\omega^2.$$For rolling without slipping, the two motions aren't independent. The contact point is instantaneously at rest, and that one fact forces a single equation linking $v_\text{cm}$ and $\omega$:
$$v_\text{cm} = R\,\omega.$$This is the rolling constraint (or no-slip constraint). It's the centerpiece of the topic. When it holds, three useful things follow: $K_\text{rot}$ can be written purely in terms of $v_\text{cm}$; energy conservation on a ramp becomes one-line algebra; and static friction at the contact does no work, because the contact isn't sliding.
The trap to watch: it's easy to treat $v_\text{cm}$ and $\omega$ as separate variables, or to compute rolling kinetic energy as $\tfrac{1}{2}Mv^2$ and forget the spin. The rest of the lesson is about not making either mistake.
The no-slip condition says the contact point doesn't slide along the ground. Pick a point on the rim and trace it as the wheel rolls: every time that point hits the ground, it momentarily stops. That single fact forces the constraint.
$$v_\text{cm} = R\,\omega \qquad (\text{rolling without slipping}).$$The center moves forward at $v_\text{cm}$. The rim, relative to the center, moves at $R\omega$ around the axis. At the top of the wheel, the two velocities add and the rim point moves at $2v_\text{cm}$. At the bottom, they cancel and the rim point is instantaneously at rest. That's what $v_\text{cm} = R\,\omega$ says geometrically.
If the wheel slips (it skids on ice, or the surface is too slick to grip), the relation drops away and $v_\text{cm}$ and $\omega$ become independent again. The wheel is then slipping: kinetic friction acts at the moving contact, and energy dissipates as heat.
A wheel of radius $R = 0.5$ m rolls along level ground without slipping. The center of mass moves forward at $v_\text{cm} = 4$ m/s. Find the angular speed $\omega$ and the velocity of the rim point at the top, the side, and the bottom of the wheel.
From $v_\text{cm} = R\,\omega$:
$$\omega = \dfrac{v_\text{cm}}{R} = \dfrac{4}{0.5} = 8 \text{ rad/s}.$$The center moves forward at $v_\text{cm}$. The rim at the top moves at $+R\omega$ relative to the center. Sum: $v_\text{cm} + R\omega = 4 + 4 = 8$ m/s, forward.
At the side, the rim's motion about the center is vertical (up or down) while the center moves horizontally. The two are perpendicular, so the rim speed is $\sqrt{v_\text{cm}^2 + (R\omega)^2} = \sqrt{16 + 16} = \sqrt{32} \approx 5.66$ m/s.
At the bottom, the rim moves at $-R\omega$ relative to the center (backward), exactly canceling $v_\text{cm}$. Sum: $v_\text{cm} - R\omega = 0$ m/s. The contact point is instantaneously at rest. This is the geometric statement that the wheel rolls without slipping.
The contact-point fact is the engine of every rolling result in this lesson. It's also why static friction does no work in pure rolling: the force acts at a point with zero velocity, so the rate of work done on the wheel is zero.
The headline problem in 6.5: release an object from rest at height $h$ on a ramp, roll it without slipping to the bottom, find $v_\text{cm}$ at the bottom. Static friction holds the no-slip constraint but does no work, so mechanical energy is conserved.
Top of the ramp, at rest: $E_i = Mgh$. Bottom, rolling: $E_f = \tfrac{1}{2}Mv_\text{cm}^2 + \tfrac{1}{2}I\omega^2$. Use the constraint $\omega = v_\text{cm}/R$ and write $I = \beta MR^2$, where $\beta$ is the dimensionless shape factor. Then:
$$K_\text{tot} = \tfrac{1}{2}Mv_\text{cm}^2 + \tfrac{1}{2}\beta MR^2 \cdot \dfrac{v_\text{cm}^2}{R^2} = \tfrac{1}{2}Mv_\text{cm}^2 (1 + \beta).$$Setting $Mgh = \tfrac{1}{2}Mv_\text{cm}^2(1 + \beta)$ and solving:
$$v_\text{cm} = \sqrt{\dfrac{2gh}{1 + \beta}}.$$Two things pop out. First, $M$ cancels: heavier objects don't win the race. Second, $R$ cancels: bigger objects don't either. The only thing that matters is $\beta$, which depends only on shape. Lower $\beta$, larger $v_\text{cm}$, faster finish.
Three objects roll from rest down identical ramps of height $h = 1$ m: a solid sphere ($\beta = \tfrac{2}{5}$), a uniform solid disk ($\beta = \tfrac{1}{2}$), and a thin hoop ($\beta = 1$). All have the same mass $M$ and radius $R$. Find $v_\text{cm}$ at the bottom for each. Use $g = 10$ N/kg.
$v_\text{cm} = \sqrt{\dfrac{2(10)(1)}{1 + 2/5}} = \sqrt{\dfrac{20}{7/5}} = \sqrt{\dfrac{100}{7}} \approx 3.78$ m/s. $K_\text{trans} : K_\text{rot}$ splits as $5/7 : 2/7$. Of $mgh = 10$ J, about $7.14$ J becomes $K_\text{trans}$ and $2.86$ J becomes $K_\text{rot}$.
$v_\text{cm} = \sqrt{\dfrac{20}{1 + 1/2}} = \sqrt{\dfrac{40}{3}} \approx 3.65$ m/s. Split: $2/3 : 1/3$. About $6.67$ J in translation, $3.33$ J in rotation.
$v_\text{cm} = \sqrt{\dfrac{20}{1 + 1}} = \sqrt{10} \approx 3.16$ m/s. Split: $1/2 : 1/2$. Half the $mgh$ goes to translation, half to rotation. The hoop is rotational dead weight in disguise.
Order of finish = order of $\beta$: smaller first. A frictionless sliding block (no rotation, $\beta = 0$) would beat all three at $\sqrt{2gh} = \sqrt{20} \approx 4.47$ m/s, since none of its energy budget is hiding in spin.
Why static friction does no work. Friction is what holds the constraint $v_\text{cm} = R\omega$ in place. Without it, the wheel would slide instead of rolling. But the static friction force acts at the contact point, and the contact point has zero velocity (the geometric statement of rolling without slipping). Work equals force times the velocity of the point of application, so the work done on the wheel is zero. Static friction in pure rolling supplies the torque that spins the wheel up, but it doesn't drain energy from the mechanical-energy budget. That's what lets the energy-conservation argument work.
Rolling motion misleads students who apply the rolling condition too freely, forget the spin's share of the energy, or assume friction can only drain it. You will meet all three in your diagnostic.
The constraint $v_\text{cm} = R\omega$ is so neat that it gets used everywhere, on every wheel, regardless of context. But $v_\text{cm} = R\omega$ isn't a law. It's a condition, valid only when the contact point is at rest, i.e., when the wheel rolls without slipping. A wheel that slips (a car on ice, a bowling ball thrown so hard it skids before it rolls, a yo-yo on a long string) breaks the constraint: $v_\text{cm}$ and $\omega$ become independent.
Fix. Check whether the problem says "rolling without slipping" or, equivalently, whether the contact point has zero velocity. If yes, apply $v_\text{cm} = R\omega$ freely. If the wheel slips, treat $v_\text{cm}$ and $\omega$ as separate unknowns; kinetic friction at the moving contact dissipates energy as heat.
Topic 3.1 said $K = \tfrac{1}{2}mv^2$ for a moving object, and that formula has worked for every block and ball you've seen so far. But a rolling object does two things at once: its center of mass translates and the body rotates about that center. Both store kinetic energy.
Fix. For any rigid body that rolls, $K_\text{tot} = K_\text{trans} + K_\text{rot} = \tfrac{1}{2}Mv_\text{cm}^2 + \tfrac{1}{2}I\omega^2$. Both terms, every time. For a rolling solid sphere, $K_\text{rot}$ is $2/7$ of the total; for a hoop, half. Reporting only $\tfrac{1}{2}Mv^2$ misses real, substantial energy.
The rule from Topic 2.7 and Unit 3 ("friction does negative work, so mechanical energy is lost to heat") is true for kinetic friction on a sliding surface. For static friction in pure rolling, it's wrong. The static friction at the contact point holds the no-slip constraint and supplies the torque that spins the wheel up. Without it the wheel would slide. But it does no work on the wheel, because the contact point has zero velocity. No work means no energy transfer.
Fix. In pure rolling, static friction is a constraint force: it connects translation to rotation, but it doesn't drain energy. Mechanical energy is conserved. The opposite extreme matters too: when a wheel slips, the contact point moves, kinetic friction acts there, and kinetic friction does dissipate energy. The rule flips across the no-slip boundary.