Exercises — Rolling without slipping — v = Rω condition

What we need: link spin $\omega$ to centre speed $v$. That is the velocity form. Why this form: we are given $\omega$ and want $v$ — no time-change, no distance, so use $v=R\omega$. $v=R\omega =0.5\times 4=2\text{ m/s}.$

What we need: the point-speed rule "bottom zero, centre vee, top twice". Why: rolling combines translation ($+v$ everywhere) with rotation (backward $R\omega$ at bottom, forward $R\omega$ at top). Since $R\omega =v$:

• Top $=v+R\omega =v+v=2v=6\text{ m/s}$.
• Contact $=v-R\omega =v-v=0\text{ m/s}$.

What we need: the acceleration form, because both quantities are rates-of-change of speed. Why this form: given $\alpha$, want $a$ — differentiate the constraint once ⇒ $a=R\alpha$. $a=R\alpha =0.30\times 6=1.8{\text{ m/s}}^2.$

(a) Rearranging $v=R\omega$ to solve for spin: $\omega =\frac{v}{R}=\frac{5}{0.35}=14.29\text{ rad/s}.$ Why divide: we know $v$ and $R$, want $\omega$; $R$ multiplies $\omega$, so undo it by dividing. (b) Top speed $=2v=10\text{ m/s}$ (translation $v$ plus forward rotation $R\omega =v$).

What we need: the displacement form $x_{cm}=R\theta$, because we are given an angle turned and want a distance. Why radians: the constraint uses $\theta$ in radians. One revolution $=2\pi$ rad, so $\theta =2\times 2\pi =4\pi \text{ rad}.$ $x_{cm}=R\theta =0.20\times 4\pi =2.513\text{ m}.$ Sanity check: distance $=$ number of turns $\times$ circumference $=2\times 2\pi R=2\times 2\pi (0.20)=2.513\text{ m}$. ✓ Same answer, as it must be.

Step 1 — linear acceleration: $a=\frac{\Delta v}{t}=\frac{4-0}{5}=0.8{\text{ m/s}}^2$. Why: uniform acceleration means $a$ is constant $=$ (change in speed)$/$(time). Step 2 — convert via the constraint: since it never slips, $a=R\alpha$, so $\alpha =\frac{a}{R}=\frac{0.8}{0.10}=8{\text{ rad/s}}^2.$

What we do: add the two velocities as vectors at the 3 o'clock point — this is exactly the amber right-triangle drawn in the figure.

• Translation (cyan arrow, pointing right): $+v$, horizontal (forward).
• Rotation about centre (white arrow, pointing up): the rim point is a distance $R$ from the centre; its rotational speed is $R\omega =v$. At the 3 o'clock point, rotation carries it straight up (tangent to the circle there).

Why perpendicular: rotational velocity is always tangent to the circle, i.e. perpendicular to the line from centre to point. At 3 o'clock that line is horizontal, so the tangent is vertical — that is why the cyan and white arrows meet at a right angle in the figure.

Combine: the two arrows are the two legs of a right triangle; the amber arrow is its hypotenuse (the resultant). By Pythagoras: $v_{3\text{ o’clock}}=\sqrt{v^2+v^2}=v\sqrt{2}\approx 1.414v,$ pointing up-and-forward at $45^{\circ }$ above horizontal — the slope of the amber arrow.

Why this matters: it shows point speed varies smoothly from $0$ (bottom) through $v\sqrt{2}$ (side) to $2v$ (top) — the wheel is not a rigid block moving at one speed.

The trick: since the contact point is instantaneously at rest, the whole wheel can be treated as pure rotation about that contact point (the Instantaneous axis of rotation) — that amber pivot dot at the bottom of the figure. Why allowed: a rotating rigid body's point speed $=\omega \times (\text{distance from the axis})$. The contact point has zero speed, so it is the axis for that instant.

• Centre: distance from contact axis $=R$ (the white double-arrow in the figure), so speed $=\omega R=v$. ✓ (matches what we know) — shown by the short white velocity arrow.
• Top: distance from contact axis $=2R$ (the taller cyan double-arrow — the top is diametrically opposite the contact point). So speed $=\omega (2R)=2\omega R=2v$. ✓ — shown by the longer cyan velocity arrow, exactly twice the length of the white one.

The figure makes the "twice the distance ⇒ twice the speed" reasoning visible: same $\omega$, double the radius arm.

What we need: rolling KE $=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$ — translational plus rotational. Why the constraint appears: we know $v$, not $\omega$, but no-slip gives $\omega =v/R$, so we can write everything in $v$. $KE=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{R}^2\right){\left(\frac{v}{R}\right)}^2=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2=\frac{7}{10}m{v}^2.$ Why the $R$ cancels: the $R^2$ in $I$ meets the $1/R^2$ from ${\omega }^2$ — a hallmark of rolling problems. $KE=\frac{7}{10}(2)(3^2)=0.7\times 2\times 9=12.6\text{ J}.$

(a) Use $v^2=u^2+2ad$ with start speed $u=0$: $a=\frac{v^2}{2d}=\frac{6^2}{2\times 9}=\frac{36}{18}=2{\text{ m/s}}^2.$ Why this kinematic equation: it links speed, distance and acceleration with no time — exactly the three quantities we have. (b) No slipping, so the differentiated constraint holds: $\alpha =\frac{a}{R}=\frac{2}{0.30}=6.667{\text{ rad/s}}^2.$

(a) Test the constraint: if rolling, $v$ should equal $R\omega =0.30\times 30=9\text{ m/s}$. But the car moves at only $6\text{ m/s}$. Since $6\ne 9$, it is slipping — $v=R\omega$ does NOT hold. (b) Contact-point velocity $=v_{cm}-R\omega$ (translation forward, rotation drags it back): $v_{contact}=6-9=-3\text{ m/s}.$ The minus sign means the contact patch is moving backward at $3\text{ m/s}$ relative to the ground — the tyre is spinning faster than the road passes under it, so it skids backward (kinetic friction now acts, per Static vs kinetic friction).

(a) $\alpha =\frac{{\omega }_1-{\omega }_0}{t}=\frac{10-2}{4}=2{\text{ rad/s}}^2.$ (b) No slip ⇒ $a=R\alpha =0.4\times 2=0.8{\text{ m/s}}^2.$ (c) Total angle turned: since $\alpha$ is constant, the angular speed rises in a straight line from ${\omega }_0$ to ${\omega }_1$. For any quantity that changes at a constant rate, the average value is simply the midpoint of start and end, $\bar{\omega }=\frac{{\omega }_0+{\omega }_1}{2}$, and distance covered $=$ average rate $\times$ time. That is why constant angular acceleration lets us use $\theta =\bar{\omega }t$ (it is the exact area under a straight-line $\omega$-vs-$t$ graph — a trapezium): $\theta =\frac{{\omega }_0+{\omega }_1}{2}t=\frac{2+10}{2}\times 4=24\text{ rad}.$ $x_{cm}=R\theta =0.4\times 24=9.6\text{ m}.$ Cross-check via linear speeds: $v_0=R{\omega }_0=0.8\text{ m/s}$, $v_1=R{\omega }_1=4\text{ m/s}$, average $=2.4\text{ m/s}$, times $4\text{ s}=9.6\text{ m}$. ✓

Set up each energy using $\omega =v/R$ so $\frac{1}{2}I{\omega }^2=\frac{1}{2}I{v}^2/R^2$.

• Hoop: rotational $=\frac{1}{2}(m{R}^2)(v^2/R^2)=\frac{1}{2}m{v}^2$. Translational $=\frac{1}{2}m{v}^2$. Total $=m{v}^2$. Rotational fraction $=\frac{\frac{1}{2}m{v}^2}{m{v}^2}=\frac{1}{2}=50\%$.
• Disc: rotational $=\frac{1}{2}(\frac{1}{2}m{R}^2)(v^2/R^2)=\frac{1}{4}m{v}^2$. Translational $=\frac{1}{2}m{v}^2$. Total $=\frac{3}{4}m{v}^2$. Rotational fraction $=\frac{\frac{1}{4}m{v}^2}{\frac{3}{4}m{v}^2}=\frac{1}{3}\approx 33.3\%$. Answer: the hoop stores a larger fraction ($50\%$ vs $33.3\%$) in rotation, because its mass sits far from the axis, giving a bigger Moment of inertia.