Exercises — Relation to linear quantities - v = rω, a_t = rα, a_c = rω²

We want speed along the circle, so the tool is $v=r\omega$ (power one). $v=r\omega =0.15\times 40=6\text{m/s}$ $\omega$ is already in rad/s, so we multiply directly — no conversion needed.

Direction of velocity is changing, so we need $a_c=r{\omega }^2$ (power two). $a_c=r{\omega }^2=0.15\times 40^2=0.15\times 1600=240{\text{m/s}}^2$

Tangential acceleration changes the magnitude of speed, and it uses $\alpha$: the tool is $a_t=r\alpha$. $a_t=r\alpha =0.3\times 8=2.4{\text{m/s}}^2$ Here $\alpha >0$, so $a_t>0$: the point is speeding up, and $a_t$ points forward along $\vec{v}$.

Step 1 — WHAT & WHY: convert rev/min to rad/s. The formula $v=r\omega$ demands $\omega$ in radians per second, because it descends from the arc-length relation $s=r\theta$ (with $s$ the arc length defined above), which itself only holds when $\theta$ is in radians (see Radian measure and arc length). One revolution $=2\pi$ radians, and one minute $=60$ s. $\omega =33.333\times \frac{2\pi }{60}=3.491\text{rad/s}$ Step 2 — apply the bridge. $v=r\omega =0.12\times 3.491=0.419\text{m/s}$

Step 1 — find $\alpha$. Angular acceleration is the rate of change of $\omega$: $\alpha =\frac{\Delta \omega }{\Delta t}=\frac{60-0}{12}=5{\text{rad/s}}^2$ Step 2 — apply $a_t=r\alpha$. $a_t=0.4\times 5=2{\text{m/s}}^2$

Because the body is rigid, both ants share the same $\omega$. So in $v=r\omega$ only $r$ differs, and speed is proportional to radius: $\frac{v_B}{v_A}=\frac{r_B\omega }{r_A\omega }=\frac{r_B}{r_A}=\frac{0.20}{0.05}=4$ Ant B is 4 times faster.

(a) Centripetal (toward centre): $a_c=r{\omega }^2=0.5\times 20^2=200{\text{m/s}}^2$ (b) Tangential (along the circle): $a_t=r\alpha =0.5\times 5=2.5{\text{m/s}}^2$ (c) These two arrows are perpendicular ($a_c$ radial, $a_t$ tangential), so combine with Pythagoras: $a=\sqrt{a_c^2+a_t^2}=\sqrt{200^2+2.5^2}\approx 200.016{\text{m/s}}^2$ Notice $a_c$ utterly dominates — at high spin the pull-to-centre dwarfs the speed-up. Look at the figure below: the total (green) arrow leans only a hair off the radial direction.

Setting up the geometry first. The total acceleration is the green arrow in the figure above: it is the diagonal of a right triangle whose two legs are the radial leg $a_c$ (pointing inward) and the tangential leg $a_t$ (pointing along the motion). Call the angle between the total arrow and the radial (inward) leg by the name $\varphi$ — that is the tilt we are asked for. On this triangle, $a_t$ is the side opposite $\varphi$ and $a_c$ is the side adjacent to $\varphi$, so "opposite over adjacent" gives the tangent of $\varphi$: $\tan \varphi =\frac{a_t}{a_c}=\frac{2.5}{200}=0.0125$ To undo the tangent (find "which angle has this tan?") we use arctan: $\varphi =\arctan (0.0125)\approx 0.716°$ Barely off radial — as expected when $a_c\gg a_t$.

"Steady speed" means speed is not changing, so $a_t=0$. But direction is changing, so there is centripetal acceleration. Here we are handed $v$ directly, so the handy form is $a_c=v^2/r$: $a_c=\frac{v^2}{r}=\frac{15^2}{50}=\frac{225}{50}=4.5{\text{m/s}}^2$ Total acceleration $=4.5{\text{m/s}}^2$, pointing to the centre. (See Centripetal force and circular motion for the force that supplies this.)

(a) Invert the bridge $v=r\omega$ to get $\omega =v/r$: $\omega =\frac{v}{r}=\frac{10}{0.25}=40\text{rad/s}$ (b) WHY this formula. We are given a change in $\omega$ over an angle turned $\theta$ (not over a time), so we need a relation linking $\omega$, $\alpha$ and $\theta$ with time eliminated. For constant $\alpha$ that relation is ${\omega }^2={\omega }_0^2+2\alpha \theta$. Where does it come from? It is the exact angular twin of the linear kinematics formula $v^2=v_0^2+2as$: replace distance $s$ by angle $\theta$, speed $v$ by $\omega$, and linear acceleration $a$ by $\alpha$. (You can derive it by taking $\omega ={\omega }_0+\alpha t$, solving for $t$, and substituting into $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$.) With ${\omega }_0=0$ (starts from rest): ${\omega }^2=2\alpha \theta \Rightarrow \alpha =\frac{{\omega }^2}{2\theta }=\frac{40^2}{2\times 8}=\frac{1600}{16}=100{\text{rad/s}}^2$ (c) Bridge to linear tangential accel: $a_t=r\alpha =0.25\times 100=25{\text{m/s}}^2$

(a) From $a_c=r{\omega }^2$, solve for $\omega$: $\omega =\sqrt{\frac{a_c}{r}}=\sqrt{\frac{90}{0.10}}=\sqrt{900}=30\text{rad/s}$ (b) From $a_t=r\alpha$: $\alpha =\frac{a_t}{r}=\frac{30}{0.10}=300{\text{rad/s}}^2$ (c) Bridge for speed: $v=r\omega =0.10\times 30=3\text{m/s}$

(a) Speed grows uniformly from $0$ to $12\text{m/s}$ in $3\text{s}$, so $a_t=\frac{\Delta v}{\Delta t}=\frac{12}{3}=4{\text{m/s}}^2$ (b) From $a_t=r\alpha$: $\alpha =\frac{a_t}{r}=\frac{4}{0.8}=5{\text{rad/s}}^2$ (c) At $t=3\text{s}$, $v=12$, so use $a_c=v^2/r$: $a_c=\frac{12^2}{0.8}=\frac{144}{0.8}=180{\text{m/s}}^2$ (d) Perpendicular combine: $a=\sqrt{a_c^2+a_t^2}=\sqrt{180^2+4^2}=\sqrt{32416}\approx 180.044{\text{m/s}}^2$

(a) Rolling without slipping ties the centre’s speed to the spin: $v_{\text{cm}}=R\omega$, so $\omega =\frac{v_{\text{cm}}}{R}=\frac{7}{0.35}=20\text{rad/s}$ (b) The top point has the rolling velocity ($v_{\text{cm}}$ forward) plus the rotational velocity of the rim about the centre ($R\omega$ forward at the top). They add: $v_{\text{top}}=v_{\text{cm}}+R\omega =7+7=14\text{m/s}$ (c) At the contact point the rotational velocity ($R\omega$ backward) cancels the forward $v_{\text{cm}}$: $v_{\text{bottom}}=v_{\text{cm}}-R\omega =7-7=0\text{m/s}$ The bottom point is instantaneously at rest — that is what "without slipping" means. (More in Rolling motion.)

The tilt of the total-acceleration arrow off the radial direction is $\tan \varphi =a_t/a_c=r\alpha /(r{\omega }^2)=\alpha /{\omega }^2$. As $\omega \to \infty$, $\alpha /{\omega }^2\to 0$, so $\varphi \to 0$: the arrow snaps toward the centre. That confirms fast spinners feel an almost purely inward pull. They are equal when $a_c=a_t$: $r{\omega }^2=r\alpha \Rightarrow {\omega }^2=\alpha \Rightarrow \omega =\sqrt{\alpha }=\sqrt{10}\approx 3.162\text{rad/s}$ (Neat: the equal-point depends only on $\alpha$, not on $r$, since $r$ cancels.)

Every bridge formula carries a factor $r$, and here $r=0$: $v=r\omega =0,a_t=r\alpha =0,a_c=r{\omega }^2=0$ The pivot point does not move at all — it sits on the axis. Angular quantities $\omega ,\alpha$ are huge, but linear quantities all vanish because there is no circle to trace. This is the boundary case that shows why the axis is special: the whole disk shares $\omega$, yet the axis point has zero speed.

(a) From $a_c=r{\omega }^2$, solve for $\omega$: $\omega =\sqrt{\frac{a_c}{r}}=\sqrt{\frac{500}{0.20}}=\sqrt{2500}=50\text{rad/s}$ (b) Rim speed: $v=r\omega =0.20\times 50=10\text{m/s}$ (c) Reaching $50\text{rad/s}$ from rest in $2.5\text{s}$ needs, from $\alpha =\Delta \omega /\Delta t$: $\alpha =\frac{50-0}{2.5}=20{\text{rad/s}}^2,a_t=r\alpha =0.20\times 20=4{\text{m/s}}^2$

$v=r\omega$ (multiply once) ::: linear speed from spin $a_t=r\alpha$ (tangential, uses $\alpha$; sign follows $\alpha$) ::: how the speed changes $a_c=r{\omega }^2=v^2/r$ (radial, always inward, never reverses) ::: how the direction changes combine $a_t$ and $a_c$ ::: $a=\sqrt{a_t^2+a_c^2}$ because they are perpendicular rolling without slipping ::: $v_{\text{cm}}=R\omega$, bottom point at rest