A spinning wheel has one angular velocity, yet its parts move at different speeds. Linking the two kinds of quantity explains why, and it ties turning to centripetal acceleration.
Why it works
A point at radius $r$ that turns through $\Delta\theta$ travels the arc $\Delta s = r\Delta\theta$. Divide by the time: its speed is $v = \dfrac{\Delta s}{\Delta t} = r\dfrac{\Delta\theta}{\Delta t} = r\omega$. If $\omega$ changes, $v$ changes $r$ times as fast: the tangential acceleration is $a_t = r\alpha$. And moving in a circle at speed $v$ always needs the centripetal acceleration $a_c = \dfrac{v^2}{r} = r\omega^2$, toward the center.
Key idea
For a point at distance $r$ from the axis (angles in radians): speed $v = r\omega$, tangential acceleration $a_t = r\alpha$ (along the path), centripetal acceleration $a_c = r\omega^2$ (toward the center). All points share one $\omega$ and $\alpha$, but $v$, $a_t$ and $a_c$ grow with $r$.
Worked examples
Worked example: Speed on a merry-go-round
A merry-go-round turns at $0.80\ \text{rad/s}$. How fast does a child $2.5\ \text{m}$ from the center move? And one $1.0\ \text{m}$ from the center?
- $$v = r\omega = (2.5\ \text{m})(0.80\ \text{rad/s}) = 2.0\ \text{m/s}.$$
- $v = (1.0)(0.80) = 0.80\ \text{m/s}$. Same $\omega$, but the outer child covers a bigger circle in the same time.
Answer: $2.0\ \text{m/s}$ and $0.80\ \text{m/s}$
Worked example: Accelerations of a rim point
A disk of radius $0.20\ \text{m}$ turns at $5.0\ \text{rad/s}$ and is speeding up at $8.0\ \text{rad/s}^2$. Find the tangential and centripetal accelerations of a rim point.
- $a_t = r\alpha = (0.20)(8.0) = 1.6\ \text{m/s}^2$, along the direction of motion.
- $a_c = r\omega^2 = (0.20)(5.0)^2 = 5.0\ \text{m/s}^2$, toward the center.
- They are at right angles, so the total is $\sqrt{1.6^2 + 5.0^2} = 5.2\ \text{m/s}^2$.
Answer: $a_t = 1.6\ \text{m/s}^2$, $a_c = 5.0\ \text{m/s}^2$
Worked example: Linked by a belt
A belt joins a $5.0\ \text{cm}$ motor pulley turning at $1200$ rpm to a $20\ \text{cm}$ pulley. How fast does the big pulley turn?
- A belt that doesn't slip moves at one speed, so both rims have the same $v$: $r_1\omega_1 = r_2\omega_2$.
- $\omega_2 = (1200\ \text{rpm})\dfrac{5.0}{20} = 300$ rpm.
Answer: $300$ rpm
Common mistake
Thinking every point on a spinning wheel moves at the same speed. Every point has the same angular velocity, but its speed is $v = r\omega$: the hub barely moves while the rim races. Check with extremes: the point on the axle ($r = 0$) has $v = 0$.