Connecting Linear and Rotational Motion
Introduced| English | Meaning |
|---|---|
| tangential velocity/tænˈdʒenʃl vəˈlɒsɪti/ | tangential velocity |
| tangential acceleration/tænˈdʒenʃl əkˌseləˈreɪʃn/ | The component of acceleration along the circular path, changing the speed. |
| rigid body/ˈrɪdʒɪd ˈbɒdi/ | An idealised object whose shape and distances between its particles do not change. |
The edge outruns the middle
- Stand at the centre of a spinning merry-go-round and you barely move.
- Stand at the edge and you are flung along fast.
- Same spin, yet wildly different speeds — it all depends on how far out you are.
- One simple link connects the turning to the travelling.
Linear speed from spin
- A point at radius $r$ moves with tangential velocity 切向速度 $v = \omega r$.
- Farther from the axis means faster — the outer edge wins.
- Every point shares the same $\omega$, but their linear speeds differ.
As the radius turns through an angle, a point moves along an arc at speed v
A wheel of radius $0.5\ \text{m}$ spins at $\omega = 8\ \tfrac{\text{rad}}{\text{s}}$. The linear speed of a rim point (in m/s)?
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$v = \omega r = 8 \times 0.5 = 4\ \tfrac{\text{m}}{\text{s}}$.
On a spinning disk, a point farther from the axis has a higher linear speed.
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$v = \omega r$ grows with $r$, so outer points move faster.
A turntable spins at $\omega = 10\ \tfrac{\text{rad}}{\text{s}}$. A point at $r = 0.2\ \text{m}$ has what linear speed (in m/s)?
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$v = \omega r = 10 \times 0.2 = 2\ \tfrac{\text{m}}{\text{s}}$.
Two kinds of acceleration
- Speeding up the spin gives a tangential acceleration 切向加速度 $a_t = \alpha r$, along the motion.
- Turning in a circle also gives a centripetal acceleration $v^2/r$, pointing inward.
- A point that is both circling and speeding up has both at once.
The acceleration along the direction of motion of a rim point is $a_t = ____\, r$.
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Tangential acceleration is $a_t = \alpha r$, mirroring $v = \omega r$.
A rigid body
- A rigid body 刚体 turns all together — every point keeps its distance from the axis.
- So all points share one $\omega$ and one $\alpha$.
- Their linear speeds simply scale with their radius $r$.
Linear from rotational
A point on a spinning body moves faster the farther it is from the axis: v = r times omega.
Two points on the same rigid spinning disk, at different radii, share the same...
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A rigid body turns as one, so every point has the same $\omega$. Linear speed differs with radius.
Where it shows up
- Rolling wheels: the contact point, the axle, and the top all move at different linear speeds.
- Pulleys: the rope's linear speed equals $\omega r$ of the pulley's rim.
- Gears: teeth mesh at equal linear speed, so a small gear spins faster.
Select all true statements for a point on a spinning rigid body.
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$v = \omega r$ and all points share $\omega$; linear speeds differ with radius, so the third is false.
A merry-go-round turns at $\omega = 2\ \tfrac{\text{rad}}{\text{s}}$.
- A child at $r = 3\ \text{m}$ moves at $v = \omega r = 6\ \tfrac{\text{m}}{\text{s}}$.
- A child at $r = 1\ \text{m}$ moves at only $2\ \tfrac{\text{m}}{\text{s}}$ — same spin, three times slower.
Do not confuse angular and linear speed. Two riders on the same turntable have the same $\omega$ but different $v$. Only the linear speed depends on the radius; the angular speed is shared by the whole body.
A point on a spinning rigid body has tangential velocity $v = \omega r$ and tangential acceleration $a_t = \alpha r$. Every point shares one $\omega$ and $\alpha$, but linear speed grows with radius — the outer edge moves fastest. This links wheels, pulleys, and gears.
Write from a rim speed of 3 m/s
- A disk of radius 0.50 m spins at 6 rad/s.
- Find the speed of a point on the rim.
A disk of radius 0.50 m spins at 6 rad/s. How fast is a point on the rim moving, in m/s?
6 times 0.50 is 3.
What is the rim speed?
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Angular speed times radius is 6 times 0.50. Other options — 4 m/s (the lesson rim speed); 6 m/s (the angular speed used as a linear speed).