Connecting Linear and Rotational Motion
Đã giới thiệuOn a merry-go-round, the edge whips past
- Stand near the centre of a spinning merry-go-round and you drift slowly.
- Stand at the edge and you are whipped around fast — yet everyone shares the same spin.
- The whole platform turns together, so everyone has the same angular velocity.
- But each rider's linear speed depends on how far out they sit.
The bridge equations
- Arc length links to angle: $s = r\theta$.
- Linear speed links to angular speed: $v = r\omega$.
- Tangential acceleration links to angular acceleration: $a_t = r\alpha$.
- In every case, multiply the angular quantity by the radius $r$.
Radius sets the speed
Change the radius and see how the arc length (and so the linear speed) scales with r.
A point is $0.3\ \text{m}$ from a wheel's axis, spinning at $\omega = 10\ \tfrac{\text{rad}}{\text{s}}$. What is its linear speed, in $\tfrac{\text{m}}{\text{s}}$?
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$v = r\omega = 0.3 \times 10 = 3\ \tfrac{\text{m}}{\text{s}}$.
The arc length swept is $s = r\_\_$ (fill in the angular quantity).
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$s = r\theta$ — radius times the angle in radians.
Select all correct linear–rotational bridge equations.
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Each links an angular quantity to a linear one by multiplying by $r$. $v = \omega/r$ is wrong.
Farther out means faster
- Because $v = r\omega$, a larger radius gives a larger linear speed for the same spin.
- The rim of a wheel moves faster than a point near the hub.
- This is why a long lever tip, a fan blade edge, or the outer horse on a carousel moves quickest.
- Same $\omega$ everywhere on the object; different $v$ at different radii.
On a rigid spinning disk, how do a rim point and a point near the hub compare?
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All points share $\omega$; the rim has a larger radius, so $v = r\omega$ is larger there.
For the same wheel above, what is the linear speed at $0.6\ \text{m}$ from the axis, in $\tfrac{\text{m}}{\text{s}}$?
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$v = r\omega = 0.6 \times 10 = 6\ \tfrac{\text{m}}{\text{s}}$ — double the radius, double the speed.
Rolling without slipping
- When a wheel rolls without slipping, its centre moves at $v = r\omega$.
- The contact point is momentarily at rest; the top moves at $2v$.
- This neat condition ties a wheel's spinning to its travelling.
- We will use it again for rolling energy in the next topic.
A wheel rolling without slipping has its centre moving at $v = r\omega$.
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Rolling without slipping ties travel to spin exactly by $v = r\omega$.
Every point on a rigid spinning object shares the same $\omega$, but not the same $v$. Do not assume the edge and the centre move at the same linear speed — the edge is much faster because its radius is larger ($v = r\omega$).
A point sits $0.3\ \text{m}$ from the axis of a wheel turning at $\omega = 10\ \tfrac{\text{rad}}{\text{s}}$.
- $v = r\omega = 0.3 \times 10 = 3\ \tfrac{\text{m}}{\text{s}}$.
A point at $0.6\ \text{m}$ would move at $6\ \tfrac{\text{m}}{\text{s}}$ — twice as fast, for the same spin.
Linear and angular motion are linked by the radius: $s = r\theta$, $v = r\omega$, $a_t = r\alpha$. Every point on a rigid body shares the same $\omega$, but points farther from the axis have larger $v$. A wheel rolling without slipping obeys $v = r\omega$.
Write from a rim speed of 3.2 m/s
- A point is 0.80 m from the axis of a wheel spinning at 4 rad/s.
- Find the linear speed.
A point is 0.80 m from a wheel's axis. The wheel spins at 4 rad/s. What is the linear speed of that point, in m/s?
0.80 times 4 is 3.2.
What is the linear speed?
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Radius times angular speed is 0.80 times 4. Other options — 6 m/s (the lesson doubled radius); 4 m/s (the angular speed used as a linear speed).