Centripetal force
Introduced| English | Meaning |
|---|---|
| tangent/ˈtændʒənt/ | A straight line that touches a curve at a point and has the same direction there. |
| centripetal acceleration/senˈtrɪpɪtl əkˌseləˈreɪʃn/ | The acceleration towards the centre of a circular path, caused by the changing direction of velocity. |
| centripetal force/senˈtrɪpɪtl fɔːs/ | The resultant force towards the centre of a circular path. |
| perpendicular/ˌpɜːpənˈdɪkjʊlə/ | Meeting at a right angle of 90 degrees. |
| tension/ˈtenʃn/ | A pulling force transmitted through a stretched string, cable or similar object. |
| banked/bæŋkt/ | Describing a curved road or track tilted so that the outside edge is higher than the inside edge. |
Let go and it flies off
- Swing a ball on a string in a circle, then let go.
- It shoots off in a straight line — not outwards, but along the tangent 切线.
- So while it circled, something must have pulled it inward.
Centripetal acceleration 向心加速度
- Constant speed, but ever-changing direction, means the velocity changes — an acceleration.
- It points to the centre: $a = \dfrac{v^{2}}{r} = r\omega^{2}$ (and, since $v = r\omega$, also $a = v\omega$).
Centripetal force and speed
F = mv²/r
For circular motion the force needed grows with the square of the speed — double v, four times the force.
An object moving round a circle at constant speed is still accelerating because:
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Velocity is a vector. Even at constant speed, the changing direction means a changing velocity — an acceleration toward the centre.
An object moves at $4.0\ \dfrac{\text{m}}{\text{s}}$ round a circle of radius $2.0\ \text{m}$. Find its centripetal acceleration.
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$a = \dfrac{v^{2}}{r} = \dfrac{4.0^{2}}{2.0} = 8.0\ \dfrac{\text{m}}{\text{s}^2}$.
Centripetal force 向心力
- By $F = ma$: $F = \dfrac{mv^{2}}{r} = mr\omega^{2}$, pointing to the centre.
- It is always perpendicular 垂直 to the velocity, so it does no work and the speed stays constant.
A spinning fairground ride needs a centripetal force toward the centre
The centripetal force points toward the centre of the circle.
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Yes — always toward the centre, perpendicular to the velocity.
A $2.0\ \text{kg}$ ball moves at $4.0\ \dfrac{\text{m}}{\text{s}}$ round a circle of radius $2.0\ \text{m}$. What centripetal force is needed?
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$F = \dfrac{mv^{2}}{r} = \dfrac{2.0 \times 4.0^{2}}{2.0} = 16\ \text{N}$.
Not a new force
- "Centripetal force" is not a new kind of force.
- It is the net result of the real forces — tension 张力, gravity, friction, a normal force.
The velocity points along the tangent; the force and acceleration point to the centre
The centripetal force is:
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It is whatever real force(s) happen to point to the centre — not a separate force of its own.
Where it comes from
- Ball on a string → tension. Car on a flat corner → friction.
- Planet or satellite → gravity. Banked 倾斜 track → the inward part of the normal force.
On a banked track the horizontal part of the road's force provides the centripetal force
Match each circular motion to the force that provides the centripetal force.
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Different situations, different real forces — but each points to the centre and provides $\dfrac{mv^{2}}{r}$.
Worked example: how fast round the corner?
A $1200\ \text{kg}$ car takes a flat bend of radius $50\ \text{m}$. The largest friction force the tyres can provide is $8200\ \text{N}$.
- Friction is the centripetal force: $F = \dfrac{mv^{2}}{r}$, so the fastest safe speed has $\dfrac{1200\,v^{2}}{50} = 8200$.
- Solve: $v^{2} = \dfrac{8200 \times 50}{1200} = 342$, so $v = 18\ \dfrac{\text{m}}{\text{s}}$ (about $65\ \dfrac{\text{km}}{\text{h}}$).
- Tighter bend: halve $r$ and $v_{\text{max}}$ falls by $\sqrt{2}$ — the required force grows as $\dfrac{1}{r}$.
- Check: on a wet road the available friction drops, so the safe speed drops too, which is what the physics says and what road signs assume.
An $800\ \text{kg}$ car rounds a flat bend of radius $40\ \text{m}$. The maximum friction force available is $6400\ \text{N}$. What is the maximum safe speed, in $\dfrac{\text{m}}{\text{s}}$?
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$\dfrac{mv^{2}}{r} = F$, so $v^{2} = \dfrac{6400 \times 40}{800} = 320$ and $v = 17.9\ \dfrac{\text{m}}{\text{s}}$.
Vertical circles
- Going round an upright loop, the speed changes (gravity does work).
- At the top, weight and tension both point to the centre: $T + mg = \dfrac{mv^{2}}{r}$. At the bottom they oppose: $T - mg = \dfrac{mv^{2}}{r}$.
- The slowest speed at the top with the string just tight is $v_{\text{min}} = \sqrt{gr}$ (set tension $= 0$).
Going round a vertical circle, gravity helps at the top and opposes at the bottom — so the string tension is largest at the bottom
At the top of a vertical loop, the slowest speed (string just tight) is:
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With tension $= 0$, gravity alone is the centripetal force: $mg = \dfrac{mv^{2}}{r}$, so $v_{\text{min}} = \sqrt{gr}$.
Worked example: tension top and bottom
A $0.50\ \text{kg}$ ball on a $1.0\ \text{m}$ string is whirled in a vertical circle. At the bottom its speed is $8.0\ \dfrac{\text{m}}{\text{s}}$.
- Bottom: $T - mg = \dfrac{mv^{2}}{r}$, so $T = 0.50 \times 9.81 + \dfrac{0.50 \times 8.0^{2}}{1.0} = 4.9 + 32 = 37\ \text{N}$.
- Speed at the top: energy conservation over a rise of $2r$: $v_{\text{top}}^{2} = 8.0^{2} - 2g(2.0) = 64 - 39 = 25$, so $v_{\text{top}} = 5.0\ \dfrac{\text{m}}{\text{s}}$.
- Top: $T + mg = \dfrac{mv^{2}}{r}$, so $T = \dfrac{0.50 \times 25}{1.0} - 4.9 = 12.5 - 4.9 = 7.6\ \text{N}$.
- Check: the tension is far larger at the bottom, which is where a string breaks. At the top $v_{\text{top}} > \sqrt{gr} = 3.1\ \dfrac{\text{m}}{\text{s}}$, so the string stays taut.
Never draw "centripetal force" as an extra arrow on a force diagram — it is the resultant of the real forces, and there is no outward "centrifugal" force in this course. At the top of a loop the weight helps ($T + mg$); at the bottom it opposes ($T - mg$). And in a vertical circle the speed is not constant, so use energy conservation to move between top and bottom.
At the top of a vertical circle the string tension is smallest, because there the ____ of the ball also acts towards the centre.
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At the top $T + mg = \dfrac{mv^{2}}{r}$: the weight supplies part of the centripetal force, so the string needs to supply less.
You've got it
- circular motion needs an inward (centripetal) acceleration $a = \dfrac{v^{2}}{r} = r\omega^{2} = v\omega$
- centripetal force $F = \dfrac{mv^{2}}{r}$ — the net of real forces, toward the centre; it does no work
- vertical loop: $T + mg = \dfrac{mv^{2}}{r}$ at the top, $T - mg = \dfrac{mv^{2}}{r}$ at the bottom; $v_{\text{min}} = \sqrt{gr}$ at the top