Centripetal force · 向心力
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| tangent/ˈtændʒənt/ | 切线 | qiè xiàn |
| centripetal acceleration/senˈtrɪpɪtl əkˌseləˈreɪʃn/ | 向心加速度 | xiàng xīn jiā sù dù |
| centripetal force/senˈtrɪpɪtl fɔːs/ | 向心力 | xiàng xīn lì |
| perpendicular/ˌpɜːpənˈdɪkjʊlə/ | 垂直 | chuízhí |
| tension/ˈtenʃn/ | 张力 | zhāng lì |
| banked/bæŋkt/ | 倾斜 | qīng xié |
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.
松手,它就飞出去
- 用绳子拴着球在圆周上甩,然后松手。
- 它沿 直线 飞出去——不是向外,而是沿着切线(tangent)。
- 所以它转圈的时候,一定有什么东西把它 向内 拉。
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 $a = v\omega$).
向心加速度
- 速率恒定,但 方向 不断改变,意味着速度在变——这就是加速度。
- 它指向圆心:$a = \dfrac{v^{2}}{r} = r\omega^{2}$(由于 $v = r\omega$,也有 $a = v\omega$)。
Centripetal force and speed · 向心力与速度
F = mv²/r
For circular motion the force needed grows with the square of the speed — double v, four times the force. · 对于圆周运动，所需的力随速度的平方增长——速度 v 加倍，力变为四倍。
An object moving round a circle at constant speed is still accelerating because: · 一个以恒定速率绕圆运动的物体仍在加速,因为:
Velocity is a vector. Even at constant speed, the changing direction means a changing velocity — an acceleration toward the centre. · 速度是矢量。即使速率不变,方向改变也意味着速度改变——一个指向圆心的加速度。
An object moves at $4.0\ \dfrac{\text{m}}{\text{s}}$ round a circle of radius $2.0\ \text{m}$. Find its centripetal acceleration. · 一个物体以 $4.0\ \dfrac{\text{m}}{\text{s}}$ 绕半径 $2.0\ \text{m}$ 的圆运动。求它的向心加速度。
$a = \dfrac{v^{2}}{r} = \dfrac{4.0^{2}}{2.0} = 8.0\ \dfrac{\text{m}}{\text{s}^2}$. · $a = \dfrac{v^{2}}{r} = \dfrac{4.0^{2}}{2.0} = 8.0\ \dfrac{\text{m}}{\text{s}^2}$。
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
向心力
- 由 $F = ma$:$F = \dfrac{mv^{2}}{r} = mr\omega^{2}$,指向圆心。
- 它始终与速度 垂直,所以 不做功,速率保持恒定。
旋转的游乐设施需要指向圆心的向心力
The centripetal force points toward the centre of the circle. · 向心力指向圆心。
Yes — always toward the centre, perpendicular to the velocity. · 是的——总是指向圆心,垂直于速度。
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? · 一个 $2.0\ \text{kg}$ 的球以 $4.0\ \dfrac{\text{m}}{\text{s}}$ 绕半径 $2.0\ \text{m}$ 的圆运动。需要多大的向心力?
$F = \dfrac{mv^{2}}{r} = \dfrac{2.0 \times 4.0^{2}}{2.0} = 16\ \text{N}$. · $F = \dfrac{mv^{2}}{r} = \dfrac{2.0 \times 4.0^{2}}{2.0} = 16\ \text{N}$。
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
不是一种新的力
- "向心力"不是一种新的力。
- 它是真实的力——张力(tension)、重力、摩擦力、法向力——的 合力。
速度沿切线;力和加速度指向圆心
The centripetal force is: · 向心力是:
It is whatever real force(s) happen to point to the centre — not a separate force of its own. · 它就是恰好指向圆心的那个(些)真实力——不是它自己单独的一种力。
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
它从哪里来
- 绳子上的球 → 张力。平路上转弯的汽车 → 摩擦力。
- 行星或卫星 → 重力。倾斜(banked)的赛道 → 法向力向内的分量。
在倾斜赛道上,路面作用力的水平分量提供向心力
Match each circular motion to the force that provides the centripetal force. · 把每种圆周运动与提供向心力的力配对。
Different situations, different real forces — but each points to the centre and provides $\dfrac{mv^{2}}{r}$. · 不同的情形,不同的真实力——但每个都指向圆心,提供 $\dfrac{mv^{2}}{r}$。
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.
例题:转弯能多快?
一辆 $1200\ \text{kg}$ 的汽车驶过半径 $50\ \text{m}$ 的平坦弯道。轮胎能提供的最大摩擦力为 $8200\ \text{N}$。
- 摩擦力就是向心力: $F = \dfrac{mv^{2}}{r}$,所以最快的安全速率满足 $\dfrac{1200\,v^{2}}{50} = 8200$。
- 求解: $v^{2} = \dfrac{8200 \times 50}{1200} = 342$,所以 $v = 18\ \dfrac{\text{m}}{\text{s}}$(约 $65\ \dfrac{\text{km}}{\text{h}}$)。
- 更急的弯: $r$ 减半,$v_{\text{max}}$ 降为原来的 $\dfrac{1}{\sqrt{2}}$——所需的力按 $\dfrac{1}{r}$ 增长。
- 检查: 湿路面上可用的摩擦力减小,所以安全速率也减小——这既是物理的结论,也是路牌的假设。
A $800\ \text{kg}$ 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 m/s? · 一辆 $800\ \text{kg}$ 的汽车驶过半径 $40\ \text{m}$ 的平坦弯道。可用的最大摩擦力为 $6400\ \text{N}$。最大安全速率是多少(单位 m/s)?
$\dfrac{mv^{2}}{r} = F$, so $v^{2} = \dfrac{6400 \times 40}{800} = 320$ and $v = 17.9\ \dfrac{\text{m}}{\text{s}}$. · $\dfrac{mv^{2}}{r} = F$,所以 $v^{2} = \dfrac{6400 \times 40}{800} = 320$,$v = 17.9\ \dfrac{\text{m}}{\text{s}}$。
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
竖直圆周
- 绕竖直的圆环运动时,速率会变化(重力做功)。
- 在 最高点,重力和张力都指向圆心:$T + mg = \dfrac{mv^{2}}{r}$。在 最低点 它们方向相反:$T - mg = \dfrac{mv^{2}}{r}$。
- 最高点绳子刚好绷直时的最小速率是 $v_{\text{min}} = \sqrt{gr}$(令张力 $= 0$)。
绕竖直圆周运动时,重力在最高点帮忙、在最低点反抗——所以绳子的张力在最低点最大
At the top of a vertical loop, the slowest speed (string just tight) is: · 在竖直圆环的顶部,最慢的速率(绳子刚好绷紧)是:
With tension $= 0$, gravity alone is the centripetal force: $mg = \dfrac{mv^{2}}{r}$, so $v_{\text{min}} = \sqrt{gr}$. · 当张力 $= 0$ 时,只有重力作为向心力:$mg = \dfrac{mv^{2}}{r}$,所以 $v_{\text{min}} = \sqrt{gr}$。
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.
例题:最高点和最低点的张力
一个 $0.50\ \text{kg}$ 的球拴在 $1.0\ \text{m}$ 长的绳子上,在竖直平面内甩动。在最低点它的速率为 $8.0\ \dfrac{\text{m}}{\text{s}}$。
- 最低点: $T - mg = \dfrac{mv^{2}}{r}$,所以 $T = 0.50 \times 9.81 + \dfrac{0.50 \times 8.0^{2}}{1.0} = 4.9 + 32 = 37\ \text{N}$。
- 最高点的速率: 上升 $2r$ 的能量守恒:$v_{\text{top}}^{2} = 8.0^{2} - 2g(2.0) = 64 - 39 = 25$,所以 $v_{\text{top}} = 5.0\ \dfrac{\text{m}}{\text{s}}$。
- 最高点: $T + mg = \dfrac{mv^{2}}{r}$,所以 $T = \dfrac{0.50 \times 25}{1.0} - 4.9 = 12.5 - 4.9 = 7.6\ \text{N}$。
- 检查: 张力在最低点大得多,那正是绳子会断的地方。在最高点 $v_{\text{top}} > \sqrt{gr} = 3.1\ \dfrac{\text{m}}{\text{s}}$,所以绳子保持绷紧。
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.
绝不要在受力图上把"向心力"画成一个额外的箭头——它是真实的力的 合力,而且本课程中不存在向外的"离心力"。在圆环的 最高点 重力 帮忙($T + mg$);在 最低点 重力 反抗($T - mg$)。另外,竖直圆周运动的速率 不 恒定,所以要用能量守恒在最高点和最低点之间换算。
At the top of a vertical circle the string tension is smallest, because there the ____ of the ball also acts towards the centre. · 在竖直圆周的最高点绳子的张力最小,因为在那里球的 ____ 也指向圆心。
At the top $T + mg = \dfrac{mv^{2}}{r}$: the weight supplies part of the centripetal force, so the string needs to supply less. · 在最高点 $T + mg = \dfrac{mv^{2}}{r}$:重力提供了一部分向心力,所以绳子需要提供的就少了。
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
你掌握了
- 圆周运动需要向内的(向心)加速度 $a = \dfrac{v^{2}}{r} = r\omega^{2} = v\omega$
- 向心力 $F = \dfrac{mv^{2}}{r}$——真实的力的合力,指向圆心;它不做功
- 竖直圆环:最高点 $T + mg = \dfrac{mv^{2}}{r}$,最低点 $T - mg = \dfrac{mv^{2}}{r}$;最高点 $v_{\text{min}} = \sqrt{gr}$