Centripetal force
| English | Chinese | Pinyin |
|---|---|---|
| tangent | 切线 | qiè xiàn |
| centripetal acceleration | 向心加速度 | xiàng xīn jiā sù dù |
| centripetal force | 向心力 | xiàng xīn lì |
| perpendicular | 垂直 | chuízhí |
| tension | 张力 | zhāng lì |
| banked | 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.
Centripetal acceleration 向心加速度
- Constant speed, but ever-changing direction, means the velocity changes — an acceleration.
- It points to the centre: $a = \dfrac{v^{2}}{r} = r\omega^{2}$.
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:
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.
$a = \dfrac{v^{2}}{r} = \dfrac{4.0^{2}}{2.0} = 8.0\ \dfrac{\text{m}}{\text{s}^2}$.
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.
A spinning fairground ride needs a centripetal force toward the centre
The centripetal force points toward the centre of the circle.
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?
$F = \dfrac{mv^{2}}{r} = \dfrac{2.0 \times 4.0^{2}}{2.0} = 16\ \text{N}$.
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:
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.
Different situations, different real forces — but each points to the centre and provides $\dfrac{mv^{2}}{r}$.
Vertical circles
- Going round an upright loop, the speed changes (gravity does work).
- At the top, the slowest speed 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:
With tension $= 0$, gravity alone is the centripetal force: $mg = \dfrac{mv^{2}}{r}$, so $v_{\text{min}} = \sqrt{gr}$.
You've got it
- circular motion needs an inward (centripetal) acceleration $a = \dfrac{v^{2}}{r} = r\omega^{2}$
- centripetal force $F = \dfrac{mv^{2}}{r}$ — the net of real forces, toward the centre
- top of a vertical loop: $v_{\text{min}} = \sqrt{gr}$